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ELECTROCHEMISTRY, 1., , SECTION (A) : GALVANIC CELL, ITS REPRESENTATION & SALT BRIDGE, , 1.1., , Introduction :, Batteries are everywhere in modern societies. They provide the electric current to start our, automobiles and to power a host of products such as pocket calculator, digital watches, heart, pacemaker, radio, and tape recorders., Electrochemistry is the area of chemistry concerned with the interconversion of chemical, energy and electrical energy. A battery is an electrochemical cell, a device for interconverting, chemical and electrical energy. A battery takes the energy released by a spontaneous chemical, reaction and uses it to produce electricity., , Electrochemical cell:, It is device for converting chemical energy in to electrical energy., Electrochemical cell are of two types, , Galvanic cells or Voltaic cell,  A spontaneous chemical reaction, generates an electric current., , , , Electrolytic cell, An electric current drives a, nonspontaneous reaction., , Thus the two types of cells are reverse of each other., , 1.2., , Construction/ Working principle, Whenever a metal strip is put in an electrolyte the process of oxidation and reduction takes place, simultaneously within the system. Due to this there is a potential difference between the metal phase, and the liquid phase., On joining the metal strips through a wire (of negligible resistance) the current flows as long as the, potential difference exists between the metal phase and the liquid phase., , I., , Anode:, Some metals (which are reactive) are found to have tendency to go into the solution phase when these, are placed in contact with their ions or their salt solution., For example: Zn rod is placed in ZnSO4 solution., , Figure : 1, , Figure : 2, , Figure : 3

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The Zn atom or metal atoms will move in the solution to form Zn+2. After some time following equilibrium, will be established., Zn(s), Zn2+ +2e–, There will be accumulation of sufficient negative charge on the rod which will not allow extra zinc ions to, move in the solution. i.e. solution will be saturated with Zn+2 ions., The extra positive charge of the solution will be more concentrated around the negatively charged rod., An electrical double layer is developed in the system and hence a potential difference is created, between the rod and the solution which is known as electrode potential., This particular electrode is known as anode :, , On anode oxidation will take place. (Release of electron)., , To act as source of electrons., , It is of negative polarity., , The electrode potential is represented by EZn(s)/Zn2  (aq), II., , Cathode :, , Figure : 1, , Figure : 2, , Figure : 3, , Some metals (Cu, Ag, Au etc.,) are found to have the opposite tendency i.e., when placed in contact, with their aqueous ions, the ions from the solution will get deposited on the metal rod., The following equilibrium will be established :, , Cu2+ + 2e–, , Cu(s)., , So rod will have deficiency of electron (positive charge). Extra negative charge will surround this, positively charged rod and form double layer. An electrical double layer is developed in the system and, hence a potential difference is created between the rod and the solution which is known as electrode, potential. This will be known as cathode., , , At cathode reduction will take place. (Gain of e– will take place), , , , To act as sink of electron., , , , Positive polarity will be developed., , , , Their electrode potential can be represented by : ECu2  (aq)/Cu(s), , Is w hereoxidation occurs, Anode : Is w hereelectronsare produced, Has a negative sign, , 1.3., , Is w here reductionoccurs, Cathode : Is w hereelectronsare consumed, Has a positive sign, , Construction of Cell :, , , It has two half-cells, each having a beaker containing a metal strip that dips in its aqueous, solution., , , , The metal strips are called electrodes and are connected by an conducting wire., , , , Two solutions are connected by a salt bridge., , , , The oxidation and reduction half reactions occur at a separate electrodes and electric current flows, through the wire.

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Selection of electrolyte for Salt Bridge :, , , , , , The electrolyte in salt bridge should be such that speed of its cation equals speed of its anion in, electrical field., For that charge and sign of the ions should be almost equal. Transport number of cation = Transport, number of anion, or, Mobility of cation = Mobility of anion, KCl is generally preferred but KNO3 or NH4NO3 can also be used., If Ag+, Hg22+, Pb2+, Tl+ ions are present in a cell then in salt bridge KCl is not used because there can be, formation of precipitate of AgCl, Hg2Cl2, PbCl2 or TlCl at mouth of tube which will prevent the migration, of ions and its functioning will stop., , Functions of Salt Bridge :, , , , , , , , , , , A salt bridge is a U–shaped inverted tube that contains a gel permeated with an inert electrolyte., It connects the solution of two half-cell to complete the circuit., It minimize the liquid junction potential. The potential difference between the junction of two liquids., It maintains the electrical neutrality of the solution in order to give continuous flow or generation of, current., "The simultaneous electrical neutrality of the anodic oxidation chamber and cathodic reduction chamber, is due to same mobility or velocity of K+ and NO3– ions taken into salt bridge., If the salt bridge is removed then voltage drops to zero., The ions of the inert electrolyte do not react with other ion in the solution and the ions are not oxidised, or reduced at the electrodes., Generally tube is filled with a paste of agar-agar powder with a natural electrolyte/generally not, common to anodic/cathodic compartment with porous plugs at each mouth of tube., It prevents mechanical mixing of two electrolytic solution., Liquid-Liquid Junction Potential :, The potential difference which arises between two solutions (during the progress of reaction) when in, contact with each other., , Shorthand Notation for Galvanic Cells, , , We require two half cells to produce an electrochemical cell, which can be represented by following few, rules;, , , , The anode half-cell is always written on the left followed on the right by cathode half-cell., The separation of two phases (state of matter) is shown by a vertical line., The various materials present in the same phase are shown together using commas., , , , , , , , , The salt bridge is represented by a double slash (||)., The significant features of the substance viz. pressure of a gas, concentration of ions etc. are indicated, in brackets immediately after writing the substance., For a gas electrode, the gas is indicated after the electrode for anode and before the electrode in case, of cathode. (i.e. Pt H2/H+ or H+/H2 Pt)

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Solved Example, Example 1., Solution:, , Write short hand notation for the following reaction, Sn2+(aq) + 2Ag+(aq)  Sn4+(aq) + 2Ag(s)., The cell consists of a platinum wire anode dipping into an Sn+2 solution and a silver cathode, dipping into an Ag+ solution therefore Pt(s) | Sn2+(aq), Sn4+ (aq) || Ag+ (aq) | Ag(s)., , Example 2., , Write the electrode reaction and the net cell reaction for the following cells. Which electrode, would be the positive terminal in each cell ?, (a), Zn | Zn2+ || Br–, Br2 | Pt, (b), Cr| Cr3+ || I–, I2 | Pt, (c), Pt | H2, H+ || Cu2+ | Cu, (d), Cd | Cd2+ || Cl–, AgCl | Ag, , Solution:, (a), , (b), , (c), , (d), , Oxidation half cell reaction, Zn  Zn2+ + 2e–, Reduction half cell reaction, Br2 + 2e–  2Br–, Net cell reaction, Zn + Br2  Zn2+ + 2Br– (Positive terminal : cathode Pt), Oxidation half reaction, [Cr  Cr3+ + 3e–] × 2, Reduction half reaction, [I2 + 2e–  2I–] × 3, Net cell reaction, 2Cr + 3I2  2Cr3+ + 6I– (Positive terminal : cathode Pt), Oxidation half reaction, H2  2H+ + 2e–, Reduction half reaction, Cu2+ + 2e–  Cu, Net cell reaction, H2 + Cu2+  Cu + 2H+ (Positive terminal : cathode Cu), Oxidation half reaction, Cd  Cd2+ + 2e–, Reduction half reaction, [AgCl + e–  Ag + Cl–] × 2, Net cell reaction, Cd + 2AgCl  Cd2+ + 2Ag + 2Cl–, (Positive terminal : cathode Ag), , 2., , SECTION (B) : ELECTROCHEMICAL SERIES & ITS APPLICATIONS, , 2.1., , Electrode Potential :, , , , , , , The driving force that pushes the negative charge electrons away from the anode and pulls them, towards the cathode is an electrical potential called electromotive force also known as cell potential, or the cell voltage. Its unit is volt, The potential difference developed between metal electrode and its ions in solution in known as, electrode potential., Electrode potential depends upon :, , Concentration of the solution., , Nature of the metal., , Nature of the electrolyte., , Pressure temperature conditions.

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, , The potential difference developed between metal electrodes and the solution of its ions at 1 M, concentration at 1 bar pressure and at a particular temperature is known as standard electrode, potential., , Oxidation Potential (O.P.), The electrode potential for oxidation half, reaction., Tendency to get oxidised., Greater the O.P. then greater will be, tendency to get oxidised., , 2.2., , , , , , 2.3., , , , , , , Reduction Potential (R.P.), The electrode potential for reduction half, reaction., Tendency to get reduced., Greater the R.P. greater will be tendency to, get reduced., , Type of Electrode, , Electrode reaction in standard condition, , Representation, , Reduction : Zn2+ + 2e–  Zn(s), , 1, , Metal electrode, (Zn electrode,, Cu electrode etc.), , 0, ( SRP), EZn, 2, / Zn(s ), , Oxidation : Zn(s)  Zn2+ + 2e–, , 0, (SOP), EZn, (s ) / Zn2, , Reduction : 2e– + 2H+ + H2O2  2H2O, , 2, , Hydrogen peroxide, electrode, , EH0 2O2 / H2O, , Oxidation : H2O2  O2 + 2H+ + 2e–, , EH0 2O2 / O2, , 3, , Redox electrode, , Reduction : MnO4– + 8H+ + 5e–  Mn2+ + 4H2O, , 0, EMnO, , / Mn2 , , Reduction : AgCl(s) + e  Ag(s) + Cl, , 0, EAgCl, (s ) / Ag (s ) / Cl , , 4, , Metal-Metal, insoluable, electrode, , –, , salt, , Oxidation : Ag(s) +Cl–  AgCl(s) + e–, , –, , 4, , 0, EAg, (s ) / AgCl (s ) / Cl , , Reference electrode :, The potential of a single electrode cannot be determined what were the potential difference between, two electrodes can be accurately measured using a reference electrode., An electrode is chosen as a reference with respect to which all other electrodes are valued., Standard Hydrogen Electrode (SHE) is taken as standard reference electrode. Its electrode potential is, arbitrarily assumed to be 0.00 volt., , Standard Hydrogen Electrode (SHE) consists of a platinum electrode in contact with H2 gas and, aqueous H+ ions at standard state conditions (1 atm H2 gas, 1 M H+ (aq),)., 2H+(aq, 1M) + 2e–H2(g, 1 atm), E° = 0 V, H2(g, 1atm)2H+ (aq, 1M) + 2e–, E° = 0 V, , Cell potential :, The difference in electrode potentials of the two half cell reactions (oxidation half cell and reduction half, cell) is known as emf of the cell or cell potential., The emf of the cell or cell potential can be calculated from the values of electrode potential of the two, half cell constituting the cell. The following three method are in use:, When oxidation potential of anode and reduction potential of cathode are taken into account:, E°cell = oxidation potential of anode + reduction potential of cathode, = E°ox(anode) + E°red(cathode), When reduction potential of both electrodes are taken into account :, E°cell = Reduction potential of cathode – Reduction potential of anode, = E°cathode – E°anode both are reduction potential., When oxidation potential of both electrodes are taken into account :, E°cell = oxidation potential of anode – Oxidation potential of cathode, = E°ox (anode) – E°ox (cathode)

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, , , The standard cell potential E° is the cell potential when both reactants and products are in their, standard states–solutes at 1 M concentration, gases at a partial pressure of 1 atm, solids and liquids in, pure from, with all at a specified temperature, usually 25° C., E°cell is intensive property so on multiplying/Dividing cell reaction by any number, the E° cell value would, not change., , Calculation of electrode potential :, , At Anode, 2H+ + 2e–, , H2(g), , , , , Oxidation potential O.P. = EH, , , , , , , , Under standard state,  E 0H2 ( g ) / H  (aq.) = SOP, , , , 2H+ + 2e–, 2 (g ) / H, , , , ( aq.), , At Cathode, H2(g), , Reduction Potential (R.P.) EH  / H, , 2 (g ), , = RP, ,  Under standard state.,  EH0  / H (g ) = SRP, , For SHE reference potential is taken to be zero at all temperature., , 2, , , , SOP = – SRP = 0 for SHE., , , To calculate standard potential of any other electrode a cell is coupled with standard hydrogen, electrode (SHE) and its potential is measured that gives the value of electrode potential of that, electrode., Anode, , :, , Zinc electrode, , Cathode :, Cell, , :, , SHE, Zinc electrode || SHE, , Cell potential :, Ecell = EH  / H, , 2 (g ), , – E°Zn2+/Zn, , = 0.76 V (at 298 K experimentaly), So, EºZn2+/Zn = – 0.76 V (SRP), EºZn/Zn2+(aq) = 0.76 V(SOP), , , So, w.r.t. H2, Zn has greater tendency to get oxidised., In similar manner reduction potentials (SRP) at 298 K for many other electrodes are calculated and are, arranged in a series increasing order known as electrochemical series.

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Electrochemical Series :, Electrode, *Li, K, Ba, Ca, Na, , Reaction, Li+ + e–  Li(s), K+ + e–  K (s), , SRP (at 298 K), – 3.05 V, – 2.93 V, , Ca+2 + 2e–  Ca(s), Na+ + e–  Na(s), Mg+2 + 2e–  Mg(s), , – 2.87 V, – 2.71 V, , * Electrolytes (H2O), , H2O(l) + e– , , – 0.828 V, , *Zn, Cr, *Fe, Cd, Co, Ni, Sn, Pb, *H2, , Zn+2 + 2e–  Zn(s), Cr+3 + 3e–  Cr(s), Fe2+ + 2e-  Fe, Cd+2 +2e–  Cd(s), , – 0.76 V, – 0.74 V, – 0.44 V, – 0.40 V, , Ni+2 + 2e–  Ni(s), Sn+2 + 2e–  Sn(s), Pb+2 + 2e–  Pb(s), , – 0.24 V, – 0.14 V, – 0.13 V, 0.00 V, , Mg, Al, , 1, 2, , H2 + OH–, , Cu, , 2H+ + 2e– H2(g), Cu2+ + 2e–  Cu(s), , 2, Fe, Hg, Ag, Hg, Br2, , Fe3+ + e–  Fe2+, Hg22+ + 2e-  Hg(l), Ag+ + e–  Ag, Hg2+  Hg(l), Br2 + 2e–  2Br–, , * Electrolytes, *, *, *, *, , 1, 2, , – 2.37 V, , 0.34 V, 0.77 V, 0.79 V, , 1.06 V, , O2 + 2H + 2e  H2O(), –, , +, , 1.23 V, , + 14H + 6e  2Cr + 7H2O, Cl2 +2e–  2 Cl–, MnO4– + 8H+ + 5e–  Mn2+ + 4H2O, F2 + 2e–  2F–, Cr2O72–, , +, , –, , +3, , 1.33 V, 1.36 V, 1.51 V, 2.87 V, , Solved Examples, Example 1., Calculate Eºcell of (at 298 K), Zn(s) / ZnSO4(aq) || CuSO4(aq) / Cu(s), Given that, EºZn/Zn2+(aq) = 0.76 V, EºCu(s) / Cu2+(aq) = – 0.34 V, Solution:, Eºcell = (S.R.P)cathode – (S.R.P)anode, = 0.34 – (– 0.76) = 1.1 V, Example 2., , Given the cell Ag AgCl(s) | NaCl (0.05 M) || Ag NO3 (0.30 M) | Ag, (a), Write half reaction occurring at the anode., (b), Write half reaction occurring at the cathode., (c), Write the net ionic equation of the reaction., (d), Calculate Eºcell at 25°C., (e), Does the cell reaction go spontaneous as written ?, (Given E°AgCl,Cl = + 0.22 volt) ; E º Ag  / Ag = + 0.80 volt), , Solution:, , (a) LHS electrode is anode and half reaction is oxidation., Ag+ + Cl– AgCl(s) + e–, ... (i)

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(b) RHS electrode is cathode and half reaction is reduction., Ag + e Ag(s), ... (ii), (c) From equation (i) and (ii) cell reaction is : Cl – (0.05 M) + Ag+ (0.30 M) AgCl(s), (d) E°cell = E°right – E°left, = (0.80 – 0.22 volt = 0.58 volt, (e) Yes, the e.m.f. value is positive, the reaction will be spontaneous as written in the cell, reaction., , 3., , SECTION (C) : CONCEPT OF G, , 3.1., , Free energy changes for cell reaction :, , , , , , , , , , , , , The free energy change G (a thermochemical quantity) and the cell potential E(an electrochemical, quantity) both measure the driving force of a chemical reaction., The values of G and E are directly proportional and are related by the equation,, , G = –nFE, where n = Number of moles of electron transferred in the reaction., F = Faraday constant = 96485 C/mole e– 96500 C/mole e–, Calculation of Electrode Potential of unknown electrode with the help of given (two) electrode., Obtain the reaction of the 3rd electrode with the help of some algebraic operations on reactions of the, given electrodes., Then calculate Gº of the 3rd reaction with the help of some algebaric operations of Gº of 1st and 2nd, reactions., Use Gº = –nF Eºelec. to calculate unknown E.P., 0, 0, is intensive property so if we multiply/Divide electrode reaction by any number the Ecell, value, Ecell, would not changed, i.e., Zn2+ + 2e–  Zn(s), Eº = – 0.76 V, Multiply by 2, 2Zn2+ + 4e–  2Zn(s), Eº = – 0.76 V (remain same), , Solved Example, Example 1., Given that EºCu2+/Cu = 0.337 V and EºCu+ /Cu2+ = – 0.153 V. Then calculate EºCu+/Cu., Solution:, Cu2+ + 2e– Cu, G1, (i)., Cu+  Cu2+ + e–, G2, After adding, Cu++ e–  Cu, , , , G1 + G2 = G3, –2F E10 – F E 20 = – F E 30, E3 = 2 E10 + E 20 = 2 × 0.337 – 0.153 = 0.674 – 0.153 = 0.521 V, Example 2., , E 0Mn2  / MnO4 = –1.51 V ;, , E 0MnO2 / Mn2 = + 1.23 V, , E 0MnO4 / MnO2 = ?, , Solution:, , 2+, , 4H2O + Mn, (i)., ,  Mn, , Mn O4, , (All in acidic medium), O4, , G1, , + 8H + 5e–, +, , + 8H + 5e  4H2O + Mn, 2e + MnO2 + 4H+  Mn2+ + 2H2O, 2H2O + Mn2+  MnO2 + 4H+ + 2e–, 4H+ + Mn O4 + 3e–  MnO2 + 2H2O, (i) + (ii) = (iii), G3 = – G1 – G2, –3E3F = 5 E10 F + 2 E 20 F, +, , –, , 2+, , , , –, , (ii)., (iii)., , , E=, , [5E1  2E2 ] [5( 1.51)  2(1.23)], =, 3, 3, , –G1, G2, –G2, , G3, , =, , [ 7.55  2.46], 5.09, =, = 1.69 V, 3, 3

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Example 3., , Will Fe2+ disproportionate or not, , anode, , cathode, 0.77 V,  0.44 V, Fe  , ,  Fe2+  ,  Fe, 3+, , Solution:, , -0.036, This is known as latimer diagram., S.R.P to right of the species greater than SRP of it's left species will undergo, disproportionation., , 4., , SECTION (D) : NERNST EQUATION & ITS APPLICATIONS (INCLUDING, CONCENTRATION CELLS), , 4.1., , Nernst Equation :, , , , , , , , , , 4.2., , , Cell potentials depend on temperature and on the composition of the reaction mixtures., It depends upon the concentration of the solute and the partial pressure of the gas, if any., The dependence upon the concentration can be derived from thermodynamics., From thermodynamics, G = G° + RT ln Q, – nFE = – nFE° + 2.303 R T log Q, 2.303RT, E = E° –, log Q, nF, Take T = 298 K, R = 8.314 J/mol K, F = 96500 C, 0.059, Now we get,, E = E° –, log Q, n, Where n = number of transfered electron, Q = reaction quotient, Nernst equation can be used to calculate cell potentials for non standard conditions also., Nernst equations can be applied to half cell reactions also., , Applications of Nernst equation, Nernst Equation for Electrode Potential, Mn+(aq) + ne–, M(s), RT,  M (s ) , 0, ERedn = Ered, –, n  n  , nF, M , 0, ERedn = Ered, –, , At 298K,, , , 2.303 RT,  M (s ) , log  n  , nF, M , , 0, ERedn = ERe, –, dn, , 0.059,  1 , log  n  , n, M , , Hydrogen Electrode, H2(g), , 2H+(aq) + 2e–,  (H  )2 , 0.0591, E = Eº –, log , , 2,  PH2 , , , , , , Metal–metal soluble salt electrode., Zn2+ + 2e–  Zn(s), 2.303 RT,  1 , 0, ERedn = ERe, log  2  at 298K, , dn,  Zn , nF

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Equilibrium in electrochemical cell, , G0 = – nF Eºcell, , G = – nF Ecell, From thermodynamics, , G = G0 + RTnQ, at chemical equilibrium G = 0, Ecell = 0 cell will be of no use, so,, G0 = – RTn Keq, at equilibrium – nF Eºcell = –2.303 RT log (Keq), nF, log Keq =, Eºcell, 2.303 RT, at 298 K and R = 8.314 J/mol K, n, log Keq =, Eºcell, 0.059, , , , , , Solved Examples, Example 1., , Cr2O72 ,Cr 3 in H2SO4   0.05M Pt, 0.01 M, 0.1 M, , Calculate Ecell of Pt(s), , Cl2 (g) Cl (aq), 0.1 atm 102 M, , 0, Given that ECr, O2 , , 0, = 1.33 V ; ECl, = –1.36 V, , /Cl, , 2 7, , / Cr 3, , 2, , 6e– + 14H+ + Cr2O72–  2Cr+3 + 7H2O, [2Cl–  Cl2 + 2e–] × 3, 14H+ + 6Cl– + Cr2O72–  3Cl2 + 2Cr+3 + 7H2O, 0, = 1.33 – (+1.36) = – 0.03, Ecell, , Sol., , Ecell = – 0.03 –, , [Cr 3  ]2 [PCl2 ]3, 0.059, 0.059  23, log, = – 0.03 –, 6, 6, [H  ]14 [Cl  ]6 [Cr2O72  ], , Ecell = – 0.26 V, Example 2., , The E°cell for the reaction Fe + Zn2+, Zn + Fe2+, is – 0.32 volt at 25°C. What will be the, 2+, equilibrium concentration of Fe , when a piece of iron is placed in a 1 M Zn2+ solution ?, , Sol., , We have the Nernst equation at equilibrium at 25°C, 0.0591, E° =, log K, ... (i), n, Since E°Cell for the given reaction is negative, therefore, the reverse reaction is feasible for, which E°cell will be + 0.32 V, Thus for, Zn, +, Fe2+, Fe, +, Zn2+, ;, E°Cell + 0.32 V, –, x, –, (1–x), 0.0591, 0.0591, [Zn 2 ], [Zn2 ], Now, E° =, log, or, 0.32, =, log, n, 2, [Fe2 ], [Fe2 ], , [Zn2 ], , – 10.829, [Fe2 ], [Fe2+] = 1.483 × 10–11 M, log =, , 4.4., , Taking antilog,, , Work done by a cell :, , (i) Let 'n' faraday charge be taken out of a cell of EMF 'E'; then work done by the cell will be calculated as :, work = Charge × Potential = nFE, (ii) Work done by cell = Decrease in free energy, so, – G = nFE, or, W max = + nFEº where Eº is standard EMF of the cell.

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Solved Examples, Example 1., , Calculate the maximum work that can be obtained from the Daniel cell given below Zn(s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s). Given that E ºZn2  / Zn = – 0.76 V and E ºCu 2  / Cu = + 0.34 V., , Solution:, , Cell reaction is :, Zn(s) + Cu2+ (aq)  Cu(s) + Zn2+ (aq), Here n = 2, Eºcell = Eºcathode – Eºanode (On the basis of reduction potential), = + 0.34 – (0.76) = 1.10 V, We know that :, W max = Gº = – nFEº, = – (2 mol) × (96500 C mol) × (1.10 V) = – 212300 C.V. = – 212300 J, or, W max = – 212300 J, , 4.5., , , , , , Concentration cells :, A concentration cell consists of two electrodes of the same material, each electrode dipping in a, solution of its own ions and the solution being at different concentrations., The two solutions are separated by a salt bridge., e.g. Ag(s) | Ag+ (a1) || Ag+ (a2) | Ag(s)  (a1 < a2) a1, a2 are concentrations of each half cell, At LHS electrode anode :, Ag (s) Ag+(a1) + e–, At RHS electrode cathode :, Ag+(a2) + e– Ag(s), +, The net cell reaction is : Ag (a2) Ag+ (a1), The nernst eq. is, a, 0.059, Ecell = –, log 1 (Here n = 1, Temp, 298 K), a2, n, Likewise, the e.m.f. of the cell consisting of two hydrogen electrodes operating at different pressure P 1, and P2 (P1 > P2) and dipping into a solution HCl is :, P, 0.059, Ecell =, log 1 (at 298 K), P2, 2, , 5., , SECTION (E) : ELECTROLYSIS, , 5.1., , Electrolysis & Electrolytic cell :, Electrolysis :, , , , , , , , , , , , 5.2., , , , Electrolyte is a combination of cations and anions which in fused state or in aqueous solution can conduct, electricity., This is possible due to the movement of ions from which it is made of., The process of using an electric current to bring about chemical change is called electrolysis., Electrolysis is a process of oxidation and reduction due to current in the electrolytic solution., The product obtained during electrolysis depends on following factors., The nature of the electrolyte, The concentration of electrolyte, The charge density flowing during electrolysis., The nature of the electrode, , Active vs Inactive electrodes :, The metal electrodes in the cell that are active, because the metals themselves are components of the, half reactions., As the Daniel cell operates, the mass of the zinc electrode gradually decreases, and the [Zn 2+] in the, anode half – cell increases. At the same time, the mass of the copper electrode increases and the, [Cu2+] in the cathode half – cell decreases; we say that the Cu2+ "plates out" on the electrode.

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2H2O() O2 + 2H+ + 4e–, Cu(s)  Cu2+ + 2e–, , , Electrolytic refining, AgNO3(aq) using Cu cathode & Ag anode., Cathode :, Ag+ + e– Ag(s), 2H2O() + 2e– H2(g) + 2OH–, Anode :, NO3– X (No reaction), 2H2O() O2 + 4H+ + 4e–, Ag(s) Ag+(aq) + e–, , Eº = 1.23 V, Eº = –0.34 V, , Eº = 0.8 V, Eº = – 0.83 V, Eº = – 1.23 V, Eº = – 0.80 V, , 6., , SECTION (F) : FARADAY LAWS & ITS APPLICTIONS, , 6.1., , Faraday's Law of Electrolysis :, , , , , , 1st Law : The mass deposited/released/produced of any substance during electrolysis is proportional to, the amount of charge passed into the electrolyte., WQ, W = ZQ, Z – electrochemical equivalent of the substance., mass, Unit of Z =, = Kg/C or g/C, coulomb, Z = Mass deposited when 1 C of charge is passed into the solution., Equivalent mass (E) : mass of any substance produced when 1 mole of e– are passed through the, solution during electrolysis., Molar mass, E=, , no. of e involved in oxidation / reduction, M, e.g., Ag+ + e– Ag, E=, 1, M, Cu2+ + 2e- Cu(s), E=, 2, M, 3+, –, Al + 3e Al(s), E=, 3, 1 mole of e– = 1 Faraday of charge.,  96500 C – Charge deposit E gram metal charge, E,  E , , 1C  , g, Z=,  96500 , 96500, , , , W=, , Molar mass, EQ, Q, =, ×, , 96500, (no. o f e involved) 96500, , Molar mass, it, ×, 96500, (no. of e  involved), nd, 2 Law : When equal charge is passed through 2 electrolytic cells and this cells are connected in, series then mass deposited at electrode will be in the ratio of their electrochemical equivalents or in the, ratio of their equivalent masses., EQ, W = ZQ =, 96500, W1, z, E, = 1 = 1 (Q = same), W2, z2, E2, ,  dQ = i  dt, , , , Q = it, , W=

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Current Efficiency :, charge actually used in electricity, × 100, charge passed, mass actually produced, Current efficiency =, × 100, mass that should have been produced, , Current efficiency =, , Solved Examples, Example 1., Sol., , Example 2., Sol., Example 3., , Sol., , Example 4., , Calculate volume of the gases liberated at STP if 1 L of 0.2 molar solution of CuSO 4 is, electrolysed by 5.79 A current for 10000 seconds., 5.79  10000, 579, No. of moles of e– =, =, = 0.6, 96500, 965, Cathode :, Cu2+ + 2e– Cu(s), 0.2 mole 0.4 mole, 2H2O() + 2e– H2 + 2OH–, 0.2 mole of e– 0.1 mole of H2 at S.T.P., Anode :, 2H2O()  O2 + 4H+ + 4e–, 4 mole of e– 1 mole of O2, 0.6 mole of e– 0.15 mole of O2, so, total moles = 0.25 mole, Total volume = 5.6 Ltr., The electrochemical equivalent of copper is 0.0003296 g coulomb–1. Calculate the amount of, copper deposited by a current of 0.5 ampere flowing through copper sulphate solution for 50, minutes., According to Faraday's first law, W = Zit, W = 0.5 × 50 × 60 × 0.003296 = 0.4944 g, An electric current is passed through three cells connected in series containing ZnSO 4,, acidulated water and CuSO4 respectively. What amount of Zn and H2 are liberated when 6.25 g, of Cu is deposited? Eq. wt. of Cu and Zn are 31.70 and 32.6 respectively., , Eq. of Cu = Eq. of Zn = Eq. of H2, WH2, W, 6.25, = Zn =, 1, 31.70 32.6, The cell consists of three compartments separated, by porous barriers. The first contains a cobalt, electrode in 5.00 L of 0.100 M cobalt (II) nitrate; the, 0.20 M NO, 0.10 M NO, 0.10 M NO, second contains 5.00 L of 0.100 M KNO3 and third, Co, Ag, 0.10 M Co, 0.10 M K, 0.10 M Ag, contains 0.1 M AgNO3. Assuming that the current, within the cell is carried equally by the positive and, 5.0 L, 5.0 L, 5.0 L, negative ions, tabulate the concentrations of ions of, each type in each compartment of the cell after the, passage of 0.100 mole electrons., Given : Co2+ + 2e–  Co, Eº = – .28 V, Ag+ + e–  Ag, Eº = 0.80 V, Spontaneous reaction is :, 2Ag+ + Co Co2+ + 2Ag, Eº = 1.08, In the left compartment. Cobalt will be oxidized to cobalt (II) ion. In the right compartment, silver, ion will be reduced to silver. The passage of 0.100 mol electrons will cause the following, quantities of change., Compartment 1, Compartment 2, Compartment 3, Effect of electrode, + 0.0500 mol Co3+, – 0.100 mol Ag+, Positive ion movement –0.0250 mol Co2+, + 0.0250 mol Co2+, + 0.0500 mol K+, – 0.0500 mol K+, –, Negative ion movement +0.0500 mol NO3, – 0.0500 mol NO3–, –, 3, , 2+, , Sol., , –, , –, 3, , 3, , +, , +

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Example 2., A lead storage cell is discharged which causes the H2SO4 electrolyte to change from a, concentration of 34.6% by weight (density 1.261 g ml –1 at 25°C) to one of 27% by weight. The, original volume of electrolyte is one litre. Calculate the total charge released at anode of the, battery. Note that the water is produced by the cell reaction as H2SO4 is used up. Over all, reaction is., Pb(s) + PbO2(s) + 2H2SO4() 2PbSO4(s) +2H2O(), Before the discharge of lead storage battery,, Mass of solution = 1000 × 1.261 = 1261 g, 1261  34.6, Mass of H2SO4 =, = 436.3 g., 100, Mass of water = 1261 – 436.3 = 824.7 g, After the discharge of lead storage battery,, Let the mass of H2O produce as a result of net reaction during discharge, (Pb + PbO2 + 2H2SO4  2PbSO4 + 2H2O) is x g, x, Moles of H2O produced =, = moles of H2SO4 consumed, 18, x, Mass of H2SO4 consumed =, × 98, 18, 98 x, Now, mass of solution after discharge = 1261 –, +x, 18, Mass of H2SO4 left, % by the mass of H2SO4 after discharge =, × 100 = 27, Mass of solution after discharge, 98 x, 436.3 , 18, =, × 100 = 27, x = 22.59 g, 98 x, 1261 , x, 18, , Sol., , 8., , SECTION (H) : ELECTRICAL CONDUCTANCE, , 8.1., , Electrolytic Conductance :, , 8.2., , Factors Affecting Conductance & Resistance :, 1., , 2., , Solute : Solute interactions (Inter–Ionic force of attraction) Greater the force of attraction,, greater will be the resistance., Force  Charge, Solute : Solvent Interaction (Hydration/Solvation of Ions), Greater the solvation, 1, Solvation  Charge , greater will be resistance, size

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Li+ (Hydrated largest) Cs+ (Hydrated smallest), Resistance of LiCl > Resistance of CsCl, Solvent – Solvent interaction (Viscosity): greater the viscosity greater will be resistance, Temperature, T, R, Nature of electrolyte, Weak electrolyte – High resistance, Strong electrolyte – Low resistance, , 3., 4., 5., , Resistance :, R=, R=, , V, , , (Ohm's law ()), , , A, , – resistivity / specific resistance, – resistance of unit length wire of unit area of cross section = constant = (  m), RA, =, Resistivity of a solution is defined as the resistance of the solution between two electrodes of 1 cm 2, area of cross section and 1 cm apart., or, Resistance of 1 cm3 of solution will be it's resistivity., , Conductance :, 1, = mho =  –1, R, = S (Siemens), Conductivity/specific conductance, 1, =, =, =, unit –1  cm–1, , RA, A, = conductivity of 1 cm 3 of solution,  concentration of ions, 1, 1, =, C=, , R,   (no. of ions) no. of charge carriers, Since conductivity or resistivity of the solution is dependent on its concentration, so two more type of, conductivities are defined for the solution., C=, , , , , , , , , 8.3., , Molar conductivity/molar conductance ( m) :, , , , Conductance of a solution containing 1 mole of an electrolyte between 2 electrodes which are unit, length apart., Let the molarity of the solution 'C', C moles of electrolyte are present in 1 Lt. of solution., so molar conductance = m,   1000,   1000, m = V, m =, , m =, molarity, C, Its units are Ohm –1 cm2 mol–1, , , , Equivalent conductance : Conductivity of a solution containing 1 g equivalent of the electrolyte., , , , eq – equivalent conductivity/conductance.,   1000, eq =, Normality, Its units are Ohm –1 cm2 eq–1, , , , , , , 

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8.4., , , Ionic Mobility, Ionic Mobility = speed of the ion per unit electrical field, , , , speed, speed, =, electrical field, potential gradient, Its units are V–1 cm2 sec–1, 0, M, 0, Ionic mobility = u =, = M, 96500, F, , , , Transport Number, , , , =, , Transport Number of any ion is fraction of total current carried by that ion., 0, M, Transport Number of cation = 0, Melectrolyte, Solved Examples, If resistivity of 0.8 M KCl solution is 2.5 × 103 cm calculate m of the solution.,  = 2.5 × 10–3  cm, , Example 1., Sol., , K=, , 8.5., , , , m =, , 4  102  1000  10, = 5 × 105 –1 cm2 mole–1, 0.8, , Variation of conductivity and molar conductivity with concentration, , , Conductivity always decreases with the decrease in concentration both for weak and strong, electrolytes., The number of ions per unit volume that carry the current in a solution decreases on dilution., Molar conductivity increases with decreases in concentration. This is because the total volume,, V of solution containing one mole of electrolyte also increases., Molar conductivity is the conductance of solution., When concentration approaches zero, the molar conductivity is known as limiting molar, conductivity and is represented by the symbol º., , , , , , , , , 8.6., , 103, = 4 × 102, 2.5, , Strong Electrolytes :, , , , Example :, , , For strong electrolytes.  increases slowly with dilution and can be represented by the equation,  = º – A C1/2, The value of the constant 'A' for a given solvent and temperature depends on the type of, electrolyte i.e. the charges on the cations and anion produced on the dissociation of the, electrolyte in the solution., Thus NaCl, CaCl2, MgSO4 are known as 1-1 , 2-1 and 2-2 electrolyte respectively., All electrolytes of a particular type have the same value for 'A'.

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8.7., , Weak electrolytes, , , , Weak electrolytes like acetic acid have lower degree of dissociation at higher concentration and hence, for such electrolytes, the change in  with dilution is due to increases in the number of ions in total, volume of solution that contains 1 mol of electrolyte., At infinite dilution (i.e. concentration c zero) electrolyte dissociates completely (= 1), but at such, low concentration the conductivity of the solution is so low that it cannot be measured accurately., Molar conductivity versus c1/2 for acetic acid, (weak electrolyte) and potassium chloride, (strong electrolyte in aqueous solutions., , , , , 9., , SECTION (I) : KOHLRAUSCH LAW AND ITS APPLICATIONS, , 9.1., , Kohlarausch's Law :, , , , , , , , "At infinite dilution, when dissociation is complete, each ion makes a definite contribution, towards equivalent conductance of the electrolyte irrespective of the nature of the ion with, which it is associated and the value of equivalent conductance at infinite dilution for any, electrolyte is the sum of contribution of its constituent ions." i.e.,  = + + –, At infinite dilution or near zero concentration when dissociation is 100%, each ion makes a, definite contribution towards molar conductivity of electrolyte irrespective of the nature of the, other ion. (because interionic forces of attraction are zero), 0, 0, º m electrolyte = 0m    m, +   m, + = no. of cation in one formula unit of electrolyte, – = no. of anions in one formula unit of electrolyte, For NaCl, + = 1 – = 1, For Al2(SO4)3, + = 2 – = 3, ,  0eq electrolyte =  , eq +  eq , 0, m, Al3 , 3, ,  0eq =, , 0, m, charge on the cation, ,  0eq .Al3+ =, ,  0eq =, , 0, m, charge on the anion, ,  0eq , electrolyte =, ,  0eq Al2(SO4)3 =  0eq Al3+ +  0eq SO42 , 0, m, Al3 ,  0 SO24, + m, 3, 2, 0, 0, 2mAl, 3   3, mSO24 , 0,  eq Al2(SO4)3 =, 6, , =, , 0, m, electrolyte, total  ve charge on cation, or, total  ve charge on anion

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, , equivalent conduc tan ce at a given concentration, equivalent conduc tan ce at at inf inite dilution, , =, , , , Dissociation constant of weak electrolyte:, C 2, KC =, ;,  = degree of dissociation, C = concentration, 1 , , The degree of dissociation then it can be approximated to the ratio of molar, conductivity c at the concentration c to limiting molar conductivity, º. Thus we have :,  = /º, But we known that for a weak electrolyte like acetic acid., C 2, c 2, c 2, Ka =, =, =, (1   )  1   / ,     , , , , ,  eq, , , , , , , , =, , 0, eq, , , , Solubility(s) and KSP of any sparingly soluble salt., Sparingly soluble salt = Very small solubility, Solubility = molarity = 0, so, solution can be considered to be of zero conc or infinite dilution., K  1000, K  1000, , , m, saturated =  M =, S=, KSP = S2 (for AB type salt),  M0, So lub ility, Solved Example, Example 1., If conductivity of water used to make saturated solution of AgCl is found to be 3.1 x 10–5–1 cm–1, and conductance of the solution of AgCl = 4.5 × 10–5–1 cm–1, If  M0 AgNO3 = 200  –1 cm2 mole–1,  M0 NaNO3 = 310  –1 cm2 mole–1, Calculate KSP of AgCl, Solution:, Total conductance = 10–5,  M0 AgCl = 140, S=, Example 2., Solution:, , 140  4  105  1000, 1.4  104, =, ; S = 5.4 × 10–4 ;, 140, 14, , S2 = 1 × 10–8, , To calculate KW of water, H2O() + H2O()  H2O+(aq) + OH–(aq), 0, =  M0 H+ +  M0 OH–, m = M,H, 2O, K  1000, - Concentration of water molecules 100% dissociated Ask, molarity, K  1000, Molarity = [H+] = [OH–] =, M, , =, ,  K  1000 , KW = [H ][OH ] = , , 0,  M, , +, , , , , , , –, , 2, , Ka or Kb =, , [H  ][OH  ], H2O, , Variation of ,  m & eq of solutions with Dilution,   conc. of ions in the solution. In case of both strong and weak electrolytes on dilution the, concentration of ions will decrease hence  will decrease., , m or eq, (  C) strong electrolyte, 1000  , , m =, (  K a C ) weak electrolyte., molarity, 1000  , , eq =, normality, , C, For strong electrolyte, m , = constant, , C, C, K aC, 1, , For weak electrolyte, m , , , C, C, C

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SUMMARY, Electrochemistry is the area of chemistry concerned with the interconversion of chemical and, electrical energy. Chemical energy is converted to electrical energy in a galvanic cell, a device in which, a spontaneous redox reaction is used to produce an electric current. Electrical energy is converted to, chemical energy in an electrolytic cell, a cell in which an electric current drives a nonspontaneous, reaction. It’s convenient and reduction occur at separate electrodes., The electrode at which oxidation occurs is called the anode, and the electrode at which reduction, occurs is called the cathode. The cell potential E (also called the cell voltage or electromotive force is, an electrical measure of the driving force of the cell reaction. Cell potentials depend on temperature, ion, concentrations, and gas pressure. The standard cell are in their standard states. Cell potentials are, related to free-energy changes by the equations G = –nFE and Gº = –nFEº, where F = 96,500 C/mol, e– is the faraday, the charge on 1 mol of electrons., The standard reduction potential for the a half-reaction is defined relative to an arbitrary value of 0 V for, the standard hydrogen electrode (S.H.E.) :, , 2H+(aq, 1 M ) + 2 e–  H2(g, 1 atm), , Eº = 0 V, , Tables of standard reduction potentials--- are used to arrange oxidizing and reducing agents in order of, increasing strength, to calculate Eº values for cell reactions, and to decide whether a particular redox, reaction is spontaneous., Cell potential under nonstandard-state conditions can be calculated using the Nernst equation,, E = Eº –, , 0.0592, log Q in volts, at 25ºC, n, , where Q is the reaction quotient. The equilibrium constant K and the standard cell potential Eº are, related by the equation Eº =, , 0.0592, log K in volts, at 25ºC, n, , A battery consists of one or more galvanic cells. A fuel cell differs from a battery in that the reactants, are continuously supplied to the cell. Corrosion of iron (rusting) is an electrochemical process surface, and oxygen is reduced in a cathode region. Corrosion can be prevented by covering iron with another, metal, such as zinc, in the process called galvanizing, or simply by putting the iron in electrical contact, with a second metal that is more easily oxidized, a process called cathodic protection., Electrolysis, the process of using an electric current to bring about chemical change, is employed to, produce sodium, chlorine, sodium hydroxide, and aluminum (Hall-Heroult process) and is used in, electrorefining and electroplating. The product obtained at an electrode depends on the reduction, potentials and overvoltage. The amount of product obtained is related to the number of moles of, electrons passed through the cell, which depends on the current and the time that the current flows.

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Electrochemistry, , relates, , Electrore fining, and, electroplating, , Electrical energy, , Chemical energy, Batteries, , used in, used in, , interconverted in, , represented, using, , Shorthand, notation, , Galvanic cells, (Voltaic cells), , Electroytic cells, , process, called, , Electrolysis, , reaction are, , reaction are, having, , producing, related to, 1) Free energy, o, change, G, 2) Equilibrium, constant, K, , G = –, E=+, , Standard, reduction, potentials, , E = E° –, , requiring, G = (+), E = (–), , Electric current, , oxidation half-reaction, occurring at the, Anode, , calculate from, , Nonspontaneous, oxidation-reduction, , Spontaneous, oxidation-reduction, , Standard, o, cell potential, E, , reducing half-reaction, occurring at the, Cathode, , used in, Nernst, equation, , (Pr oducts), .0592, log, (Reac, tan ts), n, , , , E = E° –, , (Anodic ion concentration), .0592, log, (Cathodic, ion concentration), n

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MISCELLANEOUS SOLVED PROBLEMS (MSPS), Problem 1, , Solution:, , , , , Problem 2, , Na-amalgam is prepared by electrolysis of NaCl solution using liquid Hg as cathode. How long, should the current of 10 amp. is passed to produce 10% Na–Hg on a cathode of 10 g Hg., (Atomic mass of Na = 23)., (A) 7.77 min, (B) 9.44 min., (C) 5.24 min., (D) 11.39 min., (A), 90 g Hg has 10 g Na, 10, 10, , 10 g Hg =, × 10 =, g Na, 9, 90, M, it, , Weight of Na =, ×, 96500, n, 10, 23, 10  t, =, ×, [ Na+ + e  Na], 9, 96500, 1, 10  96500, , t=, = 7.77 min, 9  10  23, We have taken a saturated solution of AgBr. Ksp of AgBr is 12 × 10–14. If 10–7 mole of AgNO3, are added to 1 litre of this solution then the conductivity of this solution in terms of 10 –7 Sm–1, units will be, [Given  º(Ag ) = 4 × 10–3 Sm2 mol–1, º(Br  ) = 6 × 10–3 S m2 mol–1,  º(NO ) = 5 × 10–3 Sm2 mol–1], 3, , Solution:, , (A) 39, (B) 55, (C) 15, (A), The solubility of AgBr in presence of 10–7 molar AgNO3 is 3 × 10–7 M., Therefore [Br–] = 3 × 10–4 m3, [Ag+] = 4 × 10–4 m3 and [NO3–] = 10–4 m3, Therefore total =  Br  +  Ag +  NO = 39 Sm–1, , (D) 41, , 3, , Problem 3, , Solution:, , A hydrogen electrode X was placed in a buffer solution of sodium acetate and acetic acid in the, ratio a : b and another hydrogen electrode Y was placed in a buffer solution of sodium acetate, and acetic acid in the ratio b : a. If reduction potential values for two cells are found to be E 1, and E2 respectively w.r.t. standard hydrogen electrode, the pKa value of the acid can be given, as, E, E  E2, E  E2, E  E1, (A) 1, (B) – 1, (C) 1  0.118, (D) 2, E2, 0.118, 0.118, 0.118, (B), 1, H+ + e–  H2(g), 2, 1, E1 = 0 – 0.0591 log , (H )1, E1 = 0 + 0.0591 log [H+]1 = –0.0591 pH1, E2 = – 0.0591 pH2, Salt, pH1 = pka + log, Acid, a, pH1 = pka + log, .............. (1), b, b, pH2 = pka + log, a, a, pH2 = pka – log, .............. (2), b, Add (1) & (2) pH1 + pH2 = 2 pka, E1, E2,  E  E2 , 2pka = –, , , pka = –  1, , , 0.0591 0.0591,  0.118 

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Problem 4, , At what, , [Br  ], [CO 32  ], , does the following cell have its reaction at equilibrium?, , Ag(s)  Ag2 CO3(s)  Na2CO3 (aq)  KBr(aq) AgBr(s)  Ag(s), KSP = 8 × 10–12 for Ag2 CO3 and KSP = 4 × 10–13 for AgBr, (A) 1 × 10–7, (B) 2 × 10–7, (C) 3 × 10–7, Solution:, , (B), , (D), , 4 × 10–7, , Ag(s) ,  Ag+ (aq) + 1e–, , Anode :, , Cathode :, Ag+(aq) + 1e– ,  Ag, ––––––––––––––––––––––––––––––––––––, , , 1e, , Ag(AgBr ) ,  Ag(Ag, 2CO3 ), , Net :, , 0=0+, , 0.059, log, 1, ,  K SP AgBr , , ,  [Br  ] , K SP Ag2CO3, , , , K SP, , AgBr, , [Br  ], , [CO32 ], , , , , , 4  10 13, 8  10 12, , =, , [Br  ], [CO 32  ], , , , [Br  ], [CO 32  ], , =, , =, , K SP Ag2CO3, [CO32 ], , 2 × 10–7, , Problem 5, , Solution:, , Problem 6, , Solution:, , , , Problem 7, , A resistance of 50 is registered when two electrodes are, suspended into a beaker containing a dilute solution of a strong, electrolyte such that exactly half of the them are submerged, into solution. If the solution is diluted by adding pure water, (negligible conductivity) so as to just completely submerge the, electrodes, the new resistance offered by the solution would be, (A) 50 , (B) 100 , (C) 25 , (D) 200 , (A), 1, R=, k A, The k is halved while the A is doubled. Hence R remains 50 ., Calculate the cell EMF in mV for, Pt  H2 (1atm)  HCl (0.01 M)  AgCl(s)  Ag(s), at 298 K, If Gf° values are at 25°C., kJ, kJ, – 109.56, for AgCl(s) and – 130.79, for (H+ + Cl–) (aq), mol, mol, (A) 456 mV, (B) 654 mV, (C) 546 mV, (A), G0cell reaction = 2 (– 130.79) – 2 (– 109.56) = – 42.46 kJ/mole, , (D) None of these, , (for H2 + 2AgCl 2Ag + 2H+ + 2Cl– ), 42460, E0cell =, , = + 0.220 V, 2  96500, 1, 0.059, Now Ecell = + 0.220 +, log, = 0.456 V = 456 mV., 2, (0.01)4, Consider the cell Ag(s) | AgBr(s)|Br(aq) || AgCl(s) | Cl(aq)| Ag(s) at 25ºC. The solubility, product constants of AgBr & AgCl are respectively 5 × 1013 & 1 × 1010. For what ratio of the, concentrations of Br & Cl ions would the emf of the cell be zero ?, (A) 1 : 200, (B) 1 : 100, (C) 1 : 500, (D) 200 : 1

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Solution:, , (A), , 0, EBr, , , AgBr Ag, , and, , E0Cl, , AgCl Ag, , = E0Ag, = E0Ag, , Ag, , Ag, , 0.059, log KSP AgBr = E0Ag, 1, 0.059, +, log KSP AgCl = E0Ag, 1, +, , Now cell reaction is, Ag + Br–  AgBr + 1e–, AgCl + 1e–  Ag + Cl–, Br– + AgCl  Cl– + AgBr, 0.059, [Br  ], 0 = (0.7257 – 0.59) +, log, 1, [Cl ], , , , [Br  ], [Cl ], , Ag, , Ag, , – 0.7257, – 0.59, , = 0.005, , Problem 8, , The conductivity of a solution may be taken to be directly proportional to the total concentration, of the charge carries (ions) present in it in many cases. Using the above find the percent, decrease in conductivity (k) of a solution of a weak monoacidic base BOH when its 0.1 M, solution is diluted to double its original volume. (Kb = 10–5 for BOH) (Take 50 = 7.07) (Mark, the answer to nearest integer),, , Solution:, , Initially [OH–] = 10 5  0.1 = 10–3, [ions]total = 2 × 10–3 M, 1, later, [OH–] = 10 5 , = 50 × 10–4 M, 20, , , , , , [ions]total = 2 50 × 10–4 M, , , , , , % change on [ions]total =, , Problem 9, , Solution:, , At 0.04 M concentration the molar conductivity of a solution of a electrolyte is 5000 –1 cm2, mol–1 while at 0.01 M concentration the value is 5100 –1 cm2 mol–1. Making necessary, assumption (Taking it as strong electrolyte) find the molar conductivity at infinite dilution and, also determine the degree of dissociation of strong electrolyte at 0.04 M., , From the graph we can see the  M, value of 5200 –1 cm2 mol–1. Hence, , , , =, , 5000, = 0.9615  0.96, 5200, , 2 50  20, × 100 = – 29.29%, 20, , Ans. 96, , Ans. 29

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 Marked Questions may have for Revision Questions., , PART - I : SUBJECTIVE QUESTIONS, Section (A) : Galvanic cell, its Representation & salt bridge, Commit to memory :, Notation for Galvanic cell :, , , , LOAN  left oxidation- Anode-negative, , A-1. In the galvanic cell Cu | Cu2+ || Ag+ | Ag, the electrons flow from Cu-electrode to Ag-electrode. Answer, the following questions regarding this cell :, (a) Which is the anode ?, (b) Which is the cathode ?, (c) What happens at anode-reduction or oxidation ?, (d) What happens at cathode-oxidation or reduction ?, (e) Which electrode loses mass ?, (f) Which electrode gains mass ?, (g) Write the electrode reactions., (h) Write the cell reaction, (i) Which metal has greater tendency to loss electron-Cu or Ag ?, (j) Which is the more reactive metal-Cu or Ag ?, (k) What is the function of salt bridge represented by the symbol || ?, A-2., , A-3., , Write cell reaction of the following cells :, (a) Cu | Cu2+(aq) | | Ag+(aq) | Ag, (c) Pt, Cl2 | Cl–(aq) | | Ag+(aq) | Ag, , (b) Pt | Fe2+, Fe3+ | | MnO4–, Mn2+, H+ | Pt, (d) Cd | Cd2+(aq) | | H+(aq) | H2 | Pt, , Write cell notation of each cell with following cell reactions :, (a) Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g), (b) 2Fe3+(aq) + Sn2+(aq)  2Fe2+(aq) + Sn4+(aq), (c) Pb(s) + Br2(I)  Pb2+(aq) + 2Br–(aq), , Section (B) : Electrochemical series & its Applications, Commit to memory :, SRP  Oxidising power , B-1., , 1, 1,  Non-metallic character , reducing power, Metallic character, , The reduction potential values are given below, Al3+/Al = –1.67 volt, Mg2+/Mg = –2.34 volt, Cu2+/Cu = +0.34 volt, 2/– = +0.53 volt., Which one is the best reducing agent ?, , B-2. The standard reduction potential value of the three metallic cations X, Y and Z are 0.52, –3.03 and, –1.18 V respectively. Write the decreasing order of reducing power of the corresponding metals :, B-3. (i), (ii), (iii), , Which of the following oxides is reduced by hydrogen ?, MgO, CuO and Na2O, Which of the following oxides will decompose most easily on heating ?, ZnO, CuO, MgO and Ag2O, The value of EºOX for electrode reactions,, Fe Fe2+ + 2e–,, Cu Cu2+ + 2e–, and, Zn  Zn2+ + 2e–, are 0.444, –0.337 and 0.763 volt respectively. State which of these metals can replace the, other two from the solution of their salts ?

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D-3. The standard reduction potential of Cu2+ / Cu couple is 0.34 V at 25°C. Calculate the reduction potential, at pH = 14 for this couple., (Given : Ksp, Cu (OH)2 = 1.0 × 10–19)., D-4. The EMF of the cell M |Mn+ (0.02 M) || H+ (1M) | H2 (g) (1 atm), Pt at 25°C is 0.81V. Calculate the, valency of the metal if the standard oxidation potential of the metal is 0.76V., D-5., , Consider the following electrochemical cell :, (a) Write a balanced net ionic equation for the spontaneous reaction that take place in the cell., (b) Calculte the standard cell potential Eº for the cell reaction., (c) If the cell emf is 1.6 V, what is the concentration of Zn2+ ?, (d) How will the cell potential be affected if KI is added to Ag+ half-cell ?, , D-6. NO3  NO2 (acid medium), Eº = 0.790 V, NO3  NH3OH+ (acid medium), Eº = 0.731 V., At what pH, the above two will have same E value? Assume the concentration of all other species, NH3OH+ except [H+] to be unity., D-7., , The standard oxidation potential of Zn referred to SHE is 0.76V and that of Cu is –0.34V at 25ºC. When, excess of Zn is added to CuSO4, Zn displaces Cu2+ till equilibrium is reached. What is the approx value, [Zn2 ], of log, at equilibrium?, [Cu2 ], , Section (E) : Electrolysis, Commit to memory :, Higher SOP means higher tendency of oxidation., Higher SRP means higher tendency of reduction., SOP order : SO42– < NO3– < Cl– < H2O < Br– < Ag < I– < OH– < Cu.....< Li, SRP order : Follow ECS, E-1., 1, 2, 3, 4, 5, 6, 7, , ELECTROLYTE, NaCl (Molten) with Pt electrode, NaCl (aq) with Pt electrode, Na2SO4 (aq) with Pt electrode, NaNO3 (aq) with Pt electrode, AgNO3 (aq) with Pt electrode, CuSO4 (aq) with Inert electrode, CuSO4 (aq) with Copper electrode, , Section (F) : Faraday laws & its Applictions, Commit to memory :, Faraday's law of electrolysis :, EQ, Ist law, W = ZQ =, 96500, Q = it, , ANODE Product CATHODE Product

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W1, Z, E, = 1 = 1 (Q = same), W2, Z2, E2, , 2nd law, , Current efficiency () =, , actual amount of product, × 100, theortical amount of product, ,  EQ  , W actual = ,  96500  100, , F-1., , Find the number of electrons involved in the electrodeposition of 63.5 g of copper from a solution of, copper sulphate is :, , F-2. A current 0.5 ampere when passed through AgNO3 solution for 193 sec. deposited 0.108 g of Ag . Find, the equivalent weight of Ag :, F-3. A certain metal salt solution is electrolysed in series with a silver coulometer. The weights of silver and, the metal deposited are 0.5094 g and 0.2653g. Calculate the valency of the metal if its atomic weight is, nearly that of silver., F-4., , 3A current was passed through an aqueous solution of an unknown salt of Pd for 1Hr. 2.977g of Pd +n, was deposited at cathode. Find n. (Given Atomic mass of Pd = 106.4), , F-5., , How long a current of 2A has to be passed through a solution of AgNO 3 to coat a metal surface of, 80cm2 with 5m thick layer? Density of silver = 10.8g/cm 3., , F-6. A certain electricity deposited 0.54g of Ag from AgNO 3 Solution. What volume of hydrogen will the, same quantity of electricity liberate at STP (Vm = 22.4 L/mol)., F-7. A current of 3.7A is passed for 6hrs. between Ni electrodes in 0.5L of 2M solution of Ni(NO 3)2. What will, be the molarity of solution at the end of electrolysis?, F-8., , Cd amalgam is prepared by electrolysis of a solution of CdCl 2 using a mercury cathode. How long, should a current of 5A be passed in order to prepare 12% Cd-Hg amalgam when 2 g Hg is used as, cathode (atomic weight of Cd = 112.4), , F-9. Electrolysis of a solution of HSO4– ions produces S2O82–. Assuming 75% current efficiency, what current, should be employed to achieve a production rate of 1 mole of S2O82– per hour ?, , Section (G) : Commercial Cells & Corrosion, Commit to memory :, At STP, Vm (molar volume of the gas) = 22.4 L/mol, Volume of gas required at STP = moles of gas × 22.4, G-1. A fuel cell uses CH4(g) and forms CO32– at the anode. It is used to power a car with 80 Amp. for 0.96 hr., How many litres of CH4(g) (STP) would be required ? (Vm = 22.4 L/mol) (F = 96500). Assume 100%, efficiency., G-2., , Find Eº of cell formed for rusting of iron ?, 0, EFe, = +0.44 V, / Fe2 , 0, = –1.23 V, EH, O|O |H, 2, , 2, , Section (H) : Electrical Conductance, Commit to memory :, Conductivity () = C ×, where,, , A, , =, , 1, ×, R, A, , = cell constant, C = conductance, R = resistance., A, A = surface area of electrodes, = distance between electrodes.

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  1000, S cm2 mol–1, M,   1000, Equivalent conductance (eq) =, S cm2 eq–1, N, where, M = molarity, N = normality and N = M × valence factor, Molar conductance (m) =, , H-1., , The resistance of a M/10 KCl solution in 245 ohms. Calculate the specific conductance and the molar, conductance of the solution if the electrodes in the cell are 4 cm apart and each having an area of 7.0, sq. cm., , H-2. The equivalent conductance of 0.10 N solution of MgCl 2 is 97.1 mho cm2 eq.–1 at 25°C. A cell with, electrodes that are 1.50 cm 2 in surface area and 0.50 cm apart is filled with 0.1N MgCl 2 solution. How, much current will flow when the potential difference between the electrodes is 5 volts ?, H-3., , The specific conductance of a N/10 KCl solution at 18°C is 1.12 × 10–2 mho cm–1. The resistance of the, solution contained in the cell is found to be 65 ohms. Calculate the cell constant., , Section (I) : Kohlrausch law and its applications, Commit to memory :, Kohlrausch law : At infinite dilution, ºm, electrolyte =   ºm +   ºm, where, , , + = number of cations in one formula unit of electrolyte., – = number of anions in one formula unit of electrolyte., At infinite dilution equivalent conductance :, ºeq, electrolyte = ºeq+ + ºeq–,  eq, m, Degree of dissociation (D.O.D.) =  =, = 0,  ºm,  eq, For weak electrolyte, dissociation constant (Ka) =, Solubility (s) =, , C 2, , where, C = concentration of electrolyte., 1 , ,   1000, and Ksp = S2 for AB type salt.,  ºm, , I-1., , The molar conductance of an infinitely dilute solution of NH4Cl is 150 and the ionic conductances of, OH– and Cl– ions are 198 and 76 respectively. What will be the molar conductance of the solution of, NH4OH at infinite dilution. If the molar conductance of a 0.01 M solution NH4OH is 9.6, what will be its, degree of dissociation?, , I-2., , Given the molar conductance of sodium butyrate, sodium chloride and hydrogen chloride as 83, 127, and 426 mho cm2 mol–1 at 25°C respectively. Calculate the molar conductance of butyric acid at infinite, dilution., , I-3., , Calculate Ka of acetic acid if its 0.05 N solution has equivalent conductance of 7.36 mho cm 2 at 25°C., , (  CH,  390.7) ., 3 COOH, , I-4., , The specific conductance of a saturated solution of AgCl at 25°C after subtracting the specific, conductance of conductivity of water is 2.28 × 10–6 mho cm–1. Find the solubility product of AgCl at, 25°C. (  AgCl  138.3 mho cm2 ), , Section (J) : Conductometric Titration, Commit to memory :, H+ and OH– ions are highly conducting., J-1. Draw approximate titration curve for following –, (1) HCl(aq) is titrated with NaOH, (2) CH3COOH(aq) is titrated with NaOH, (3) Equimolar mixture of HCl and HCN titrated withNaOH, (4) NH4Cl(aq) is titrated withNaOH

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PART - II : ONLY ONE OPTION CORRECT TYPE, * Marked Questions are having more than one correct option., Section (A) : Galvanic cell, its Representation & salt bridge, A-1., , In a galvanic cell, (A) Chemical reaction produces electrical energy (B) electrical energy produces chemical reaction, (C) reduction occurs at anode, (D) oxidation occurs at cathode, , A-2. Which of the following is/are function(s) of salt-bridge ?, (A) It completes the electrical circuit with electrons flowing from one electrode to the other through, external wires and a flow of ions between the two compartments through salt - bridge, (B) it minimises the liquid - liquid junction potential, (C) both correct, (D) none of these, A-3., , Salt bridge contains :, (A) calomel, , (B) sugar, , (C) H2O, , (D) agar-agar paste, , A-4. The emf of the cell, Ni | Ni (1.0 M) || Ag (1.0M) | Ag [E° for Ni / Ni = –0.25 volt, E° for Ag+/Ag = 0.80, volt] is given by (A) –0.25 + 0.80 = 0.55 volt, (B) –0.25 – (+0.80) = –1.05 volt, (C) 0 + 0.80 – (–0.25) = + 1.05 volt, (D) –0.80 – (–0.25) = – 0.55 volt, 2+, , +, , 2+, , Section (B) : Electrochemical series & its Applications, B-1. Eº for F2 + 2e  2F– is 2.8 V, Eº for, (A) 2.8 V, B-2., , (B) 1.4 V, , 1, F2 + e  F– is, 2, (C) – 2.8 V, , (D) – 1.4 V, , º, º, Consider the cell potentials = EMg, – 2.37 V and EFe, = – 0.04 V. The best reducing agent would, 3, 2, | Fe, | Mg, , be, (A) Mg2+, , (B) Fe3+, , (C) Mg, , (D) Fe, , B-3. If a spoon of copper metal is placed in a solution of ferrous sulphate (A) Cu will precipitate out, (B) iron will precipitate, (C) Cu and Fe will precipitate, (D) no reaction will take place, B-4. The position of some metals in the electrochemical series in decreasing electropositive character is, given as Mg > Al > Zn > Cu > Ag. What will happen if a copper spoon is used to stir a solution of, aluminium nitrate ?, (A) The spoon will get coated with aluminium, (B) An alloy of aluminium and copper is formed, (C) The solution becomes blue, (D) There is no reaction, B-5., , For Zn2+ / Zn, E° = –0.76 V, for Ag+/Ag E° = 0.799 V. The correct statement is (A) the reaction Zn getting reduced Ag getting oxidized is spontaneous, (B) Zn undergoes reduction and Ag is oxidized, (C) Zn undergoes oxidation Ag+ gets reduced, (D) No suitable answer, , B-6. Electrode potential data are given below., Fe3+ (aq) + e– Fe2+ (aq);, Eº = + 0.77, Al3+ (aq) + 3e–  Al (s);, Eº = – 1.66 V, Br2 (aq) + 2e–  2Br– (aq) ;, Eº = + 1.08 V, Based one the data given above, reducing power of Fe2+, Al and Br– will increase in the order :, (A) Br– < Fe2+ < Al, (B) Fe2+ < Al < Br–, (C) Al < Br– < Fe2+, (D) Al < Fe2+ < Br–, B-7. KCl can be used in salt bridge as electrolyte in which of the following cells?, (A) Zn | ZnCl2 || AgNO3 | Ag, (B) Pb | Pb(NO3)2 || Cu(NO3)2 | Cu, (C) Cu | CuSO4 || AuCl3 | Au, (D) Fe | FeSO4 || Pb(NO3)2 | Pb

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B-8. Consider the following Eº values :, 0, EFe, = + 0.77 V, ;, 3, / Fe2 , , E0Sn2  / Sn = – 0.14 V, , Under standard conditions the potential for the reaction is, Sn (s) + 2 Fe3+ (aq)  2Fe2+ (aq) + Sn2+ (aq), (A) 1.68V, (B) 1.40 V, (C) 0.91 V, , (D) 0.63 V, , Section (C) : Concept of G, C-1., , C-2., , Given standard electrode potentials :, Fe3+ + 3e–  Fe ;, E° = –0.036 volt, Fe2+ + 2e–  Fe;, E° = –0.440 volt, The standard electrode potential E° for Fe3+ + e–  Fe2+, (A) –0.476 volt, (B) –0.404 volt, (C) 0.440 volt, , (D) 0.772 volt, , Cu+ + e–  Cu, E° = x1 volt; Cu2+ + 2e–  Cu, E° = x 2 volt, then for Cu2+ + e–  Cu+, E° (volt), will be (A) x1 – 2x2, (B) x1 + 2x2, (C) x1 – x2, (D) 2x2 – x1, , C-3. Which of the following statements about the spontaneous reaction occurring in a galvanic cell is always, true?, (A) Eºcell > 0, Gº < 0, and Q < K, (B) Eºcell > 0, Gº < 0, and Q > K, (C) Eºcell > 0, Gº > 0, and Q > K, (D) Ecell > 0, G < 0, and Q < K, , Section (D) : Nernst equation & its Applications (including concentration cells), D-1. The standard emf for the cell reaction Zn + Cu2+  Zn2+ + Cu is 1.10 volt at 25°C. The emf for the cell, reaction when 0.1 M Cu2+ and 0.1 M Zn2+ solutions are used at 25°C is, (A) 1.10 volt, (B) 0.110 volt, (C) –1.10 volt, (D) –0.110 volt, D-2., , D-3., , D-4., , D-5., , Consider the cell, , Ag, H2 (Pt) H3 O (aq), Ag. The measured EMF of the cell is 1.0 V. What is the, 1 atm pH  5.03 x M, , value of x ? E0Ag ,Ag = + 0.8 V. [T = 25°C] ;, , E0Ag ,Ag = + 0.8 V. [T = 25°C], , (A) 2 × 10–2 M, , (C) 1.5 × 10–3 M, , (B) 2 × 10–3 M, , (D) 1.5 × 10–2 M, , Zn | Zn2+ (C1)|| Zn2+ (C2)|Zn. for this cell G is negative if (A) C1 = C2, (B) C1 > C2, (C) C2 > C1, , (D) None, , H, (p1 ) (1 M), spontaneous if :, (A) p1 = p2, , (D) p1 = 1 atm, , Pt, , H2, , H2, H, Pt, (1 M) (p2 ), , (B) p1 > p2, , (where p1 and p2 are pressures) cell reaction will be, , (C) p2 > p1, , Pt | (H2) | pH = 1 || pH = 2 | (H2)Pt, 1 atm, 1 atm, The cell reaction for the given cell is :, (A) spontaneous, (B) non - spontaneous (C) equilibrium, , (D) none of these, , D-6. The EMF of a concentration cell consisting of two zinc electrodes, one dipping into, , M, sol. of the same salt at 25C is, 16, (B) 0.0250 V, (C) 0.0178 V, , M, sol. of zinc, 4, , sulphate & the other into, (A) 0.0125 V, , (D) 0.0356 V, , Section (E) : Electrolysis, E-1. In an electrolytic cell of Ag/AgNO3/Ag, when current is passed, the concentration of AgNO3, (A) Increases, (B) Decreases, (C) Remains same, (D) None of these

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E-2., , If 0.224 L of H2 gas is formed at the cathode, the volume of O2 gas formed at the anode under identical, conditions, is, (A) 0.224 L, (B) 0.448 L, (C) 0.112 L, (D) 1.12 L, , E-3. The two aqueous solutions, A (AgNO3) and B (LiCl) were electrolysed using Pt. electrodes. The pH of, the resulting solutions will, (A) increase in A and decrease in B, (B) decrease in both, (C) increase in both, (D) decrease in A and increase in B., E-4., , In the electrolysis of aqueous CuBr2 using Pt electrodes :, (A) Br2 gas is not evolved at the anode, (B) Cu (s) is deposited at the cathode, (C) Br2 gas is evolved at anode and H2 gas at cathode, (D) H2 gas is evolved at anode., , E-5. During electrolysis of CuSO4 using Pt-electrodes, the pH of solution, (A) increases, (B) decreases, (C) remains unchanged (D) cannot be predicted, , Section (F) : Faraday laws & its Applictions, F-1. How many faradays are required to reduce one mol of MnO4¯ to Mn2+ (A) 1, (B) 2, (C) 3, F-2., , (D) 5, , Three faradays of electricity was passed through an aqueous solution of iron (II) bromide. The mass of, iron metal (at. mass 56) deposited at the cathode is (A) 56 g, (B) 84 g, (C) 112 g, (D) 168 g, , F-3. A current of 2 A was passed for 1 h through a solution of CuSO4 0.237 g of Cu2+ ions were discharged, at cathode. The current efficiency is, (A) 42.2%, (B) 26.1%, (C) 10%, (D) 40.01%, F-4., , A current of 9.65 ampere is passed through the aqueous solution NaCl using suitable electrodes for, 1000 s. The amount of NaOH formed during electrolysis is, (A) 2.0 g, (B) 4.0 g, (C) 6.0 g, (D) 8.0 g, , F-5. Salts of A (atomic mass 15), B (atomic mass 27) and C (atomic mass 48) were electrolysed using same, amount of charge. It was found that when 4.5 g of A was deposited, the mass of B and C deposited, were 2.7g and 9.6 g. The valencies of A, B and C respectively., (A) 1, 3 and 2, (B) 3, 1 and 3, (C) 2, 6 and 3, (D) 3, 1 and 2, , Section (G) : Commercial Cells & Corrosion, G-1. During discharge of a lead storage cell the density of sulphuric acid in the cell :, (A) Increasing, (B) decreasing, (C) remians unchanged, (D) initially increases but decrease subsequently, G-2. In H2O2 fuel cell the reaction occuring at cathode is :, (A) 2 H2O + O2 + 4 e  4 OH, (B) 2H2 + O2  2H2O (l), 1, (C) H+ + OH  H2O, (D) H+ + e  H2 ., 2, G-3., , Which is not correct method for prevention of iron from Rusting (A) Galvanisation, (B) Connecting to sacrificial electrode of Mg, (C) Making medium alkaline, (D) Making medium acidic

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Section (H) : Electrical Conductance, H-1. Which of the following curve represents the variation of M with, , (A), , (B), C, , (C), C, , H-2. Which has maxmium conductivity :, (A) [Cr(NH3)3 Cl3], (B) [Cr(NH3)4 Cl2]Cl, H-3., , C for AgNO3 ?, , (D), C, , (C) [Cr(NH3)5Cl]Cl2, , C, , (D) [Cr(NH3)6]Cl3, , Resistance of decimolar solution is 50 ohm. If electrodes of surface area 0.0004 m 2 each are placed at, a distance of 0.02 m then conductivity of solution is :, (A) 1 s cm–, (B) 0.01 s cm–, (C) 0.001 s cm–, (D) 10 s cm–, , Section (I) : Kohlrausch law and its applications, I-1., , The ionization constant of a weak electrolyte (HA) is 25 × 10–6 while the equivalent conductance of its, 0.01 M solution is 19.6 S cm 2 eq–1. The equivalent conductance of the electrolyte at infinite dilution, (in S cm2 eq–1 ) will be, (A) 250, (B) 196, (C) 392, (D) 384, , I-2., , The conductivity of a saturated solution of BaSO4 is 3.06 × 10–6 ohm–1 cm–1 and its equivalent, conductance is 1.53 ohm–1 cm2 equiv–1. The Ksp for BaSO4 will be, (A) 4 × 10–12, (B) 2.5 × 10–13, (C) 25 × 10–9, (D) 10–6, , I-3., , Molar conductance of 0.1 M acetic acid is 7 ohm 1 cm2 mol 1. If the molar cond. of acetic acid at infinite, dilution is 380.8 ohm 1 cm2 mol 1, the value of dissociation constant will be :, (A) 226  10 5 mol dm 3, (B) 1.66  10 3 mol dm 1, 2, 3, (C) 1.66  10 mol dm, (D) 3.442  10 5 mol dm 3, , I-4., , The conductivity of a solution of AgCl at 298 K is found to be 1.382 × 10–6 –1 cm–1. The ionic, conductance of Ag+ and Cl– at infinite dilution are 61.9 –1 cm2 mol–1 and 76.3 –1 cm2 mol–1,, respecitvley. The solubility of AgCl is, (A) 1.4 × 10–5 mol L–1 (B) 1 × 10–2 mol L–1, (C) 1 × 10–5 mol L–1, (D) 1.9 × 10–5 mol L–1, , I-5., , Molar conductances of BaCl2, H2SO4 and HCl at infinite dilutions are x 1, x2 and x3, respectively., Equivalent conductance of BaSO4 at infinite dilution will be :, [x  x 2  x3 ], [x  x 2  x3 ], [x  x2  2x3 ], (A) 1, (B) 1, (C) 2 (x1 + x2 – 2x3), (D) 1, 2, 2, 2, , J-1. 20 ml KOH solution was titrated with 0.2 mol/l H2SO4 solution in, conductivity cell. Concentration of KOH solution was –, (A) 0.3 M, (B) 0.15, (C) 0.12, (D) None of these, , Conductivity, , Section (J) : Conductometric Titration, , 15, , 30, , 0 Volume of H2 SO 4 (ml)

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J-2., , Following curve for conductometric titration is obtained when –, , Conduction, , , x = Equivalence point, x, Volume of Solutions , , (A) NaOH solution is added in to HCl solution, (B) NaOH solution is added in to CH3COOH solution, (C) NH4OH solution is added in to HCl solution, (D) NH4OH solution is added in to CH3COOH solution, , PART - III : MATCH THE COLUMN, 1., , Match the column, Column I, (A), (B), (C), , (D), , 2.^, , Column II, , Zn | Zn+2 | | Mg2+ | Mg, c1, , c2, , (c1 = c2), , Zn | Zn+2 | | Ag+ | Ag at. equilibrium, Ag | Ag+ | | Ag+ | Ag, c1, , c2, , (c1 = c2), , Fe | Fe+2 | | Ag | Ag+, c1, , c2, , (c1 = c2), , (p), , Ecell = 0, , (q), , E0cell 0, , (r), , E0cell = +ve, , (s), , E0cell = –ve, , Match Matrix ( E0Ag / Ag = 0.8)., Column I, (A), , Column II, , Pt | H2 (0.1 bar) | H+ (0.1 M) || H+ (1 M) | H2 (0.01 bar) | Pt, +, , –9, , +, , –2, , (p), , Concentration cell, , (B), , Ag | Ag (10 M) || Ag (10 M) | Ag, , (q), , Ecell > 0, , (C), , Cu | Cu2+ (0.1 M) || Cu2+ (0.01 M) | Cu, , (r), , Eºcell = 0, but cell is working., , (D), , Pt | Cl2 (1bar) | HCl (0.1 M) || NaCl (0.1M) | Cl2 | Pt (1 bar), , (s), , non working condition, , , ,  Marked Questions may have for Revision Questions., , PART - I : ONLY ONE OPTION CORRECT TYPE, 1., , Given : Eº(Cu2+ | Cu) = 0.337 V and Eº (Sn2+ | Sn) = – 0.136 V. Which of the following statements is, correct?, (A) Cu2+ ions can be reduced by H2(g), (B) Cu can be oxidized by H+, 2+, (C) Sn ions can be reduced by H2(g), (D) Cu can reduce Sn2+

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2. Using the standard potential values given below, decide which of the statements I, II, III, IV are correct., Choose the right answer from (a), (b), (c) and (d), Fe2+ + 2e– = Fe,, Eº = – 0.44 V, Cu2+ + 2e– = Cu,, Eº = + 0.34 V, Ag+ + e– = Ag,, Eº = + 0.80 V, I. Copper can displace iron from FeSO4 solution, II. Iron can displace copper from CuSO4 solution, III. Silver can displace Cu from CuSO4 solution, IV. Iron can displace silver from AgNO3 solution, (A) I and II, (B) II and III, (C) II and IV, (D) I and IV, 3., , Red hot carbon will remove oxygen from the oxide AO and BO but not from MO, while B will remove, oxygen from AO. The activity of metals A, B and M in decreasing order is, (A) A > B > M, (B) B > A > M, (C) M > B > A, (D) M > A > B, , 4., , Which statement is correct., (A) In SHE, the pressure of dihydrogen gas should be low and pH of solution should be zero., (B) In the reaction H2O2 + O3  2H2O + 2O2, H2O2 is oxidised to H2O., (C) The absolute value of electrode potential cannot be determined., (D) According to IUPAC conventions, the standard electrode potential pertains to oxidation reactions, only., , 5., , The electrode oxidation potential of electrode M(s)  Mn+(aq) (2M) + ne– at 298 K is E1. When, temperature (in ºC) is doubled and concentration is made half, then the electrode potential becomes E2., Which of the following represents the correct relationship between E1 and E2 ?, (A) E1 > E2, (B) E1 < E2, (C) E1 = E2, (D) Cann’t be predicted, , 6., , A galvanic cell is composed of two hydrogen electrodes, one of which is a standard one. In which of the, following solutions should the other electrode be immersed to get maximum emf ?, Ka(CH3COOH) = 2 × 10–5, Ka(H3PO4) = 10–3., (A) 0.1 M HCl, (B) 0.1 M CH3COOH, (C) 0.1 M H3PO4, (D) 0.1 M H2SO4, , 7., , Two weak acid solutions HA1 and HA2 each with the same concentration and, having pKa values 3 and 5 are placed in contact with hydrogen electrode (1, atm, 25ºC) and are interconnected through a salt bridge. The emf of the cell, is :, (A) 0.21 V, (B) 0.059 V, (C) 0.018 V, (D) 0.021 V, , 8., , A hydrogen electrode placed in a buffer solution of CH3COONa and CH3COOH in the ratios of x : y and, y : x has electrode potential values E1 volts and E2 volts, respectively at 25ºC. The pKa values of acetic, acid is (E1 and E2 are oxidation potentials), E  E2, E  E1, E  E2, E  E2, (A) 1, (B) 2, (C) – 1, (D) 1, 0.118, 0.118, 0.118, 0.118, , 9., , What is the emf at 25° C for the cell, Ag,, , AgBr (s), Br¯, a  0.34, , Fe3 , , Fe2, a  0.1 a  0.02, , Pt, , The standard reduction potentials for the half-reactions AgBr + e–  Ag + Br– and Fe3+ + e–  Fe2+, are + 0.0713 V and + 0.770 V respectively., (A) 0.474 volt, (B) 0.529 volt, (C) 0.356 volt, (D) 0.713 volt, 10., , When the sample of copper with zinc impurity is to be purified by electolysis, the appropriate electrode, are, (A) pure zinc as cathode and pure copper as anode, (B) impure sample as cathode and pure copper as anode, (C) impure zinc as cathode and impure sample as anode, (D) pure copper as cathode and impure sample as anode

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11., , Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of, water. The total volume of the two gases (dry and at STP) produced will be approximately (in litres), (A) 22.4, (B) 44.8, (C) 67.2, (D) 89.4, , 12., , Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of MnO 2., Passing a current of 27A for 24 hours gives 1kg of MnO2. The current efficiency in this process is :, (A) 100%, (B) 95.185%, (C) 80%, (D) 82.951%, , 13., , During the preparation of H2S2O8 (per disulphuric acid) O2 gas also releases at anode as byproduct,, When 9.72 L of H2 releases at cathode and 2.35 L O2 at anode at STP, the weight of H2S2O8 produced, in gram is, (A) 87.12, (B) 43.56, (C) 83.42, (D) 51.74, , 14., , When the electric current is passed through a cell having an electrolyte, the positive ions move towards, cathode and negative ions towards the anode. If the cathode is pulled out of the solution, (A) the positive and negative ions will move towards anode, (B) the positive ions will start moving towards the anode while negative ions will stop moving, (C) the negative ions will continue to move towards anode while positive ions will stop moving, (D) the positive and negative ions will start moving randomly, , 15., , When iron is rusted, it is :, (A) reduced, (B) oxidised, , (C) evaporated, , (D) decomposed, , PART - II : NUMERICAL VALUE QUESTIONS, 1., , H4XeO6 + 2H+ + 2e–  XeO3 + 3H2O, F2 + 2e–  2F–, O3 + 2H+ + 2e–  O2 + H2O, Ce4+ + e–  Ce3+, 2HClO + 2H+ + 2e–  Cl2 + 2H2O, ClO4– + 2H+ + 2e–  ClO3– + H2O, , Eº = 3 V, Eº = 2.87 V, Eº = 2.07 V, Eº = 1.67 V, Eº = 1.63 V, Eº = 1.23 V, , ClO– + H2O + 2e–  Cl¯ + 2OH–, BrO– + H2O + 2e–  Br¯ + 2OH–, ClO4– + H2O + 2e–  ClO3– + 2OH–, , Eº = 0.89 V, Eº = 0.76 V, Eº = 0.36 V, , [Fe(CN)6]3– + e–  [Fe(CN)6]4–, Eº = 0.36 V, Based on the above data, how many of the following statements are correct ?, (A) F2 is better oxidizing agent than H4X6O6., (B) Ozone can oxidize Cl2, (C) ClO4– is better oxidizing agent in basic medium than in acidic medium, (D) Ferrocyanide ion can be easily oxidized by ClO–, Ce4+, Li+, BrO–, (E) ClO– can oxidize Br– and ClO3– in basic medium, (F) Ce4+ can oxidize Cl2 in acidic medium under standard conditions., 2., , A hydrogen gas electrode is made by dipping platinum wire in a solution of NaOH of pH = 10 and by, passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of, 2.303 RT, electrode is 10x milivolt. Find x ? (Take, = 0.059), F, , 3., , Estimate the cell potential of a Daniel cell having 1.0M Zn2+ and originally having 1.0M Cu2+ after, sufficient NH3 has been added to the cathode compartment to make NH3 concentration 2.0M at, equilibrium. Given Kf for [Cu(NH3)4]2+ = 1 × 1012, E° for the reaction, Zn + Cu2+  Zn2+ + Cu 1.1V., 2.303 RT, (Take, = 0.06, log 6.25 = 0.8) Respond as 10  your answer., F

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4., , Molar conductivity of 0.04 MgCl2 solution at 298 k is 200 Scm2mole–1. A conductivity cell which is filled, with MgCl2 have area of cross-section of electrode 4cm 2 & distance between electrode is 8 cm. If, potential difference between electrode is 10V then find current flow in miliampere., , 5., , o, The conductivity of a solution which is 0.1 M in Ba(NO 3)2 and 0.2 M in AgNO3 is 5.3 Sm–1. If (Ag ) =, , o, o, 6 × 10–3 Sm2 mol–1 & (Ba2 ) = 13 × 10–3 Sm2 mol–1, determine (NO3– ) in same unit. Report your answer, , after multiplying by 1000., 6., , , (weak mono basic HA acid) = 390.7 S cm 2 mol–1, m, , m of HA at 0.01 M is 3.907 S cm 2 mol–1, Find pH of 0.01 M HA ?, 7., , For a saturated solution of AgCl at 25°C,  = 3.4 × 10–6 ohm–1 cm–1 and that of H2O() used is, 2.02 × 10–6 ohm–1 cm–1. m for AgCl is 138 ohm–1 cm2 mol–1 then the solubility of AgCl in mili moles per, m3 will be :, , 8., , At 298 K, the conductivity of pure water is 5.5 × 10–6 S m–1. Calculate the ionic product of water using, the following data :, ºm values (in S m2 mol–1) : Ba(OH)2 = 5.3 × 10–2, HCl = 4.25 × 10–2, BaCl2 = 2.8 × 10–2 ., , Does your answer match with experimental value. Write 20 for yes & 40 for No., 9., , , How many of the following comparisons are correct with respect to their  m, ?, , (A) K+ > Na+, (E) H3O+ > Mg2+, , (B) K+ > H3O+, (F) K+ > Mg2+, , (C) Ca2+ > Na+, , (D) Mg2+ > NH4+, , PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE, 1., , , , , , Given EAg /Ag = 0.80V, E Mg2 / Mg = –2.37V, E Cu2 / Cu = 0.34V, EHg2 / Hg = 0.79 V., , Which of the following statements is/are correct, (A) AgNO3 can be stored in copper vessel, (B) Mg(NO3)2 can be stored in copper vessel, (C) CuCl2 can be stored in silver vessel, (D) HgCl2 can be stored in copper vessel, 2., , Any redox reaction would occur spontaneously, if :, (A) the free energy change ( G) is negative, (C) the cell e.m.f. (Eº) is negative, , 3., , Consider an electrolytic cell E being powered by a galvenic cell G, as shown in the figure. Then :, , (A) Anode of E is connected to cathode of G, (C) Cathode of E is connected to anode of G, 4., , (B) the Gº is positive, (D) the cell e.m.f. is positive, , (B) Anode of E is connected to anode of G, (D) Cathode of E is connected to cathode of G, , On electrolysis, in which of the following, O2 would be liberated at the anode ?, (A) dilute H2SO4 with Pt electrodes, (B) aqueous AgNO3 solution with Pt electrodes, (C) dilute H2SO4 with Cu electrodes, (D) aqueous NaOH with a Fe cathode & a Pt anode

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5., , A current of 2.68 A is passed for one hour through an aqueous solution of CuSO 4 using copper, electrodes. Select the correct statement(s) from the following :, (A) increase in mass of cathode = 3.174 g, (B) decrease in mass of anode = 3.174 g, (C) no change in masses of electrodes, (D) the ratio between the change of masses of cathode and anode is 1 : 2., , 6., , Three moles of electrons are passed through three solutions in succession containing AgNO 3, CuSO4, and AuCl3, respectively. The molar ratio of amounts of cations reduced at cathode will be, 1 1 1, (A) 1 : 2 : 3, (B) :, :, (C) 3 : 2 : 1, (D) 6 : 3 : 2, 1 2 3, , 7., , If same quantity of electricity is passed through three electrolytic cells containing FeSO4, Fe2(SO4)3 and, Fe(NO3)3, then, (A) the amount of iron deposited in FeSO4 and Fe2(SO4)3 are equal, (B) the amount of iron deposited in FeSO4 is 1.5 times of the amount of iron deposited in Fe(NO3)3., (C) the amount of iron deposited in Fe2(SO4)3 and Fe(NO3)3 are equal, (D) the same amount of gas is evolved in all three cases at the anode., , 8., , When a lead storage battery is discharged then :, (A) SO2 is evolved, (B) lead sulphate is produced at both electrodes, (C) sulphuric acid is consumed, (D) water is formed, , 9., , Mark out the correct statement(s) regarding electrolytic molar conductivity., (A) It increase as temperature increases., (B) It experiences resistance due to vibration of ion at the mean position., (C) Increase in concentration decreases the electrolytic molar conductivity of both the strong as well as, the weak electrolyte., (D) Greater the polarity of solvent, greater is the electrolytic molar conduction., , 10., , On increasing dilution following will increase :, (A) Equivalent conductivity, (C) Molar conductivity, , (B) Conductivity, (D) All of these, , 11., , The resistances of following solutions of KCl were measured using conductivity cells of different cell, constants, at same temperature. (Consider that at concentration less than 0.1 M, the specific, conductivity of solution is directly proportional to the concentration of solution.), Concentration of Solution, Cell Constant, 1., 0.1 M, 1 cm–1, 2., 0.01 M, 10 cm–1, 3., 0.005 M, 5 cm–1, 4., 0.0025 M, 25 cm–1, Which of the following comparisons between their conductances (G) is/are correct ?, (A) G1 is maximum, (B) G4 is minimum, (C) G3 >> G2, (D) G4 is maximum, , 12., , Identify correct statements :, (A) Kohlraush law is applicable only on weak electrolyte., (B) On increasing dilution conductance, molar conductivity, equivalent conductivity increases but, conductivity decreases., K, (C) m =, following formula has units m – dm2/mol, K  – dm—1, C  mol/., C, (D) Equation m = m  b C is applicable on weak as well as strong electrolyte.

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13., , Select the correct option(s):, ºeq SO 2– , ºeq Al3 , º, 4, (A), = ºm Al 3 &, = m SO24– , 2, 3, º, ºm SO 2 , , 3, , º, º, m Al , 4, (B)  eq Al3  =, &  eq SO 24  =, 3, 2, ºm Al3  ºm SO 2 , º, 4, , (C) eq (Al2(SO4) 3) =, +, 3, 2, º, (D) ºm (Al2(SO4) 3) = 6 ×  eq (Al2(SO4) 3), ,  , , 14., , Which of the following order is correct related to their mobility in solution:, , , , , , 2, , , (A) Csaq > Rbaq > K aq > Naaq > Liaq, (B) Beaq > Liaq > Csaq, , , 2, , 2, (C) Haq > Liaq > Beaq > Naaq > Mg aq, , 15., , , , , 2, (D) Haq  > Naaq  > Liaq  > Beaq , , For strong electrolyte M increases slow with dilution and can be represented by the equation, o,  M = M, – AC½, Select correct statement, o, (A) Plot of M against C½ is obtain a straight line with intercept M, & and slope ‘–A’, (B) Value of A depends upon temperature solvent and nature of electrolyte., (C) NaCl and KCl have different value of constant ‘A’, (D) NaCl and MgSO4 have different value of constant ‘A’, , PART - IV : COMPREHENSION, Read the following passage carefully and answer the questions., Comprehension # 1, If an element can exist in several oxidation states, it is convenient to display the reduction potentials, corresponding to the various half reactions in diagrammatic form, known as Latimer diagram. The, Latimer diagram for chlorine in acid solution is, 1.20 V, 1.18 V, 1.60 V, 1.67 V, 1.36 V, ClO4– ,  ClO3–  HClO2  HClO ,  Cl2 ,  Cl–, in basic solution., 0.37 V, 0.30 V, 0.68 V, 0.42 V, 1.36 V, ClO4– ,  ClO3– ,  ClO2– ,  ClO– ,  Cl2 ,  Cl–, The standard potentials for two nonadjacent species can also be calculated by using the concept that, G° as an additive property but potential is not an additive property and G° = – nFx0. If a given, oxidation state is a stronger oxidising agent than in the next higher oxidation state, disproportionation, can occur. The reverse of disproportionation is called comproportionation. The relative stabilities of the, oxidation state can also be understood by drawing a graph of G°/F against oxidation state, known as, Frost diagram, choosing the stability of zero oxidation state arbitrarily as zero. The most stable, oxidation state of a species lies lowest in the diagram. Disproportionation is spontaneous if the species, lies above a straight line joining its two product species., 1., , Which of the following couple have same value of potential at pH = 0 and pH = 14?, ClO4, Cl, ClO2, ClO, (A), (B), (C), (D) 2, , Cl2, Cl2, ClO3, Cl, , 2., , What is the potential of couple, (A) 1.78 V, , 3., , ClO , , Cl, (B) – 0.94 V, , at pH = 14 ?, (C) 0.89 V, , (D) – 0.89 V, , Which of the following statement is correct ?, (A) Cl2 undergoes disproportionation into Cl– and ClO– both at pH = 0 and pH = 14., (B) Cl2 undergoes disproportionation into Cl– and ClO– at pH = 14 but not at pH = 0., (C) Cl2 undergoes disproportionation into Cl– and ClO– at pH = 0 but not at pH = 14., (D) None of these

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4., , For a hypothetical element, the Frost diagram is shown in figure?, , Which of the following oxidation state is least stable?, (A) – 1, (B) 0, (C) + 2, 5., , (D) + 3, , Which of the following statement is correct ? According to Q.4, (A) A+1 undergoes disproportionation into A and A2+., (B) A2+ undergoes disporportionation in A and A3+., (C) A undergoes comporportionation in A+1 and A–1., (D) All of the above., , Comprehension # 2, The molar conductance of NaCl varies with the concentration as shown in the following table and all, values follows the equation, C, , = m, –b C, m, , C, Where  m, = molar specific conductance, , = molar specific conductance at infinite dilution, m, , C = molar concentration, Molar Concentration of NaCl, 4 × 10–4, 9 × 10–4, 16 × 10–4, , Molar Conductance in ohm–1 cm2 mole–1, 107, 97, 87, , When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the, cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl – and SO4–2 are 80 ohm–1 cm2, mole–1 and 160 ohm–1 cm2 mole–1 respectively., 6., , What is the molar conductance of NaCl at infinite dilution ?, (A) 147 ohm–1 cm2 mole–1, (B) 107 ohm–1 cm2s mole–1, –1, 2, –1, (C) 127 ohm cm mole, (D) 157 ohm–1 cm2 mole–1, , 7., , What is the cell constant of the conductivity cell (C), (A) 0.385 cm–1, (B) 3.85 cm–1, (C) 38.5 cm–1, , (D) 0.1925 cm–1, , 8., , If the cell (C) is filled with 5 x 10–3 (N) Na2SO4 the observed resistance was 400 ohm. What is the molar, conductance of Na2SO4., (A) 19.25 ohm–1 cm2 mole–1, (B) 96.25 ohm–1 cm2 mole–1, –1, 2, –1, (C) 385 ohm cm mole, (D) 192.5 ohm–1 cm2s mole–1, , 9., , If a 100 mL solution of 0.1M HBr is titrated using a very concentrated solution of NaOH, then the, conductivity (specific conductance) of this solution at the equivalence point will be (assume volume, change is negligible due to addition of NaOH) . Report your answer after multipling it with 10 in Sm –1., [Given  º(Na ) = 8 × 10–3 Sm2 mol–1 ,  º(Br  ) = 4 × 10–3 S m2 mol–1], (A) 6, , (B) 12, , (C) 15, , (D) 24

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Comprehension # 3, Answer 10, Q.11 and Q.12 by appropriately matching the information given in the three columns of the, following table., The curves in Column 1 shows the variation of conductivity during different titrations. The analyte and titrants, has been listed in Column 2 & Column 3 respectively., Column-1, (I), , (II), , Conductivity, decreases, initially then increases slowly, then increases rapidly, Conductivity, decreases, initially then increases, , Column-2 (Titrate), , Column-3 (Titrant), , (i), , (C2H5)2NH, , (P), , HCl, , (ii), , CH3COOH, , (Q), , NaOH, , (III), , Conductivity, decreases, initially, then, remains, approximately same, , (iii), , HBr, , (R), , CH3COOH, , (IV), , Conductivity increases initially, then remains approximately, same, , (iv), , NaOH, , (S), , NH4OH, , 10., , Which of the following is an incorrect combination of curves in Column 1., (A) (II) (iii) (Q), (B) (I) (i) (P), (C) (I) (iii) (S), (D) (I) (ii) (Q), , 11., , The correct combination for a titration in which conductance at equivalent point is lower than initial, (A) (I) (ii) (Q), (B) (I) (iii) (S), (C) (III) (iv) (R), (D) (IV) (ii) (S), , 12., , Select the correct combination, (A) (I) (iii) (Q), (B) (IV) (ii) (S), , (C) (I) (iii) (S), , (D) (I) (iv) (R), , PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS), * Marked Questions are having one or more than one correct options., 1., , Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox, titration. Some half cell reactions and their standard potentials are given below :, [JEE 2002, 3/84], –, +, –, 2+, MnO4 (aq.) + 8H (aq.) + 5e  Mn (aq.) + 4H2O(), Eº = 1.51 V, Cr2O72–(aq.) + 14H+(aq.) + 6e–  2Cr3+(aq.) + 7H2O(), Eº = 1.38 V, Fe3+(aq.) + e–  Fe2+(aq.), Eº = 0.77 V, Cl2(g) + 2e–  2Cl–(aq.), Eº = 1.40 V, Identify the only INCORRECT statement regarding the quantitative estimation of aqueous Fe(NO3)2 :, (A) MnO4– can be used in aqueous HCl., (B) Cr2O72– can be used in aqueous HCl., (C) MnO4– can be used in aqueous H2SO4., (D) Cr2O72– can be used in aqueous H2SO4., , 2., , Two students use the same stock solution of ZnSO4 and different solutions of CuSO4 to make Daniel, cell. The emf of one cell is 0.03 V higher than the other. The concentration of CuSO 4 solution in the cell, with higher emf value is 0.5 M. Find out the concentration of CuSO 4 solution in the other cell., 2.303 RT, , ,  0.06 ., [JEE 2003, 2/60],  Given :, , F, , 3., , The emf of the cell, Zn | Zn2+(0.01 M) || Fe2+(0.001 M) | Fe at 298 K is 0.2905 V. Then the value of, equilibrium constant for the cell reaction is :, [JEE 2004, 3/84], 0.32, 0.0295, (A) e, , 0.32, 0.0295, (B) 10, , 0.26, 0.0295, (C) 10, , 0.32, 0.059, (D) 10

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4., , Find the equilibrium constant at 298 K for the reaction :, Cu2+(aq) + In2+(aq), Cu+(aq) + In3+(aq), Given that EºCu2+ / Cu+ = 0.15 V, EºIn3+ / In+ = – 0.42 V, EºIn2+ / In+ = – 0.40 V., , [JEE 2004, 4/60], , 5., , The half cell reactions for rusting of iron are :, 1, 2H+ + O2 + 2e– H2O; Eº = +1.23 V & Fe2+ + 2e– Fe ; Eº = – 0.44 V, 2, Gº (in kJ/mol) for the overall reaction is :, [JEE 2005, 3/84], (A) – 76, (B) – 322, (C) – 122, (D) – 176, , Comprehension # 1, Tollen’s reagent is used for the detection of aldehyde. When a solution, with NH4OH, then gluconic acid is formed., Ag+ + e– Ag ;, C6H12O6 + H2O  C6H12O7 (Gluconic acid) + 2H+ + 2e– ;, Ag(NH3)2+ + e– Ag(s) + 2NH3 ;, RT, F, [Use 2.303 ×, = 0.0592 and, = 38.92 at 298 K], F, RT, Now answer the following three questions :, 6., , 2Ag+ + C6H12O6 + H2O  2Ag(s) + C6H12O7 + 2H+, Find ln K of this reaction :, 2Ag+ + C6H12O6 + H2O  2Ag(s) + C6H12O7 + 2H+, (A) 66.13, (B) 58.38, (C) 28.30, , of AgNO 3 is added to glucose, Eºred = 0.8 V, Eºoxd = – 0.05 V, Eº = – 0.337 V, , [JEE 2006, 5/184], (D) 46.29, , 7., , When ammonia is added to the solution, pH is raised to 11. Which half-cell reaction is affected by pH, and by how much :, [JEE 2006, 5/184], (A) Eoxd will increase by a factor of 0.65 for Eºoxd (B) Eoxd will decrease by a factor of 0.65 for Eºoxd, (C) Ered will increase by a factor of 0.65 for Eºred (D) Ered will decrease by a factor of 0.65 for Eºred, , 8., , Ammonia is always is added in this reaction. Which of the following must be INCORRECT :, (A) NH3 combines with Ag+ to form a complex., [JEE 2006, 5/184], (B) Ag(NH3)2+ is a weaker oxidising reagent than Ag+., (C) In absence of NH3, silver salt of gluconic acid is formed., (D) NH3 has affected the standard reduction potential of glucose/gluconic acid electrode., , We have taken a saturated solution of AgBr. Ksp of AgBr is 12 × 10–14. If 10–7 mole of AgNO3 are added, to 1 litre of this solution, find conductivity (specific conductance) of this solution in terms of 10–7 Sm–1., Given : º(Ag+) = 6 × 10–3 Sm2mol–1, º(Br–) = 8 × 10–3 Sm2mol–1, º(NO3–) = 7 × 10–3 Sm2mol–1., [JEE 2006, 6/184], Comprehension # 2, Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules, (approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with its, atomic/molecular masses. To handle such large number conveniently, the mole concept was, introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry,, electrochemistry and radiochemistry. The following example illustrates a typical case, involving, chemical/electrochemical reaction, which requires a clear understanding of the mole concept., A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads, to the evolution of chlorine gas at one of the electrodes., (Given : Atomic masses : Na = 23, Hg = 200 ; 1 Faraday = 96500 coulombs), Now answer the following three questions :, 9., , 10., , The total number of moles of chlorine gas evolved is :, (A) 0.5, (B) 1.0, (C) 2.0, , [JEE 2007, 4/162], (D) 3.0, , 11., , If the cathode is Hg electrode, the maximum weight (in g) of amalgam formed from this solution is :, [JEE 2007, 4/162], (A) 200, (B) 225, (C) 400, (D) 446, , 12., , The total charge (in coulombs) required for complete electrolysis is :, (A) 24125, (B) 48250, (C) 96500, , [JEE 2007, 4/162], (D) 193000

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Comprehension # 3, Redox reactions play a pivoted role in chemistry and biology. The values of standard redox potential, (Eº) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example, is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below is a set of halfcell reactions (acidic medium) along with their Eº values with respect to normal hydrogen electrode., Using this data, obtain the correct explanations to questions 13 - 14., , I2 + 2e– 2I–, , Eº = 0.54 V, , Cl2 + 2e– 2Cl–, , Eº = 1.36 V, , 2+, , + e Mn, , Eº = 1.50 V, , 2+, , Fe + e Fe, , Eº = 0.77 V, , O2 + 4H+ + 4e– 2H2O, , Eº = 1.23 V, , 3+, , Mn, , 3+, , –, , –, , 13., , Among the following, identify the correct statement :, (A) Chloride ion is oxidised by O2, (B) Fe2+ is oxidised by iodine, (C) Iodine ion is oxidised by chlorine, (D) Mn2+ is oxidised by chlorine, , 14., , While Fe3+ is stable, Mn3+ is not stable in acid solution, because :, [JEE 2007, 4/162], (A) O2 oxidises Mn2+ to Mn3+, (B) O2 oxidises both Mn2+ to Mn3+ and Fe2+ to Fe3+, (C) Fe3+ oxidises H2O to O2, (D) Mn3+ oxidises H2O to O2, , 15., , Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The, time required to liberate 0.01 mole of H2 gas at the cathode is : (1 Faraday = 96500 C), [JEE 2008, 3/163], 4, 4, 4, (A) 9.65 × 10 sec, (B) 19.3 × 10 sec, (C) 28.95 × 10 sec, (D) 38.6 × 104 sec, , 16.*, , For the reduction of NO3– ion in an aqueous solution, Eº is +0.96 V. Values of Eº for some metal ions, are given below :, V2+(aq) + 2e– V, Eº = –1.19 V, Fe3+(aq) + 3e–  Fe, Au3+(aq) + 3e–  Au, , [JEE 2007, 4/162], , Eº = –0.04 V, Eº = +1.40 V, , Hg2+(aq) + 2e– Hg, Eº = +0.86 V, The pair(s) of metals that is(are) oxidized by NO3– in aqueous solution is(are) :, [JEE 2009, 4/160], (A) V and Hg, (B) Hg and Fe, (C) Fe and Au, (D) Fe and V, Comprehension # 4, The concentration of potassium ions inside a biological cell is at least twenty times higher than the, outside. The resulting potential difference across the cell is important in several processes such as, transmission of nerve impulses and maintaining the ion balance. A simple model for such a, concentration cell involving a metal M is :, M(s) | M+(aq ; 0.05 molar) || M+(aq ; 1 molar) | M(s), For the above electrolytic cell, the magnitude of the cell potential is | Ecell | = 70 mV., Now answer the following two questions :, 17., , For the above cell :, (A) Ecell < 0 ; G > 0, , 18., , [JEE 2010, 3/163], (B) Ecell > 0 ; G < 0, , (C) Ecell < 0 ; Gº > 0, , (D) Ecell > 0 ; Gº < 0, , If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell, potential would be :, [JEE 2010, 3/163], (A) 35 mV, (B) 70 mV, (C) 140 mV, (D) 700 mV

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19., , AgNO3(aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was, measured. The plot of conductance () versus the volume of AgNO3 is :, [JEE 2011, 3/180], , (A) (P), 20., , (B) Q, , (C) (R), , (D) (S), , Consider the following cell reaction :, , [JEE 2011, 3/180], , 2Fe (s) + O2 (g) + 4H (aq)  2Fe, +, , 2+, , 2+, , (aq) + 2H2O (l), , Eº = 1.67 V, , –3, , At [Fe ] = 10 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25ºC is :, , 2.303 R (298), = 0.06), F, (A) 1.47 V, (B) 1.77 V, (Take, , (C) 1.87 V, , (D) 1.57 V, , Comprehension # 5, The electrochemical cell shown below is a concentration cell., M|M2+ (saturated solution of a sparingly soluble salt,MX2)|| M2+ (0.001 mol dm–3) |M, The emf of the cell depends on the difference in concetration of M2+ ions at the two electrodes. The emf, of the cell at 298 is 0.059 V., 21., , The solubility product (Ksp ; in mol3 dm–9) of MX2 at 298 K based on the information available in the, given concentration cell is : (Take 2.303× R × 298/F = 0.059 V), (A) 1 × 10–15, (B) 4 × 10–15, (C) 1 × 10– 12, , 22., , The value of G (in kJ mol–1) for the given cell is : (Take 1F = 96500 C mol–1 ) [IIT-JEE 2012, 3/136], (A) – 5.7, , 23., , [IIT-JEE 2012, 3/66], (D) 4 × 10–12, , (B) 5.7, , (C) 11.4, , (D) –11.4, , An aqueous solution of X is added slowly to an aqueous solution of Y as shown in list I. The variation in, conductivity of these reactions is given in List II. Match List I with List II and select the correct answer, using the code given below the lists :, [JEE(Advanced) 2013, 3/120], List I, List II, P. (C2H5)3N + CH3COOH, 1. Conductivity decreases and then increases, X, Y, Q. KI (0.1M) + AgNO3(0.01M), 2. Conductivity decreases and then does not change, X, Y, much, R. CH3COOH + KOH, 3. Conductivity increases and then does not change much, X, Y, S. NaOH +, HI, 4. Conductivity does not change much and then increases, X, Y, Codes :, P, Q, R, S, P, Q, R, S, (A), 3, 4, 2, 1, (B), 4, 3, 2, 1, (C), 2, 3, 4, 1, (D), 1, 4, 3, 2

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24., , The standard reduction potential data at 25ºC is given below., [JEE(Advanced) 2013, 3/120], Eº (Fe3+.Fe2+) = + 0.77 V;, Eº (Fe2+.Fe) = – 0.44 V;, Eº (Cu2+.Cu) = + 0.34 V;, Eº (Cu+.Cu) = + 0.52 V;, Eº (O2(g) + 4H+ + 4e–2H2O) = + 1.23 V ;, Eº (O2(g) + 2H2O + 4e– 4OH) = + 0.40 V, 3+, Eº (Cr .Cr) = – 0.74 V;, Eº (Cr2+.Cr) = – 0.91 V, Match Eº of the rebox pair in List I with the values given in List II and select the correct answer using, the code given below the lists :, List I, List II, P., Eº (Fe3+, Fe), 1., – 0.36 V, Q., –0.4 V, Eº (4H2O, 4H+ + 4OH–) 2., R., 3., –0.04 V, Eº (Cu2+ + Cu  2Cu+), S., Eº(Cr3+, Cr+2), 4., –0.83 V, Codes :, P, Q, R, S, P, Q, R, S, (A), 4, 1, 2, 3, (B), 2, 3, 4, 1, (C), 1, 2, 3, 4, (D), 3, 4, 1, 2, , 25., , In a galvanic cell, the salt bridge, (A) does not participate chemically in the cell reaction., (B) stops the diffusion of ions from one electrode to another., (C) is necessary for the occurrence of the cell reaction., (D) ensures mixing of the two electrolytic solutions., , 26., , All the energy released from the reaction X  Y, rGº = –193 kJ mol–1 is used for oxidizing M+ as M+ , M3+ + 2e– , Eº = – 0.25 V. Under standard conditions, the number of moles of M+ oxidized when one, mole of X is converted to Y is : [F = 96500 C mol –1], [JEE(Advanced) 2015, 4/168], , 27., , The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar, , [JEE(Advanced) 2014, 3/120], , conductivity of a solution of a weak acid HY (0.10 M). If 0 –  0 – , the difference in their pKa values,, X, , Y, , pKa (HX) – pKa (HY), is (consider degree of ionization of both acids to be <<1), [JEE(Advanced) 2015, 4/168], 28., , For the following electrochemical cell at 298 K,, Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) || M4+(aq) | M2+(aq) | Pt(s), Ecell= 0.092 V when, Given :, , 29., , EM0 4 / M2 = 0.151 V; 2.303, , M2  (aq) , , , = 10x., M4  (aq), , , , RT = 0.059 V, F, , The value of x is :, [JEE(Advanced) 2016, 3/124], (A) –2, (B) –1, (C) 1, (D) 2, The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a, conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm, with an area of cross section of 1 cm 2. The conductance of this solution was found to be 5 × 10–7 S. The, pH of the solution is 4. The value of limiting molar conductivity (ºm) of this weak monobasic acid in, aqueous solution is Z × 102 S cm–1 mol–1. The value of Z is, [JEE(Advanced) 2017, 3/122], , 30., , For the following cell, Zn(s) | ZnSO4(aq) || CuSO4 (aq) | Cu(s) when the concentration of Zn2+ is 10, times the concentration of Cu2+, the expression for G (in J mol–1) is :, [JEE(Advanced) 2017, 3/122], [F is Faraday constant; R is gas constant; T is temperature; Eº(cell) = 1.1 V], (A) 2.303 RT + 1.1 F, (B) 1.1 F, (C) 2.303 RT – 2.2 F, (D) –2.2 F, , 31., , For the electrochemical cell,, Mg(s) | Mg2+ (aq,1 M) || Cu2+ (aq,1M) | Cu(s), the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg2+ is changed to x M, the, F, cell potential changes to 2.67 V at 300 K. The value of x is ____. (Given,, = 11500 K V−1, where F is, R, the Faraday constant and R is the gas constant, ln(10) = 2.30)., [JEE(Advanced) 2018, 3/120], , 32., , Consider an electrochemical cell : A(s) | An+(aq, 2 M) || B2n+(aq, 1 M) | B(s). The value of H° for the cell, reaction is twice that of G° at 300 K. If the emf of the cell is zero, the S°(in J K–1mol–1) of the cell, reaction per mole of B formed at 300 K is____., (Given : ln (2) = 0.7, R(universal gas constant) = 8.3 J K–1 mol–1. H, S and G are enthalpy, entropy and, Gibbs energy, respectively.), [JEE(Advanced) 2018, 3/120]

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PART - II : JEE (MAIN) ONLINE PROBLEMS (PREVIOUS YEARS), 1., , The standard electrode potentials EMº  / M  of four metals A, B, C and D are –1.2 V, 0.6 V, 0.85 V and –, 0.76 V, respectively. The sequence of deposition of metals on applying potential is :, [JEE(Main) 2014 Online (09-04-14), 4/120], (1) A, C, B, D, (2) B, D, C, A, (3) C, B, D, A, (4) D, A, B, C, , 2., , A current of 10.0 A flows for 2.00 h through an electrolytic cell containing a molten salt of metal X. This, results in the decomposition of 0.250 mol of metal X at the cathode. The oxidation state of X in the, molten salt is : (F = 96,500 C), [JEE(Main) 2014 Online (09-04-14), 4/120], (1) 1 +, (2) 2 +, (3) 3 +, (4) 4 +, , 3., , Given :, , 4., , How many electrons would be required to deposit 6.35 g of copper at the cathode during the, electrolysis of an aqueous solution of copper sulphate ? (Atomic mass of copper = 63.5 u, N A =, Avogadro's constant)., [JEE(Main) 2014 Online (12-04-14), 4/120], NA, NA, NA, N, (1), (2), (3), (4) A, 20, 10, 5, 2, , 5., , A variable, opposite external potential (Eext) is applied to the cell Zn|Zn2+ (1 M) || Cu2+ (1M) | Cu, of, potential 1.1 V. When Eext < 1.1 V and Eext > 1.1 V respectively electrons flow from :, [JEE(Main) 2015 Online (10-04-15), 4/120], (1) Cathode to anode in both cases, (2) cathode to anode and anode to cathode, (3) anode to cathode and cathode to anode, (4) anode to cathode in both cases, , 6., , At 298 K, the standard reduction potentials are 1.51 V for MnO 4– |Mn2+, 1.36 V for Cl2|Cl– , 1.07 V for,  RT, ,  0.059 V , Br2|Br, and 0.54 V for 2|–. At pH = 3, permanganate is expected to oxidize : ,  F, , [JEE(Main) 2015 Online (11-04-15), 4/120], (1) Cl–, Br– and –, (2) Br– and –, (3) Cl– and Br–, (4) – only, , 7., , What will occur if a block of copper metal is dropped into a beaker containing a solution of 1M ZnSO 4?, [JEE(Main) 2016 Online (09-04-16), 4/120], (1) The copper metal will dissolve and zinc metal will be deposited., (2) The copper metal will dissolve with evolution of oxygen gas., (3) The copper metal will dissolve with evolution of hydrogen gas., (4) No reaction will occur., , 8., , Identify the correct statement:, [JEE(Main) 2016 Online (10-04-16), 4/120], (1) Corrosion of iron can be minimized by forming an impermeable barrier at its surface., (2) Iron corrodes in oxygen-free water., (3) Iron corrodes more rapidly in salt water because its electrochemical potential is higher., (4) Corrosion of iron can be minimized by forming a contact with another metal with a higher reduction, potential., , 9., , What is the standard reduction potential (Eº) for Fe3+  Fe ?, Given that :, [JEE(Main) 2017 Online (08-04-17), 4/120], º, 2+, –, Fe + 2e  Fe; EFe2  /Fe = – 0.47 V, , Fe3+ (aq) + e–  Fe2+ (aq) ; Eº = + 0.77 V, Al3+ (aq) + 3e–  Al(s); Eº = –1.66 V, Br2 (aq) + 2e–  2Br– ; Eº = +1.09 V, Considering the electrode potentials, which of the following represents the correct order of reducing, power?, [JEE(Main) 2014 Online (11-04-14), 4/120], (1) Fe2+ < Al < Br–, (2) Br– < Fe2+ < Al, (3) Al < Br– < Fe2+, (4) Al < Fe2+ < Br–, , º, º, EFe, Fe3+ + e–  Fe2+ ; EFe, = + 0.77 V, 2, 2, /Fe, /Fe 2 , , (1) + 0.30 V, , (2) – 0.057 V, , (3) + 0.057 V, , (4) – 0.30 V

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10., , Consider the following standard electrode potentials (Eº in volts) in aqueous solution :, Element, M3+ / M, M+ / M, [JEE(Main) 2017 Online (08-04-17), 4/120], Al, –1.66, + 0.55, Tl, +1.26, – 0.34, Based on these data, which of the following statements is correct ?, (1) Al+ is more stable than Al3+, (3) Tl+ is more stable than Al3+, , (2) Tl3+ is more stable than Al3+, (4) Tl+ is more stable than Al+, , 11., , Which of the following ions does not liberate hydrogen gas on reaction with dilute acids ?, [JEE(Main) 2017 Online (09-04-17), 4/120], (1) Mn2+, (2) Ti2+, (3) V2+, (4) Cr2+, , 12., , To find the standard potential of M3+/M electrode, the following cell is constituted : Pt / M / M3+, (0.001 mol L–1) / Ag+ (0.01 mol L–1) / Ag, The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction, M3+ + 3e– M at 298 K will be :, [JEE(Main) 2017 Online (09-04-17), 4/120], (Given EAg / Ag at 298 K = 0.80 volt), (1) 0.32 Volt, , (2) 0.66 Volt, , (3) 0.38 Volt, , (4) 1.28 Volt, , 13., , When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P was, collected at the cathode in 965 seconds. The current passed, in ampere, is :, [JEE(Main) 2018 Online (15-04-2018), 4/120], (1) 2.0, (2) 0.1, (3) 0.5, (4) 1.0, , 14.^, , When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of, p-aminophenol produced is :, [JEE(Main) 2018 Online (16-04-2018), 4/120], Note : Nitrobenzene actually convert into aniline in reduction in acidic medium., (1) 109.0 g, (2) 98.1 g, (3) 9.81 g, (4) 10.9 g, , 15., , The anodic half–cell of lead–acid battery is recharged using electricity of 0.05 Faraday. The amount of, PbSO4 electrolyzed in g during the process is : (Molar mass of PbSO 4 = 303 g mol–1), [JEE(Main) 2019 Online (09-01-2019), 4/120], (1) 15.2, (2) 22.8, (3) 7.6, (4) 11.6, , 16., , If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction, Zn(s) + Cu2+ (aq), Zn2+(aq) + Cu(s), at 300 K is approximately (R = 8 JK–1mol, F = 96000 C mol–1), [JEE(Main) 2019 Online (09-01-2019), 4/120], (1) e320, (2) e –80, (3) e160, (4) e–160, , 17., , Consider the following reduction processes :, Zn2+ + 2e–  Zn(s) ;, E0 = –0.76 V, Ca2+ + 2e–  Ca(s) ;, E0 = –2.86 V, Mg2+ + 2e–  Mg(s) ; E0 = –2.36 V, Ni2+ + 2e–  Ni(s);, E0 = –0.25 V, The reducing power of the metals increases in the order: [JEE(Main) 2019 Online (10-01-2019), 4/120], (1) Ca < Mg < Zn < Ni (2) Ni < Zn < Mg < Ca (3) Ca < Zn < Mg < Ni (4) Zn < Mg < Ni < Ca, , 18., , In the cell,, Pt(s) | H2(g,1bar) | HCl(aq) | AgCl (s) | Ag(s)| Pt(s), The cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of, 2.303RT, , ,  0.06Vat298k , (AgCl / Ag, Cl–) electrode is : Given,, F, , , [JEE(Main) 2019 Online (10-01-2019), 4/120], (1) 0.76 V, (2) 0.20 V, (3) 0.94 V, (4) 0.40 V, , 19., , The electrolytes usually used in the electroplating of gold and silver, respectively, are:, [JEE(Main) 2019 Online (10-01-2019), 4/120], (1) [Au(CN)2]– and [AgCl2]–, (2) [Au(NH3)2]+ and [Ag(CN)2]–, (3) [Au(CN)2]– and [Ag(CN)2]–, (4) [Au(OH)4]– and [Ag(OH)2]–

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20., , For the cell Zn(s) | Zn2+ (aq)||Mx+ (aq) |M(s), different half cells and their standard electrode potentials, are given below :, Mx+(aq)/M(s), , Au3+(aq)/ Au(s), , Ag+(aq)/Ag(s), , Fe3+(aq)/Fe2+(aq), , Fe2+(aq)/Fe(s), , EMx  / M/( V ), , 1.40, , 0.80, , 0.77, , –0.44, , If EZn2 / Zn = –0.76 V, which cathode will give a maximum value of Ecell per electron transferred ?, [JEE(Main) 2019 Online (11-01-2019), 4/120], (1) Au3+ / Au, (2) Fe3+ / Fe2+, (3) Fe2+ / Fe, (4) Ag+ / Ag, 21., , 22., , 23., , 24., , Given the equilibrium constant : KC of the reaction Cu(s) + 2Ag+(aq)  Cu2+ (aq) + 2Ag(s) is 10 × 1015,, RT, , , at298K  0.059 V , caluclate the E 0cell of this reaction at 298 K : 2.303, F, , , [JEE(Main) 2019 Online (11-01-2019), 4/120], (1) 0.4736 mV, (2) 0.04736 V, (3) 0.4736 V, (4) 0.04736 mV,  dE ,  for a cell are 2 V and, The standard electrode potential E and its temperature coefficient ,  dT , , , –5 × 10–4 VK–1 at 300 K respectively. The cell reaction is, Zn(s) + Cu2+ (aq)  Zn2+ (aq) + Cu(s), The standard reaction enthalpy (rH) at 300 K in kJ mol–1 is :, (Use R = 8JK–1 mol–1 and F = 96,000 Cmol–1), [JEE(Main) 2019 Online (12-01-2019), 4/120], (1) 206.4, (2) –384.0, (3) 192.0, (4) –412.8, , ºm for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm 2mol–1, respectively. If the conductivity of, 0.001 M HA is 5 × 10–5 S cm–1, degree of dissociation of HA is:, [JEE(Main) 2019 Online (12-01-2019), 4/120], (1) 0.125, (2) 0.50, (3) 0.75, (4) 0.25, , Given that] EO,  1.23V, 2 / H2O, , ES O2– / SO2  2.05V, 2 8, , , EBr, , –, 2 / Br, , E Au 3  / Au, , 4, ,  1.09V;, ,  1.4 V, , The strongest oxidizing agent is :, (1) O2, (2) Br2, 25., , (3) Au3+, , [JEE(Main) 2019 Online (08-04-19)S1, 4/120], (4) S 2O 82 –, , Calculate the standard cell potential (in V) of the cell in which following reaction takes place :, Fe2+ (aq) + Ag+ (aq)  Fe3+ (aq) + Ag(s), Given that, , E0Ag  / Ag  xV, 0, EFe,  yV, 2, / Fe, 0, EFe,  zV, 3, / Fe, , (1) x – z, , [JEE(Main) 2019 Online (08-04-19)S2, 4/120], (2) x – y, , (3) x + y – z, , (4) x + 2y – .3z, –1, , 26., , The standard Gibbs energy for the given cell reaction in kJ mol at 298 K is :, Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)., Eº = 2 V at 298 K, (Faraday's constant, F = 96000 C mol–1), [JEE(Main) 2019 Online (09-04-19)S1, 4/120], , 27., , A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using 0.1 Faraday electricity. How, many mole of Ni will be deposited at the cathode?, [JEE(Main) 2019 Online (09-04-19)S2, 4/120], (1) 0.20, (2) 0.15, (3) 0.10, (4) 0.05

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28., , Consider the statements Statement-1 and Statement-2 :, Statement-1 : Conductivity always increases with decrease in the concentration of electrolyte., Statement-2 : Molar conductivity always increase with decrease in the concentration of electrolyte., The correct option among the following is :, [JEE(Main) 2019 Online (10-04-19)S1, 4/120], (1) Both Statement-1 and Statement-2 are correct, (2) Both Statement-1 and Statement-2 are wrong, (3) Statement-1 is wrong and Statement-2 is correct, (4) Statement-1 is correct and Statement-2 is wrong, , 29., , Which one of the following graphs between molar conductivity (Am) versus C is correct?, [JEE(Main) 2019 Online (10-04-19)S2, 4/120], KCl, , Am, , Am, NaCl, , (1), , (2), , Am, KCl, , C, , NaCl, , (3), , Am, KCl, , NaCl, , C, , (4), , KCl, , NaCl, , C, , C, , 30., , Given, Co3+ + e–  Co2+ ; Eº = +1.81V ;, Pb4+ + 2e– Pb2 ; E0 = + 1.67 V, Ce4+ + e–  Ce3+ ; Eº = + 1.61 V ;, Bi3+ + 3e–  Bi ; Eº = +0.20 V, Oxidizing power of the species will increase in the order :[JEE(Main) 2019 Online (12-04-19)S1, 4/120], (1) Ce4+ < Pb4+ < Bi3+ < Co3+, (2) Bi3+ < Ce4+ < Pb4+ < Co3+, (3) Co3+ < Pb4+ < Ce4+ < Bi3+, (4) Co3+ < Ce4+ < Bi3+ < Pb4+, , 31., , The decreasing order of electrical conductivity of the following aqueous solution is :, [JEE(Main) 2019 Online (12-04-19)S2, 4/120], 0.1 M Formic acid (A) ; 0.1 M Acetic acid (B) ; 0.1 M Benzoic acid (C), (1) A > B > C, (2) A > C > B, (3) C > A > B, (4) C > B > A, , 32., , Given that the standard potentials (E°) of Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V respectively, the, E° of Cu2+/Cu+ is:, [JEE(Main) 2020 Online (07-01-20)S1, 4/120], (1) +0.158 V, (2) –0.182 V, (3) 0.182 V, (4) –0.158 V, , 33., , The equation that is incorrect is:, ,  , (3)   , 0, (1) m, ,  , –  , , NaBr, , 0, m KCl, , – m0, , NaI, , 0, m NaCl, ,  ,   ,  m0, , KBr, , 0, m KBr, ,  , –  , – m0, , NaBr, , 0, m NaBr, , [JEE(Main) 2020 Online (07-01-20)S2, 4/120], ,  , (4)   , 0, (2) m, , H2O, , 0, m NaBr, ,  , –  ,  m0, , HCl, , 0, m NaCl, ,  ,   , ,  m0, , NaOH, , 0, m KBr, ,  , –  , – m0, , NaCl, , 0, m KCl, , 34., , What would be the electrode potential for the given half cell reaction at pH = 5 ?, 2H2O O2 + 4H+ + 4e– ; E0red = 1.23 V (R = 8.314 J mol–1 K–1 ; Temp = 298 K ; oxygen under std., pressure of 1 bar), [JEE(Main) 2020 Online (08-01-20)S1, 4/120], , 35., , For an electrochemical cell, , 36., , 108 g of silver (molar mass 108 g mol –1) is deposited at cathode from AgNO3(aq) solution by a certain, quantity of electricity. The volume (in L) of oxygen gas produced at 273 K and 1 bar pressure from, water by the same quantity of electricity is _________________, [JEE(Main) 2020 Online (09-01-20)S1, 4/120], , 37., , Amongst the following, the form of water with the lowest ionic conductance at 298 K is:, [JEE(Main) 2020 Online (09-01-20)S2, 4/120], (1) distilled water, (2) sea water, (3) water from a well, (4) saline water used for intravenous injection, , [JEE(Main) 2020 Online (08-01-20)S2, 4/120], 2, [Sn, ], Sn(s) | Sn2+(aq, 1M)||Pb2+ (aq, 1M)|Pb(s) the ratio, when this cell attains equilibrium is _____, [Pb2 ], 2.303RT, , , , ,  0.06 ,  Given :ESn2 |Sn  –0.14V, EPb2 |Pb  –0.13V,, F, , 

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EXERCISE - 1, PART - I, A-1., , (a) Cu, (e) Cu, , (b) Ag, , (c) oxidation, , (d) reduction, , (f) Ag, , (g) anode-Cu Cu + 2e–; cathode-Ag+ + e–  Ag, 2+, , (h) Cu + 2Ag+ Cu2+ + 2Ag, , (i) Cu, , (j) Cu, , (k) to complete circuit and maintain electrical neutrality in solution, A-2., , (a) 2Ag+ + Cu  2Ag + Cu2+, , (b) 8H+ + MnO4– 5Fe3+ + Mn2+ + 4H2O, , (c) 2Ag+ + 2Cl–  2Ag + Cl2, , (d) Cd + 2H+  Cd2+ + H2, (c) Pb | Pb2+ | | Br– | | Br2 | Pt, , A-3., , (a) Zn | Zn2+ | | H+ | H2 | Pt, , B-1., , Mg, , B-3., , (i) CuO : Cu is below hydrogen in series, so it can reduce from CuO to Cu., , B-2., , (b) Pt | Sn2+, Sn4+ | | Fe3+, Fe2+ | Pt, , Y>Z>X, , (ii) Ag2O: Lower in series stability of oxide become lesser., (iii) Lower S.R.P. metal can displace higher S.R.P. metals ions from solution., B-4., , 1.61 V, , B-5., , 1.35 V, , C-2., , 0.756 V, , C-3., , Spontaneous,  48250 J, , D-1., , 0.059 volt, , D-2., , 1014, , D-5., , (a) The spontaneous cell reaction: Zn + 2Ag+ (aq), (b) 1.56 V, , B-6., , D-3., , 1.68 V, , – 0.2214 V, , C-1., , – 0.036 V, , C-4., , 1.14 volt, , D-4., , n=2, , Zn2+ (aq) + 2Ag (s), , (c) [Zn2+] = 4 × 10–4 M, , (d) As we add KI to cathode chamber, some Ag+ will precipitate out as:, Ag+ + I– AgI, The above reaction reducing [Ag+] from cathode chamber. This will reduce Ecell according to Nernst’s, equation., D-6., , pH = 1.5., , D-7., , log [Zn2+] / [Cu2+] = 37.22, , E-1., ELECTROLYTE, , ANODE Product CATHODE Product, , 1 NaCl (Molten) with Pt electrode, , Cl2(g), , Na, , 2 NaCl (aq) with Pt electrode, , Cl2(g), , H2(g), , 3 Na2SO4 (aq) with Pt electrode, , O2(g), , H2(g), , 4 NaNO3 (aq) with Pt electrode, , O2(g), , H2(g), , 5 AgNO3 (aq) with Pt electrode, , O2(g), , Ag, , 6 CuSO4 (aq) with Inert electrode, , O2(g), , Cu, , Cu dissolve, , Cu, , 7 CuSO4 (aq) with Copper electrode

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F-1., , 12.04 x 1023, , F-6., , F-3., , 2, , V(H2 ) = 56.0 mL. F-7., , Ni2+ = 2M, , F-8., , t = 93.65 sec., , G-1., , 8, , G-2., , 1.67 V, , H-1., , 2.332 × 10–3 mho cm–1, 23.32 mho cm2 mol–1., , H-2., , 0.1456 amp, , H-3., , 0.728 cm–1., , I-1., , 272, 0.0353, , I-3., , 1.76 × 10–5 mole/litre., , I-4., , 2.70 × 10–10 (mole/litre)2., , J-1., , (1), , Volume of NaOH, , F-4., , (3), , n=4, , Volume of NaOH, , F-5., , t = 193 sec., , F-9., ,  71.5 amp, , 382 mho cm2 mol–1., , I-2., , Conductance, , (2), , Conductance, , 108., , Conductance, , F-2., , (4), , Volume of NaOH, , PART - II, A-1., , (A), , A-2., , (C), , A-3., , (D), , A-4., , (C), , B-1., , (A), , B-2., , (C), , B-3., , (D), , B-4., , (D), , B-5., , (C), , B-6., , (A), , B-7., , (C), , B-8., , (C), , C-1., , (D), , C-2., , (D), , C-3., , (D), , D-1., , (A), , D-2., , (A), , D-3., , (C), , D-4., , (B), , D-5., , (B), , D-6., , (C), , E-1., , (C), , E-2., , (C), , E-3., , (D), , E-4., , (B), , E-5., , (B), , F-1., , (D), , F-2., , (B), , F-3., , (C), , F-4., , (B), , F-5., , (C), , G-1., , (B), , G-2., , (A), , G-3., , (D), , H-1., , (A), , H-2., , (D), , H-3., , (B), , I-1., , (C), , I-2., , (D), , I-3., , (D), , I-4., , (C), , I-5., , (D), , J-1., , (A), , J-2., , (C), , PART – III, 1., , (A - s) ; (B - p,r) ; (C - p,q) ; (D - r), , 2., , (A - p, q, r) ; (B - p, q, r) ; (C - p, s) ; (D - p, s), , EXERCISE - 2, PART - I, 1., , (A), , 2., , (C), , 3., , (C), , 4., , (C), , 5., , (B), , 6., , (B), , 7., , (B), , 8., , (A), , 9., , (D), , 10., , (D), , 11., , (C), , 12., , (B), , 13., , (B), , 14., , (D), , 15., , (B)

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PART - II, 1., , 3 (B, E & F), , 2., , 5., , o, (NO, 2, –1, –, ) = 7 Sm mol, 3, , 59, 6., , 4, , 3., , E° = 7 V., , 4., , 40, , 7., , 10, , 8., , 20, , 9., , 4, , PART - III, 1., , (BC), , 2., , (AD), , 3., , (AC), , 4., , (ABD), , 5., , (AB), , 6., , (BD), , 7., , (BCD), , 8., , (BCD), , 9., , (ACD), , 10., , (AC), , 11., , (AB), , 12., , (BC), , 13., , (BCD), , 14., , (A), , 15., , (ABD), , PART – IV, 1., , (D), , 2., , (C), , 3., , (B), , 4., , (D), , 5., , (A), , 6., , (C), , 7., , (D), , 8., , (D), , 9., , (B), , 10., , (C), , 11., , (C), , 12., , (B), , 4., , KC = 1010, , 5., , (B), , 10., , (B), , EXERCISE – 3, PART - I, 1., , (A), , 2., , 0.05 M, , 3., , (B), , –1, , 6., , (B), , 7., , (A), , 8., , (D), , 9., , 55 S m, , 11., , (D), , 12., , (D), , 13., , (C), , 14., , (D), , 15., , (B), , 16., , (ABD), , 17., , (B), , 18., , (C), , 19., , (D), , 20., , (D), , 21., , (B), , 22., , (D), , 23., , (A), , 24., , (D), , 25., , (A), , 26., , 4, , 27., , 3, , 28., , (D), , 29., , 6, , 30., , (C), , 31., , 10, , 32., , –1, , –11.62 JK mol, , –1, , PART – II, 1., , (3), , 2., , (3), , 3., , (2), , 4., , (3), , 5., , (4), , 6., , (2), , 7., , (4), , 8., , (1), , 9., , (2), , 10., , (4), , 11., , (1), , 12., , (1), , 13., , (4), , 14., , (3), , 15., , (3), , 16., , (3), , 17., , (2), , 18., , (2), , 19., , (3), , 20., , (1), , 21., , (3), , 22., , (4), , 23., , (1), , 24., , (4), , 25., , (4), , 26., , (2), , 27., , (4), , 28., , (3), , 29., , (3), , 30., , (2), , 31., , (2), , 32., , (1), , 33., , (1), , 34., , –0.93 to –0.94, , 35., , 2.13 to 2.17, , 36., , 5.66 to 5.68, , 37., , (1)

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(c), , 1.6 = 1.56 –, , or, , log, , or, , [0.1]2, , = 1.356, [Zn2 ], 0.01, [Zn2+] =, = 4.4 × 10–4 M, 22.7, , or, , D-6., , 0.059, [Zn2 ], log, 2, [Ag ]2, , 5, 4,  + 2H+ + e–  N O + H2O;, N O3, 2, , 1.356 = log, 0.01, , or, , [Zn2  ], , [Ag ]2, [Zn2 ], , = 22.7, , 0.79 volt, , 5, , NO3 + 8H+ + 6e–NH3OH+ + 2H2O;, , 0.731 volt, , 1, 1, 0.0591, 0.0591, log  2 = 0.731 –, log  8, 6, 1, [H ], [H ], 0.0591, 0.0591, 0.79 + 2 ×, log [H+] = 0.731 +,  8  log [H+], 6, 2, 8,  pH = 6 = 1.5, 0.059 = 0.059 (2 – ) pH, 6, 4, , E = 0.79 –, , D-7., , EºZn / Zn2+ = 0.76 volt, EºCu / Cu2+ = – 0.34 volt, CuSO4 + Zn ZnSO4 + Cu., O = (0.76 + 0.34) –, , F-1., , F-2., , F-3., ,  Zn2 , 0.0591, log  2 , 2,  Cu , , Cu2+ + 2e– Cu, No. of Faraday required = 2 F, 0.108 =, WAg, Wmetal, , =, , = 2 × 6.023 × 1023, , E, × 0.5 × 193, 96500, E Ag, EM, , =, , 108, 108, n, ,  Zn2  , log ,  = 37.22,  Cu2  , , , , = 12.04 × 1023, , E = 108 g/eq., , =n, , n=, , 0.5094, = 1.92, 0.2653, , n = 2., , F-4., , 3  1  60  60, 2.977, =, ., 106.4 / n, 96500, , F-5., , 10.8 × 80 × 5 × 10–4 =, , F-6., , Ag+ + e–  Ag, , F-7., , With Ni electrode, cathode reaction, Ni2+ + 2e–  Ni, Anode reaction, Ni  Ni2+ + 2e, No. Change in molarity of solution., , F-8., , Cd2+ + 2e— Cd., , , , n = 4., , 108, ×2×t, 1  96500, , , , t = 193 sec, , 0.54, = 5 × 10–3 mol, 108, 2H2O + 2e–  H2 + 2OH–, nH2 = 2.5 × 10–3, VH2 = 22.4 × 2.5 × 10–3 = 56.0 mL, , Cd required for 2g of Hg =, t=, , 12  2, = 0.2727 g., 88, , 2  96500  2727, = 93.65 Sec., 112.4  5

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2  96500, i = 71.48 Amp., 60  60  0.75, , F-9., , 2HSO4– S2 O82– + 2e– + 2H+, , G-1., , CH4 + 10OH¯  CO32– + 7H2O + 8e–, 80 x 3600 x 0.96, No. of Faradays required =, 96500, 1 80 x 3600 x 0.96, Hence mol. of CH4 required = x, 8, 96500, 80 x 3600 x 0.96, 1, VCH4 =, x, x 22.4 L = 8.356 x 0.96 = 8.02 L, 8, 96500, , i=, , 4, 1, ×, = 2.332 × 10–3 mho cm–1., 245 7, 2.33  10–3  1000, m =, = 23.32 mho cm2 mol–1., 0.1, , H-1., , K=, , H-2., , eq = 97.1 Scm2 eq–1, C = 0.1 N, A = 1.5 cm2, l = 0.5 cm, 1000  R1  A, 1000, 0.5, 1, , eq =, 97.1 =, ×, ×, C, 1.5, 0.1, R, V, 5, , R = 34.33 , i=, =, = 0.1456 amp, 34.33, R, , , , H-3., , I-1., , Ans. 8, , C = 0.1 N,, 1, K=, , R A, , , , K = 1.12 × 10–2 Scm–1, R = 65, , A, , = cell constant = 1.12 × 10–2 × 65 = 0.728 cm–1, , ºm(NH4CI) = 150 = ºm(NH4+) + ºm(CI–), ºOH– = 198,, ºCI– = 76 ,, m(NH4OH) = 9.6,, ºm(NH4OH) = 150 – 76 + 198 = 272,  m (NH4 OH), 9.6, =, =, = 0.0353,  ºm (NH4 OH), 272, , C = 0.01, , I-2., , m(NaA) = 83 S cm2 mol–1, m(NaCI) = 127 S cm 2 mol–1, m(HA) = 83 + 426 – 127 = 382 S cm2 mol–1., , I-3., , =, , I-4., , KAgCI = 2.28 × 10–6 Scm–1,, , 7.36, 390.7, , m(HCI) = 426S cm2 mol–1, , 2, , Ka = C2, , S = 1.65 × 10–5 and, ,  7.36 , = 0.05 × , = 1.77 × 10–5 mol/ lit,  390.7 , 1000  2.28  10 6, S, Ksp = (S)2 = 2.72 × 10–10 M2., 138.3 =, , PART - II, A-1., , In galvanic cell/electro chemical cell electrical energy is produced due to some chemical reaction., , A-2., , Salt bridge complete the electrical circuit and minimises the liquid-liquid junction potential., , A-3., , Agar-Agar is a gelatin, it used in salt bridge along with KCl electrolyte.

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A-4., , Ecell = E°Ni / Ni2+ + E Ag / Ag, = 0.25 + 0.80 = 1.05 Volt., , B-1., , E0 is intensive property and it does not depend on mass of F 2 taking part., , B-2., , Lowest S.R.P., highest reducing power., , B-3., , E0Cu2  / Cu = 0.34, , 0, EFe, = –0.44 volt, 2, / Fe, , So Cu can't displace Fe2+ ., B-4., , Cu can't displace Al3+ ion from aluminium nitrate., , B-5., , Lower S.R.P. containing ion can displace higher S.R.P. containing ion., , B-6., , Lowest S.R.P., highest reducing power., , B-7., , KCl can make precipitate with AgNO3, Pb(NO3)2 so can't be used along these electrolyte., , C-1., , Fe3+ + 3e–  Fe ,, – 0.036 volt, Fe  Fe2+ + 2e–, 0.44 volt, Fe3+ + e–  Fe2+, + 3 × 0.036f –2 × 0.44 × f = – 1 × E° × f, E° = 0.772 Volt, , C-2., , Cu+ + e–  Cu, E° = x1 Volt, Cu2+ + 2e–  Cu, x2 Volt, Cu  Cu+ + e– – x1Volt, Cu2+ + e–  Cu+, – 2 × x2 × f + 1 × x1 × f = –1 × Eº × f, Eº = 2x2–x1, , C-3., , For spontaneous reaction in every condition, Ecell > 0, G < 0 and Q (reaction quotient) < K (equilibrium constant)., , D-1., , E = 1.1 –, , D-2., , H2(Pt) (1 atm) | H3 O | | Ag+(xM) | Ag, , 0.0591, 0.1, log, 0.1, 2, , [H ], 0.06, log, x, 1, [H ], 0.2, –, = log, x, 0.06, 10, = pH + log x, 3, log x = – 1.7, 10 –5.5, = 1.62 × 10–4, x, x = 2 × 10–2 M, 1.0 = (0 + 0.8) –, , D-3., , Zn  Zn2C  + 2e–, 1, , Zn2C2  + 2e–  Zn, Zn2C2 , , Zn2C1, , , , E = 1.10 Volt

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E=0–, , C, 0.0591, log 1, C2, 2, , E  +ve When C1 < C2, D-4., , H2(p )  2H+ + 2e–, 1, , 2H+ + 2e–  H2(p, X2(p, , 1), ,  X 2(p, , E=0–, , 2), , 2), , p, 0.0591, log 2, p1, 2, , P2 < P1 for E  +ve, D-5., , , H2  2H(10, + 2e–, –2, M), , 2H+(10–3 M) + 2e– H2, , ,  2H(10, 2H(10, –3, –2, M), M), ,  10 –1 , 0.0591, E=0–, log  –2 , 2,  10 , , 2, , E  – ve (Non spontaneaous)., , 0.0591, 0.0591, 16, log, =–, ×2 log 2 = – 0.0591 × 0.301 = – 0.0178 Volt., 4, 2, 2, If connected in reverse direction,E = 0.0178 volt., , D-6., , E=0–, , E-1., , At anode, Ag Ag+ + e–, At cathode, Ag+ + e– Ag, +, So conc. of Ag will remain same ., , E-2., , equivalence of H2 = equivalence of O2, volume of O2, 0.224, 2=, 4, 22.4, 22.4, 0.112 litre = volume of O2., , E-5., , In presence of inert electrode, the cell reaction is Anode :, H2O  4H+ + 4e– + O2, Cathode :, {Cu++(aq) + 2e–  Cu(s)} × 2, Net cell reaction 2Cu++ aq. + 2H2O  4H+ + 2Cu(s) + O2 , Due to increases in [H+], pH decreases., , (B) is answer, , F-1., , 8H+ + 5e– + MnO4–  Mn+2 + 4H2O, (1 mole), 5 mole e– = 5 Faraday., , F-2., , Mole of Fe deposited =, , 1, × 3 = 1.5 mole, 2, , W Fe = 1.5 × 56 = 84 g., F-3., , 63.5, × 2 × 60 × 60 = 2.37 g, 2  96500, 3, % of efficiency =, ×100., 2.37, W=

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F-4., , 2H2O + 2e–  H2 + 2OH–, 9.65  1000, No. of Faraday passed =, = 0.1 F, 96500, nOH– formed = 0.1 mol, nNaOH = 0.1 mol  4 g., , F-5., , According to Faraday’s second law, mass of C, mass of A, mass of B, =, =, equivalent mass of C, equivalent mass of A equivalent mass of B, 4.5, 9.6, 2.7, =, =, 15 / n1, 27 / n2, 48 / n3, 0.3n1 = 0.1n2 = 0.2n3 = k, 10, n1 =, k, 3, n2 = 10k, n3 = 5k, 1, 10, n1 : n2 : n3 =, : 1 :, : 10 : 5 =, 3, 3, 0.3 : 0.1 : 0.2, 3:1:2, , 1, 2, , =2:6:3, , G-1., , Discharging reaction, Pb(s) + PbO2(s) + 2H2SO4 2PbSO4 + 2H2O., , G-2., , H2–O2 fuel cell, At anode : 2OH– + H2 2H2O + 2e–, At cathode : 2 H2O + O2 + 4e–  4OH–, , H-1., , For strong electrolyte,  MC =  M – b C, , H-2., , Molar conductivity  no. of ions per mole of electrolyte., , I-1., , Ka = 25×10–6 eq = 19.6 Scm2 eq–1, C = 0.01, Ka = 0.01 × 2,  = 5 × 10–2 =, , 19.6, eq, , , , =, , , , eq =, , 25  10–6, 10–2, 19.6, 5  10–2, , 1000  3.06  10 –6, Normality, Normality = 2 × 10–3 M, 2  10 –3, Molarity =, = 10–3 M, 2, Ksp = 10–6 M, , I-2., , 1.53 =, , I-3., ,  7 , Ka = C2 = 0.1× , = 3.38 × 10–5,  380.8 , , 2, , = 5 × 10–2, = 392 Scm2 eq-1.

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I-4., , K = 1.382 × 10–6 s cm–1, AgCl = 61.9 + 76.3 = 138.2 =, , I-5., , 1000  1.382  10–6, S, , , , m,BaSO4 = (x1 + x2 x – 2x3), eq.,BaSO4, ,  S = 10–5 M., , eq.,BaSO4 =, , eq.,BaSO4, n  factor, , (x  x2 – 2x3 ), = 1, 2, , PART - III, 1., , 2., , (A) Eº Zn2  / Zn  EºMg2  / Mg, , so, , Eºcell = –ve, , (B) Eº Ag / Ag  Eº Zn2  / Zn, , so, , Eºcell = +ve, , (C) Eº Ag / Ag  Eº Zn2  / Zn, , so, , Eºcell = 0 (at equilibrium), , (D) Eº Ag / Ag  EºFe2  / Fe, , so, , Eºcell = +ve, , (A) Ecell =–, , (PH2 )c [H ]a2, 0.059, log, ,, 2, (PH2 )a [H ]c2, , 0.059, 0.059, 0.01  (0.1)2, log, =,  3 = +ve, 2, 2, (0.1)  (1)2, (B) Cell reaction Ag+c (10–2)  Ag+a (10–9), 109, 0.059, 0.059, Ecell = E0cell –, log 2 = E0cell +, ×7>0, 1, 1, 10, E0cell  0 and not conc. cell, E0cell =–, , (C) Ecell = 0 –, (D) Ecell = –, , [Cu2 ]a, 0.059, 0.059, 0.1, log, =–, log, = – ve, 0.01, 2, 2, [Cu2 ]c, , [Cl ]c, 0.059, 0.059, 0.1, log, =, log, =0, 0.1, 2, 2, [Cl ]a, , And E0cell = 0., , EXERCISE # 2, , , PART - I, 1., , 0, As E oCu2   Cu = 0.337 V > EH, , /H, , 2, ,  Cu2+ can be reduced by H2., 2., , Lower standard reduction potential related metal ions can displace higher standard reduction potential, related metal ions., , 3., , (C) M is more reactive than carbon and B is more reaitive than A. Also both B and A are less reactive, than C., , 4., , Only anode or cathode can't work alone so absolute value of reduction potential can't be determined.

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PART - II, 1., , (A) H4XeO6 has more SRP, so better oxidizing agent then F2, (B) O3 has more SRP, so will oxidize Cl2, o, (C) EClO– / ClO– is more in acidic medium than in basic medium, 4, , 3, , o, (D) E[Fe(CN)6 ]3– |[Fe(CN)6 ]4– = 0.36 V, Hence [Fe(CN)6]4– can be easily oxidized by ClO¯, Ce4+ Br2O– but not by Li+, o, o, o, (E) E – – – is more than E – – and EClO– / ClO– (OH– ) so true., , ClO / Cl (OH ), , BrO / Br, , 4, , 3, , (F) Ce4+ can’t oxidize Cl2 in acidic medium., 2., , H2  2H+ + 2e–, 1 atm, 10–10, EH, , 2, , / H, , , , 2, , EH, , , , 2, , 1, , or, , 1 × 1012 =, , Now,, , ECell = E0Cell –, , [Cu2 ] (2)4, , or, ,  0.59 V, , [Cu2+] = 6.25 × 10–14., , 0.059, 0.059, 1, [Zn2 ], log, = 1.1 –, log, = 0.71 V., 2, , 2, 2, 6.25  1014, [Cu ], , K  1000, M, K  1000, 200 =, 0.04, 200 0.04, K=, = 8 x 10–3 S cm–1., 1000, , 1, K= 9 x,  , R, m , , 8 1, 8 x 10–3 =  4  x,   R, 2, 1, R=, = x 103  ., 8  103, 4, Again  V = IR, 10, V, So, I=, =  1  103  = 4 x 10–2A., R, 4, , , 5., , / H, , Cu2+ + 4NH3  [Cu(NH3)4]2+, 1, a, 0, 0, 2, 1, [Cu(NH3 )4 ]2, Kf =, [Cu2 ] [NH3 ]4, , 3., , 4., , , , 1010, 0.059, 0, log, 2, 1, , Ksol = K Ba2  + K Ag + KNO3–, 5.3 =, , o, Ba, 2, [Ba2  ], , 10–3, , +, ,  oAg [Ag ], 10–3, , +, , o, NO, –,  [NO– ], 3, , 10–3, , 3

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14., , Li+ Na+ K+ Rb+ Cs+, degree of Hydration decreases, Size of ions decreases, ionic mobility increases, Haq  has smallest size therefore show maximum mobility., , PART - IV, 1., , From given latimer diagrams.Cl2 – Cl– is independent of H+ cocentration., , 2., , G0 = G10 + G20, using this E0 =, , 0.42  1.36, V = 0.89 V, 2, , 3., , 4., , As, , Gº, is low, stability is higher., F, , 5., , As, , Gº, is low, stability is higher so, + 2 and 0 state is more stable than +1., F, , 6., , C, , = m, –b, m, , C, , C, when C1 = 4 x 10–4  m, = 107, , and when C2 = 9 x 10–4, , so 107 =  m, – b x 2 x 10–2, , m, , 97 =, –bx3x, b = 1000, , m =  m, –b C, , m = 97, ... (1), , 10–2, , ... (2), , , , = m + b C = 107 + 103 x 2 x 10–2, m, , , = 127 ohm–1 cm2 mole–1, m, , 7., , For 25 x 10–4 (M) NaCl solution, , m =  m, –b C, , m = 127 –, But, , 103, , 10–2, , x5x, K x 1000, m =, M, , , , m = 127 – 103 (25 x 10–4)1/2, m = 77, 1,  , K =  x,  a, R, , 1, 1000,  , m =   x, x,  a, M, R, m = [Cell constant] x, , 1000, R x M, ,  77 = [Cell constant] x, , 1000, , 1000 x 25 x 10 4, Cell constant = 77 x 25 x 10–4 = 0.1925 cm–1

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X, , Y, , 23., , (P) (C2H5 )3 N + CH3 COOH  CH3COO– (aq) + (C2H5)3NH+ (aq), As CH3COOH is a weak acid, its conductivity is already less. On addition of weak base, acid-base, reaction takes place and new ions are created. So conductivity increases., (Q) KI (0.1 M) + AgNO3 (0.01 M) AgI  (ppt) + KNO3 (aq)., As the only reaction taking place is precipitation of AgI and in place of Ag+, K+ is coming in the solution,, conductivity remain nearly constant and then increases., (R) CH3COOH + KOH  CH3COOK (aq) + H2O, OH– (aq) is getting replaced by CH3COO–, which has poorer conductivity. So conductivity dereases and, then after the end point, due to common ion effect, no further creation of ions take place. So,, conductivity remain nearly same., (S) NaOH + HI  NaI (aq) + H2O, As H+ is getting replaced by Na+ conductivity dereases and after end point, due to OH–, it increases., So answer of 39 is : (P) – (3) ; (Q) – (4) ; (R) – (2); (S) – (1). Answer is (D)., , 24., , (P), , EºFe3+, Fe, , , , 1 × 0.77 + 2 × (– 0.44) = 3 × x, 0.11, x=–, V ~ – 0.04 V., 3, 4H2O, 4H+ + 4OH–, 2H2O  O2 + 4H+ + 4e–, – 1.23 V, + O2 + 2H2O + 4e–  4OH– + 0.4 V, _____________________________________, 4H2O, 4H+ + 4OH–, – 0.83 V, Eº(Cu2+ + Cu  2Cu+), , , (Q), , (R), , , , x × 1 + 0.52 × 1 = 0.34 × 2, x = 0.16 V., Cu2+ + e–  Cu+, + Cu  Cu+ + e–, , 0.16 V, – 0.52 V, , ______________________________________________________________, ,  2Cu+, Cu2+ + Cu , – 0.36 V, However, in the given option, – 0.18 V is printed., (s), , Eº(Cr3+, Cr2+), , x × 1 + 2 × (– 0.91) = 3 × (– 0.74), x – 1.82 = – 2.22, , x = – 0.4 V, Hence, most appropriate is (D)., (P) – (3) ; (Q) – (4) ; (R) – (1) ; (S) – 2., 25., , Salt bridge is introduced to keep the solutions of two electrodes separate, such that the ions in, electrode do not mix freely with each other. But it cannot stop the process of diffusion., It does not participate in the chemical reaction. However, it is not necessary for occurence of cell, reaction, as we know that designs like lead accumulator, there was no salt bridge, but still reactions, takes place.

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26., , m+  m3+ + 2e–, G0 = –nFE0 For 1 mole of m+, G0 = –2 × 96500 × (–0.25) J, = + 48250 J/mole = 48.25 KJ/mole, Energy released by conversion of 1 mole of, x y , G = –193 KJ, Hence mole of m+ convert, 193, =4, 48.25, , 27., ,      , X, , Y, ,            , , , H, , HX, , H, , Now,, , , Y, , (1), , m, = ,, m , , Also, , , , , , X, , HY, , So, , , m (HX) =  m, 1, , and, , , m (HY) =  m, 2, , (Where 1 and 2are degrees of dissociation of HX and HY respectively.), Given that, m (HY) = 10 m (HX)., , , 2 = 10 ×  m, 1, m, , , , 2 = 10 1, , C 2, Ka =, ,, 1– , , but, , (2),  << 1,, , therefore Ka = C2 ., 2, , 28., , 29., , , , K a (HX), 0.0112 0.01, 1,  1, =, =, ×  =, ., 2,  10 , K a (HY), 0.1, 1000, 0.12, , , , log (Ka (HX)) – log (Ka (HY)) = –3., , [M2 ][H ]2, Ecell = E°cell – 0.059 log10, [M4 ] pH2, 2, 0.092 = 0.151 = 0.059 log10 10x, 2, x = 2, C = 0.0015M, G = 5 × 10–7s, ,  = 120 cm, a = 1 cm2, , G = ×, , a, , 1, 120, = 6 × 10–5 s cm–1,   1000 6  10 –5  1000, c, =, =, m, 0.0015, M, pH = 4, [H+] = 10–4 = c = 0.0015 , 10 –4, =, 0.0015, 5 × 10–7 = ×, , , , , , =, , c, m, o, m, , 6  10–5  1000, 10, 0.0015, , =, o, 0.0015, m, –4, , o,  m, = 6 × 102 s cm2 mole–1, , , , pKa (HX) – pKa (HY) = 3.

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(v.f.) = 4, , W = ZIt =, , E, ×I×t, F, , M, , E  4 , , , , 109  9.65  60  60, 4  96500, W = 9.81 g, W=, , 15., , PbSO4 (s)  2e –  Pb(s)  SO24 –, For 2F current passed, PbSO4 electrolyzed = 303g/mol, For 0.05F; PbSO4 electrolyzed =, , 0.05  303, = 7.6g, 2, , 16., , Gº = –RTlnK, –2 × 96000 × 2 = –8 × 300 × lnK, or, lnK = 160, or, K = e160, , 17., , Higher the SOP, higher will be reducing power., , 18., , A:, , 19., , Fact, , 20., , Ecell  EZn(s)|Zn 2  EAu3 / Au, = SOPanode + SRPcathode, = 0.76 V + 1.40 V = 2.16V, , 21., , E°cell =, , 22., , H = – nFEcell + nFT, , 23., , ºm(HA) = °m(HCl) + °m(NaA) – °(NaCl), = 425.9 + 100.5 – 126.4 = 400, K, , 1000, 5  10 –5  103, , m, , , = 50, M, 10 –3, , 1, H2 (g)  H+ (aq.) + e–, E° = 0.0 V, 2, C : AgCl (s) + e–  Ag (s) + Cl– E° = x V, 0.0591, Ecell = E°cell –, log {[H+] [Cl–]}, 1, 0.06, 0.92 = x –, log (10–12), 1, 0.92 = x + 0.72, x = 0.92 – 0.72 = 0.2 Volts, , dE, dT, = –2 × 96000 × 2 + 2 × 96000 × 300 × (–5 × 10–4), = – 384000 – 28,800 = – 412.8 kJ/mol, , , 24., , 0.0591, 0.0591, log Kc =, log(1 1016 ) = 0.4728 V, 2, n, , 50,  0.125, 400, , Strongest oxidizing agent has highest value of SRP

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35., , At Equilibrium state. Ecell = 0 ; Eºcell = 0.01 V, Sn + Pb2+  Sn2+ + Pb,  [Sn 2  ] , 0.06, 0 = 0.01 –, log , , 2,  [Pb 2  ] , 0.01 =, , 0.06  [Sn2 ] , log, , 2,  [Pb 2 ] , ,  [Sn 2  ] , [Sb 2 ], 1, = log, , = 101/3 = 2.1544, , 3,  [Pb 2  ] , [Pb 2 ], 36., , n , , Ag deposit, , , , 108,  1 mole, 108, , Ag   e– ,  Ag, 1F charge is required to deposit 1 mole of Ag, , 1, H2O ,  O2  2H   2e–, 2, 1,  mole, 2F charge deposit , 2, 1,  mole, 1F charge will deposit , 4, , nRT, P, 1 0.08314  273, = , 4, 1, VO2 = 5.674 L, VO2 , , 37., , Theory based.

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 Marked Questions may have for Revision Questions., , This Section is not meant for classroom discussion. It is being given to promote self-study, and self testing amongst the Resonance students., , PART - I : PRACTICE TEST-1 (IIT-JEE (MAIN Pattern)), Max. Marks: 100, Max. Time : 1 Hour, Important Instructions:, A., General %, 1., The test paper is of 1 hour duration., 2., The Test Paper consists of 25 questions and each questions carries 4 Marks. Test Paper consists of, Two Sections., B., 1., , 2., , Test Paper Format and its Marking Scheme:, Section-1 contains 20 multiple choice questions. Each question has four choices (1), (2), (3) and (4) out, of which ONE is correct. For each question in Section-1, you will be awarded 4 marks if you give the, corresponding to the correct answer and zero mark if no given answers. In all other cases, minus one, (–1) mark will be awarded., Section-2 contains 5 questions. The answer to each of the question is a Numerical Value. For each, question in Section-2, you will be awarded 4 marks if you give the corresponding to the correct answer, and zero mark if no given answers. No negative marks will be answered for incorrect answer in this, section. In this section answer to each question is NUMERICAL VALUE with two digit integer and, decimal upto two digit. If the numerical value has more than two decimal places truncate/round-off, the value to TWO decimal placed., SECTION-1, This section contains 20 multiple choice questions. Each questions has four choices (1), (2), (3) and (4), out of which Only ONE option is correct., , 1., , The standard electrode potentials (reduction) of Pt/Fe3+, Fe2+ and Pt/Sn4+, Sn2+ are + 0.77 V and 0.15 V, respectively at 25° C. The standard EMF of the reaction Sn4+ + 2Fe2+  Sn2+ + 2Fe3+ is, (1) – 0.62 V, (2) – 0.92 V, (3) + 0.31 V, (4) + 0.85 V, , 2., , Which is/are correct among the following?, 0, Given, the half cell emf’s E0Cu2 | Cu  0.337 , ECu,  0.521, 1, | Cu, , (1) Cu+1 disproportionates, (3) E0Cu, , | Cu2, , 0,  ECu, 1, , | Cu, , is positive, , (2) Cu and Cu2+ comproportionates., (4) (1) and (3) Both, , 3., , How many g of silver will be displaced from a solution of AgNO3 by 4 g of magnesium?, (1) 18 g, (2) 4 g, (3) 36 g, (4) 16 g, , 4., , The electrode potentials for Cu2+(aq) + e–  Cu+(aq) and Cu+(aq) + e–  Cu(s) are +0.15 V and + 0.50V, respectively. The value of EºCu2 /Cu will be :, (1) 0.500 V, , 5., , (2) 0.325 V, , (3) 0.650 V, , (4) 0.150 V, , How much will the reduction potential of a hydrogen electrode change when its solution initially at, pH = 0 is neutralised to pH = 7 at 25ºC ?, (1) Increases by 0.059 V, (2) Decreases by 0.059 V, (3) Increases by 0.41 V, (4) Decreases by 0.41 V

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6., , Consider the following Galvanic cell as shown in, figure. By what will value the cell voltage change, when concentration of ions in anodic and, cathodic compartments are both increased by, factor of 10 at 298 K, (1) + 0.591 V, (2) – 0.0591 V, (3) – 0.1182 V, (4) 0 V, , 7., , In a cell that utilise the reaction : Zn (s) + 2H+ (0.1M)  Zn2+ (aq) + H2 (g), addition of 0.1 M H2SO4 to cathode compartment will :, (1) increase the cell emf and shift equilibrium to the left., (2) lower the cell emf and shift equilibrium to the right., (3) increase the cell emf and shift equilibrium to the right., (4) lower the cell emf and shift equilibrium to the left., , 8., , The chemical reaction, 2AgCl(s) + H2 (g) 2HCl (aq) + 2Ag (s), taking place in a galvanic cell (under standard condition) is represented by the notation., (1) Pt(s) | H2 (g), 1 bar | 1 M KCl (aq) | AgCl(s) | Ag (s), (2) Pt(s) | H2 (g), 1 bar | 1 M HCl (aq) | 1 M Ag+ (aq) | Ag (s), (3) Pt(s) | H2 (g), 1 bar | 1 M HCl (aq) | AgCl (s) | Ag (s), (4) Pt(s) | H2 (g), 1 bar | 1 M HCl (aq) | Ag (s) | AgCl (s), , 9., , For the cell, Pt | H2 (g)  H+ (aq) || Cu2+ (aq) | Cu (s) ; EºCu / Cu2 = – 0.34 V., Then calculate approximate value of Keq ?, (1) 5 × 1012, (2) 2 × 1011, , (3) 2 × 10–11, , (4) 5 × 10–12, , 10., , In the given figure, the electrolytic cell contains 1 L of an aqueous 1 M, Copper (II) sulphate solution. If 0.4 mole of electrons are passed through, cell, the concentration of copper ion after passage of the charge will be :, (1) 0.4 M, (2) 0.8 M, (3) 1.0 M, (4) 1.2 M, , 11., , Cost of electricity for the production of 'X' litre H2 at NTP at cathode is Rs. X. Then cost of electricity for the, production 'X' litre O2 gas at NTP at anode will : (assume 1 mole of electrons as one unit of electricity), (1) 2X, (2) 4X, (3) 16X, (4) 32X, , 12., , A current of 0.1 A was passed for 965 second through a solution of Cu+ solution and 0.03175 g of, copper was deposited on the cathode. Calculate the current efficiency for the copper deposition., (Cu – 63.5), (1) 79%, (2) 39.5 %, (3) 63.25%, (4) 50%, , 13., , A current of 9.95 amp following for 10 minutes, deposits 3 g of a metal. Equivalent weight of the metal is :, (1) 12.5, (2) 18.5, (3) 21.5, (4) 48.5, , 14., , The specific conductance of a N/10 KCl at 25°C is 0.0112 ohm –1 cm–1. The resistance of cell containing, solution at the same temperature was found to be 55 ohms. The cell constant will be, (1) 6.16 cm–1, (2) 0.616 cm–1, (3) 0.0616 cm–1, (4) 616 cm–1, , 15., , The equivalent conductance of a N/10 NaCl solution at 25ºC is 10–2 Sm2eq–1. Resistance of solution, contained in the cell is 50 . Cell constant is:, (1) 50 m–1, (2) 50 × 10–6 m–1, (3) 50 × 10–3 m–1, (4) 50 × 103 m–1, , 16., , For an NaCl (aq.) solution, which of the following quantities go to zero as NaCl concentration goes to, zero? (Assume the solvent’s contribution to conductivity has been subtracted off)., (1) m, (2) , (3)  m Na, (4)  m C , ,  , ,  

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17., , 18., , , Find the value of eq for potashalum., , , , Given : m K  = 73.5 –1cm2 mol–1,  mAl 3  = 198  –1cm2 mol–1, mSO42  = 160  –1cm2 mol–1, (1) 145.6 –1cm2 eq–1, (2) 1165–1cm2 eq–1, (3) 532 –1cm2 eq–1, (4) 195.5–1cm2 eq–1, , A graph of molar conductivity of three electrolytes (NaCl, HCl and NH4OH) is plotted against C, m, 1, , 2, 3, C, , Which of the following options is correct ?, (1), (2), (3), (1), NaCl HCl, NH4OH, (3), HCl, NaCl NH4OH, , (2), (4), , (1), NH4OH, NH4OH, , (2), NaCl, HCl, , (3), HCl, NaCl, , 19., , 0.1 molar solution NaCl filled in different conductivity cell. Order of equivalent conductance of NaCl, solution is :, Cell – 1, Cell – 2, Cell – 3, A, 5 cm2, 6 cm2, 10 cm2, l, 2 cm, 3 cm, 4 cm2, Equivalents :, a, b, c, conductance, A = Area of cross section, l = distance between two electrode., (1) Cell – 1 > Cell – 2 > Cell – 3, (2) Cell – 1 = Cell – 2 = Cell – 3, (3) Cell – 1 > Cell – 3 < Cell – 2, (4) None of these, , 20., , Acetic acid is titrated with NaOH solution. Which of the following statement is correct for this titration?, (1) conductance increases upto equivalence point, then it decreases, (2) conductance increases upto equivalence point, then it increases, (3) first conductance increases slowly upto equivalence point and then increases rapidly, (4) first conductance increases slowly upto equivalence point and then drops rapidly ., SECTION-2, This section contains 5 questions. Each question, when worked out will result in Numerical Value., , 21., , The E° in the given figure is X. Report the answer as 10X., 22., , The standard reduction potential for Zn+2/Zn ; Ni+2/Ni ; and Fe+2/Fe are –0.76V, –0.23V, –0.44V, respectively. In how many of the following, the reaction X + Y+2  X+2 + Y will be non-spontaneous:, X, Y, (I) Ni, Fe, (II) Ni, Zn, (III) Fe, Zn, (VI) Zn, Ni, , 23., , A current is passed through 2 voltameters connected in series. The first voltameter contains XSO 4 (aq.), and second has Y2SO4 (aq.). The relative atomic masses of X and Y are in the ratio of 2 : 1. The ratio of, the mass of X liberated to the mass of Y liberated is a : b. Find a + b

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24., , The ratio of wt. deposited of metal x, y, z on passing electric charge in ratio of 1 : 2 : 3 respectively is, 3 : 2 : 1 then the ratio of equivalent weights for the above metals respectively is a : b : c. Find a + b + c, , 25., , A resistance of 50 is registered when two electrodes are suspended into a, beaker containing a dilute solution of a strong electrolyte such that exactly, half of the them are submerged into solution as shown in figure. If the, solution is diluted by adding pure water (negligible conductivity) so as to just, completely submerge the electrodes, the new resistance offered by the, solution would be :, , Practice Test-1 (IIT-JEE (Main Pattern)), OBJECTIVE RESPONSE SHEET (ORS), Que., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , Ans., Que., Ans., Que., Ans., , PART - II : JEE (MAIN) / AIEEE OFFLINE PROBLEMS (PREVIOUS YEARS), 1., , For the following cell with hydrogen electrodes at two different pressure p1and p2,, Pt | H2(g) | H+(aq) | H2 (g) | Pt, p1, 1M, p2, emf is given by :, [AIEEE 2002, 3/225], p1, p1, p2, p, RT, RT, RT, RT, loge, loge, loge, loge 2, (1), (2), (3), (4), F, p2, 2F, p2, F, p1, 2F, p1, , 2., , Which of the following reactions is possible at anode :, (1) 2 Cr3+ + 7H2O Cr2O72– + 14H+, (2) F2 2F–, 1, (3) O2 + 2H+  H2O, (4) displacement reaction, 2, , [AIEEE 2002, 3/225], , 3., , For a cell given below :, Ag | Ag+ || Cu2+ | Cu, –, +, Ag+ + e–  Ag, Cu2+ + 2e–  Cu,, The value of Eºcell is :, (1) x + 2y, (2) 2x + y, , [AIEEE 2002, 3/225], , Eº = x, Eº = y, (3) y –x, , (4) y – 2x, , 4., , For a cell reaction involving a two electron change, the standard emf of the cell is found to be 0.295 V, at 25°C. The equilibrium constant of the reaction at 25°C will be :, [AIEEE 2003, 3/225], (1) 1 × 10–10, (2) 29.5 × 10–2, (3) 10, (4) 1 × 1010, , 5., , Standard electrode potentials of three metals A, B and C are +0.5 V, –3.0 V and –1.2 V respectively., The reducing power of these metals is in the order :, [AIEEE 2003, 3/225], (1) B > C > A, (2) A > B > C, (3) C > B > A, (4) A > C > B

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6., , Consider the following Eº values :, 0, EFe, 3, 2  = + 0.77 V ;, / Fe, , E0Sn2 / Sn = – 0.14 V, , Under standard conditions, the cell potential for the reaction given below is :, Sn(s) + 2Fe3+(aq)  2Fe2+(aq) + Sn2+(aq), (1) 1.68 V, , (2) 1.40 V, , (3) 0.91 V, , [AIEEE 2004, 3/225], , (4) 0.63 V, , 7., , The limiting molar conductivities º for NaCl, KBr and KCl are 126, 152 and 150 S cm 2 mol–1, respectively. The value of º for NaBr is :, [AIEEE 2004, 3/225], (1) 128 S cm2 mol–1, (2) 176 S cm2 mol–1, (3) 278 S cm2 mol–1, (4) 302 S cm2 mol–1, , 8., , In a cell that utilizes the reaction, Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g),, addition of H2SO4 to cathode compartment will :, (1) lower the E and shift equilibrium to the left., (2) lower the E and shift the equilibrium to the right., (3) increase the E and shift the equilibrium to the right., (4) increase the E and shift the equilibrium to the left., , 9., , [AIEEE 2004, 3/225], , 0, The EM, values for Cr, Mn, Fe and Co are – 0.41, + 1.57, + 0.77 and + 1.97 V respectively. For, 3, / M2 , , which one of these metals, the change in oxidation state from +2 to +3 is easiest :[AIEEE 2004, 3/225], (1) Cr, (2) Mn, (3) Fe, (4) Co, 10., , Aluminium oxide may be electrolysed at 1000ºC to furnish aluminium metal (At.Mass of Al = 27 amu ;, 1 Faraday = 96,500 Coulombs). The cathode reaction is Al 3+ + 3e–  Al0. To prepare 5.12 kg of, aluminium metal by this method, one would require :, [AIEEE–2005, 3/225], (1) 5.49 × 107 C of electricity, (2) 1.83 × 107 C of electricity, (3) 5.49 × 104 C of electricity, (4) 5.49 × 1010 C of electricity, , 11., , 0, 0, The molar conductivities NaOAc, and HCl, at infinite dilution in water at 25°C are 91.0 and 426.2, 0, Scm2/mol respectively. To calculate HOAc, , the additional value required is :, , (1), 12., ,  H0 2O, , (2), , 0, KCl, , (1) – 8.12, , (2) + 8.612, , (Take, , 0.474, = 8.065), 0.059, (3) – 37.83, , [AIEEE–2006, 3/165], (4) – 16.13, , Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 ., The conductivity of this solution is 1.29 Sm –1. Resistance of the same cell when filled with 0.02 M of the, same solution is 520 . The molar conductivity of 0.02 M solution of the electrolyte will be : (Take, 129, = 0.248), [AIEEE–2006, 3/165], 520, (1) 124 × 10–4 Sm2mol–1, (3) 1.24 Sm2mol–1, , 14., , (4), , [AIEEE–2006, 3/165], 0, NaCl, , Given data is at 25°C :, Ag + – Ag + e– ; E° = 0.152 V, Ag Ag+ + e– ; E° = – 0.800 V, What is the value of log Ksp for AgI :, , 13., , (3), , 0, NaOH, , (2) 1240 × 10–4 Sm2mol–1, (4) 12.4 × 10–4 Sm2mol–1, , The equivalent conductances of two strong electrolytes at infinite dilution in H 2O (where ions move, freely through a solution) at 25°C are given below :, [AIEEE–2007, 3/120], 0,  0CH3COONa = 91.0 Scm2/equiv and, HCl, = 426.2 Scm2/equiv, What additional information/quantity one needs to calculate º of an aqueous solution of acetic acid :, (1) The limiting equivalent conductance of H+ ( H ), (2) º of chloroacetic acid (ClCH2COOH), (3) º of NaCl, , (4) º of CH3COOK

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15., , The cell Zn | Zn2+(1M) || Cu2+(1M) | Cu : (E°cell = 1.10V) was allowed to completely discharge at 298 K.,   Zn2  ,   is : (Take 1.1 = 18.65), The relative concentration of Zn2+ to Cu2+  , 2, , 0.059,  Cu  , , , , (1) 1037.3, 16., , (2) 9.65 × 104, , Given : E0, , Cr3 / Cr, , = – 0.72, E0, , Fe2 / Fe, , (3) antilog (24.08), , [AIEEE–2007, 3/120], , (4) 37.3, , = – 0.42 V, , The potential for the cell Cr | Cr3+(0.1 M) || Fe2+(0.01 M) | Fe at 298 K is :, 2.303 R (298), (Take, = 0.06), [AIEEE–2008, 3/105], F, (1) 0.339 V, (2) – 0.339 V, (3) – 0.26 V, (4) 0.26 V, 17., , 0, 0, Given : EFe, = – 0.036 V, EFe, = – 0.439 V, 3, 2, / Fe, / Fe, , 3, 2, The value of standard electrode potential for the change, Fe(aq), + e–  Fe(aq), will be :, , (1) 0.385V, 18., , (3) –0.270V, , (2) 0.770V, , (4) –0.072V, , The Gibbs energy for the decomposition of Al 2O3 at 500ºC is as follows :, [AIEEE–2010, 4/144], 2, 4, Al2O3 , Al + O2 ; rG = + 966 kJmol–1. The potential difference needed for electrolytic reduction, 3, 3, of Al2O3 at 500ºC is at least :, (1) 4.5 V, , (2) 3.0 V, , (3) 2.5 V, , (4) 5.0 V, , 19., , The reduction potential of hydrogen half-cell will be negative, if :, [AIEEE–2011(1), 4/120], (1) p(H2) = 1 atm and [H+] = 2.0 M, (2) p(H2) = 1 atm and [H+] = 1.0 M, (3) p(H2) = 2 atm and [H+] = 1.0 M, (4) p(H2) = 2 atm and [H+] = 2.0 M, , 20., , The standard reduction potentials for Zn2+/Zn, Ni2+/Ni and Fe2+/Fe are –0.76, – 0.23 and –0.44 V, respectively. The reaction X + Y2+  X2+ + Y will be spontaneous, when :, [AIEEE 2012, 4/120], (1) X = Ni, Y = Fe, (2) X = Ni, Y = Zn, (3) X= Fe, Y = Zn, (4) X= Zn, Y = Ni, , 21., , Given :, , 0, E0Cr 3  / Cr = –0.74 V ; EMnO, = 1.51 V, , / Mn2 , 4, , E0Cr O2 / Cr3 = 1.33 V ; E0Cl / Cl = 1.36 V, 2, , 7, , Based on the data given above, strongest oxidising agent will be :, (1) Cl, (2) Cr3+, (3) Mn2+, , [JEE(Main) 2013, 4/120], (4) MnO4–, , 22., , Resistance of 0.2 M solution of an electrolyte is 50 . The specific conductance of the solution is 1.4 S, m–1. The resistance of 0.5 M solution of the same electrolyte is 280 . The molar conductivity of 0.5 M, solution of the electrolyte in S m 2 mol–1 is :, [JEE(Main) 2014, 4/120], (1) 5  10–4, (2) 5  10–3, (3) 5  103, (4) 5  102, , 23., , The equivalent conductance of NaCl at concentration C and at infinite dilution are C and ,, respectively. The correct relationship between C and  is given as : (where the constant B is positive), [JEE(Main) 2014, 4/120], (1) C =  + (B)C, , 24., , (2) C =  – (B)C, , (3) C =  – (B), , C, , (4) C =  + (B), , C, , The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is :, [JEE(Main) 2014, 4/120], (1) Ag, (2) Ca, (3) Cu, (4) Cr

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25., , Given below are the half-cell reactions :, Mn2+ + 2e– Mn ; Eº = –1.18 V, 2(Mn3+ + e–  Mn2+) ; Eº = +1.51 V, The Eº for 3Mn2+  Mn + 2Mn3+ will be :, (1) –2.69 V ; the reaction will not occur, (3) –0.33 V ; the reaction will not occur, , [JEE(Main) 2014, 4/120], , (2) –2.69 V ; the reaction will occur, (4) –0.33 V ; the reaction will occur, , 26., , Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the, cathode is : (at. mass of Cu = 63.5 amu), [JEE(Main) 2015, 4/120], (1) 0 g, (2) 63.5 g, (3) 2 g, (4) 127 g, , 27., , Galvanization is applying a coating of :, (1) Cr, (2) Cu, , 28., , (3) Zn, , [JEE(Main) 2016, 4/120], (4) Pb, , Given, º, ECl, , 2, , / Cl–, , º, = 1.36 V, ECr, = –0.74 V, 3, / Cr, , º, º, = 1.33 V, EMnO, = 1.51 V, ECr, –, O2  / Cr 3 , / Mn2 , 2, , 7, , 4, , Among the following, the strongest reducing agent is :, (1) Mn2+, (2) Cr3+, (3) Cl–, 29., , [JEE(Main) 2017, 4/120], (4) Cr, , How long (approximate) should water be electrolysed by passing through 100 amperes current so that, the oxygen released can completely burn 27.66 g of diborane?, [JEE(Main)-2018, 4/120], (Atomic weight of B = 10.8u), (1) 3.2 hours, (2) 1.6 hours, (3) 6.4 hours, (4) 0.8 hours, , PART - III : NATIONAL STANDARD EXAMINATION IN CHEMISTRY (NSEC) STAGE-I, 1., , The increase in the equivalent conductance of a salt solution on dilution is due to increase in the, [NSEC-2000], (A) attraction between the ions, (B) degree of ionization of the salt, (C) molecular attraction, (D) association of the salt, , 2., , When 96500 coulombs of electricity are passed through a nickel sulphate solution, the amount of nickel, deposited will be, [NSEC-2000], (A) 1.0 mol, (B) 0.5 mol, (C) 0.1 mol, (D) 2.0 mol, [NSEC-2000], , 3., , When a piece of copper wire is immersed in a silver nitrate solution, the colour of the solution turns blue, due to, [NSEC-2000], (A) oxidation of silver, (B) reduction of copper, (C) oxidation of copper, (D) formation of soluble complex, , 4., , The reduction potentials of Zn, Cu, Fe and Ag are in the order :, (A) Zn,Cu,Fe,Ag, (B) Cu,Ag,Fe,Zn, (C) Ag,Cu,Fe,Zn, , 5., , The standard reduction potentials of Cu2+/Cu and Cu+/Cu are 0.339 V and 0.518 V respectively. The, standard electrode potential of Cu2+/Cu+ half cell is :, [NSEC-2001], (A) 0.16 V, (B) 0.827 V, (C) 0.184 V, (D) 0.490 V, , 6., , How many coulombs are required for oxidation of 1 mole of H2O to O2 ?, [NSEC-2001], (A) 3.86 × 105C, (B) 9.65 × 104C, (C) 1.93 × 105C, (D) 4.825 × 104C, , 7., , The metal which can not be obtained by electrolysis of its aqueous salt solution is :, (A) Au, (B) Al, (C) Ag, (D) Cu, , 8., , The units of conductivity are :, (A) Siemen–1.cm–1., (B) Siemen.cm, , (C) Siemen.cm–1, , [NSEC-2001], (D) Fe,Zn,Cu, Ag, , [NSEC-2001], , [NSEC-2001], (D) Semen.cm–2.mol–1

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9., , The calomel electrode used a reference elecrode contains :, (A) PbO2-PbSO4 mixture, (B) HgCl2, (C) Hg2Cl2, (D) ZnCl2, , [NSEC-2001], , 10., , KCl is used in a salt bridge because :, (A) it forms a good jelly with agar-agar, (B) it is strong electrolyte, (C) it is a good conductor of elelctric current, (D) the transference number of K+ and Cl– ions are almost equal, , [NSEC-2001], , 11., , During the electrolysis of fused NaCl, the reaction occurring at the anode is :, (A) reduction of Na+ ions, (B) oxidation of Cl– ions, +, (C) oxidation of Na ions, (D) reduction of Cl– ions, , [NSEC-2001], , 12., , On electrolysis, one mole of chromium ions will be deposited by :, (A) three moles of electrons, (B) two moles of electrons, (C) one mole of electrons, (D) six moles of electrons, , [NSEC-2001], , 13., , The quantity of electricity which deposits 1.08 g of silver from AgNO 3 solution is :, [NSEC-2002], (A) 96500 coulombs, (B) 9650 coulombs, (C) 965 coulombs, (D) 96.5 coulombs., , 14., , In the conductometric titration of CH3 COOH vs NaOH, the titration curve obtained will be of the type, [NSEC-2002], (A), , (B), , (C), , (D), , 15., , The standared reduction potentials at 298 K for the half reactions are:, (a) Zn2+ (aq) + 2e–  Zn(s) ; –0.762 V, (b) Cr3+ (aq) + 3e–  Cr (s) ; – 0.740 V, +, –, (c) 2H (aq) + 2e  H2(g) ; 0.000 V, (d) Fe3+ (aq) + e–  Fe2+ (aq) ; 0.770 V, Which is the strongest reducing agent?, (A) Zn (s), (B) Cr (s), (C) H2 (g), (D) Fe2+(aq)., , [NSEC-2002], , 16., , The molar conductivities of H+, Li+ and Na+ ions in aqueous solutions at infinite dilution are in the order :, [NSEC-2003], (A) H+ > Li+ > Na+, (B) H+ < Li+ < Na+, (C) H+> Na+ > Li+, (D) Na+ > H+ > Li+., , 17., , Fe2+ + 2e  Fe, .....(i), 3+, 2+, Fe + e  Fe, .....(ii), The standard potentials (in volt) corresponding to the reactions (i) and (ii) are E1 and E2 respectively., The value (in volt) of the standard potential corresponding to the reaction Fe3+ + 3e  Fe is, [NSEC-2003], (A) (E1+E2), (B) (2E1+E2)/3, (C) (E1 +2E2)/2, (D) (E1+E2)/3., , 18., , The standard reduction potentials of Cu2+, Zn2+, Sn2+ and Ag+ are 0.34, –0.76, –0.14 and 0.80 V, respectively. The storage that is possible without any reaction is for, [NSEC-2003], (A) CuSO4 solution in a zinc vessel, (B) AgNO3 solution in a zinc vessel, (C) AgNO3 solution in a tin vessel, (D) CuSO4 solution in a silver vessel., , 19., , A certain current passed through CuSO4 solution for 100 seconds deposits 0.3175 g of copper. The, current passed (in A) is, [NSEC-2004], (A) 4.83, (B) 9.65, (C) 0.963, (D) 0.483, , 20., , The salt that can be used in the salt bridge of an electrochemical cell is, (A) FeCI3, (B) AgCI, (C) CH3COONa, , [NSEC-2004], (D) KNO3.

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21., , The conductometric titration curve (of conductance vs mL of NaOH) obtained when acetic acid is, titrated against NaOH is, [NSEC-2005], , (A), , 22., , (B), , (C), , (D), , In an alkaline energy cell the overall cell reaction is as follows :, Zn(s) + 2MnO2(s) + 2H2O  Zn(OH)2(s) + 2MnO(OH)., Which of the following reactions is taking place at the cathode?, (A) 2MnO2(s) + 2H2O + 2e  Zn(OH)2(s) + 2MnO(OH)(s), (B) 2MnO2(s) + 2H2O + 2e  2MnO(OH)(s) + 2OH–(aq), (C) Zn(s) + 2OH- (aq)  Zn(OH)2(s) + 2e, (D) Zn(OH)2(s) + 2e  Zn(s) + 2OH–(aq)., , [NSEC-2005], , 23., , What is the charge on an ion of tin if 7.42 g of metallic tin is deposited by passage of 24125 coulombs, through a solution containing the ion ?, [NSEC-2005], (A) +1, (B) +3, (C) +2, (D) +4., , 24., , The cell potential (E) and free energy change (G) accompanying an electrochemical reaction, are, related by, [NSEC-2005], [NSEC-2005], (A) G = nFE, (B) G = nFE, (C) G = nFlogE, (D) G = nF logE. ., , 25., , The mass of the copper, in grams, deposited during the passage of 2.5 ampere current through a Cu(II), sulphate solution for 1 hour is, [NSEC-2006], (A) 5.96, (B) 29.8, (C) 2.98, (D) 59.6, , 26., , The standard reduction potentials of Fe2+/Fe and Cu2+/Cu electrodes are –0.44 and 0.34 volts,, respectively. The following reaction would occur, [NSEC-2006], (A) copper will reduce Fe2+ ions, (B) iron will reduce Cu2+ ions, (C) iron will oxidise copper metal, (D) Cu2+ ions will reduce Fe2+., , 27., , Rusting of iron is due to the formation of, (A) hydrated ferrous oxide, (C) only ferric oxide, , 28., , If the equilibrium constant of the disproportionation reaction, Hg22+ = Hg0 + Hg2+, at 298 K is 0.0795, the standard e.m.f. of the reaction is, (A) –0.065 V, (B) –0.212 V, (C) 0.125 V, , [NSEC-2006], (B) hydrated ferric oxide, (D) a mixture of ferric oxide and Fe(OH)3., , [NSEC-2006], (D) 0.110 V, , 29., , The voltage for the cell: Fe /Fe2+(0.001M) // Cu2+(0.10M) /Cu2+ (0.10 M) / Cu2+ (0.10 M) / Cu is 0.807 V, at 25°C. What is the value of E° ?, [NSEC-2007], (A) 0.629 V, (B) 0.689 V, (C) 0.748 V, (D) 0.866 V, , 30., , A current of 2.0 A is used to plate Ni(s) from 500mL of a 1.0 M Ni 2+ aqueous solution. What is the [Ni 2+], after 3.0 hours ?, [NSEC-2007], (A) 0.39 M, (B) 0.46 M, (C) 0.78 M, (D) 0.89 M, , 31., , Nickel metal is added to a solution containing1.0 M Pb2+(aq) and 1.0 M Cd2+(aq). Use the standard, reduction potential to determine which of the following reaction (s) will occur., [NSEC-2008], Reaction 1 : Ni(s) + Pb2+(aq)  Pb(s) + Ni2+(aq), Reaction 2 : Ni(s) + Cd2+(aq) Cd(s) + Ni2+(aq), Reactions :, Pb2+(aq) + 2e = Pb(s), E° = –0.13 V, Ni2+(aq) + 2e = Ni(s) E° = –0.23 V, Cd2+(aq) + 2e = Cd(s) E° = –0.40 V, (A) 1 only, (B) 2 only, (C) both 1 and 2, (D) neither 1 nor 2

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32., , An electrochemical cell constructed for the reaction, Cu2+ (aq) + M(s)  Cu(s) + M2+(aq) has an E° = 0.75 V., The standard reduction potential for Cu2+(aq) is 0.34 V. What is the standard reduction potential for, M2+(aq) ?, [NSEC-2008], (A) 1.09 V, (B) 0.410 V, (C) – 0410 V, (D) – 1.09 V, , 33., , An electric current is passed through a silver voltameter connected to a water voltameter. 0.324 g of, silver was deposited on the cathode of the silver voltameter. The volume of oxygen evolved at NTP is :, [NSEC-2009], (A) 5.6 cm3, (B) 16.8 cm3, (C) 11.2 cm3, (D) 22.4 cm3, , 34., , The amount of copper (At. wt. 63.54) deposited by passing 0.2 faraday of electricity through copper, sulphate is, [NSEC-2009], (A) 3.175 g, (B) 6.350 g, (C) 31.75 g, (D) 63.35 g, , 35., , When aqueous solution of sodum chloride is electrolysed using platinum electrode the cathode reaction, is,, [NSEC-2009], 1, (A) Na+ + e– Na, (B) H2O + e–  H2 + OH–, 2, (C) Na+ + OH– Na+ + OH– + e–, (D) Na+ + H2O + e– Na + H+ + OH–, , 36., , The standard electrode potential values for four metals K, L, M and N are respectively, –3.05, –1.66, –, 0.40 and +0.80V. The best reducing agent is –, [NSEC-2009], (A) L, (B) K, (C) N, (D) M, , 37., , 10Cl– (aq) + 2MnO4– (aq) + 16H+ (aq) 5Cl2 (g) + 2Mn2+ (aq) + 8H2O (l), [NSEC-2009], The value of Eº for the above reaction at 25ºC is 0.15V. Hence, the value of K for this reaction is:, (A) 2.4 × 1025, (B) 4.9 × 1012, (C) 1.2 × 105, (D) 3.4 × 102, , 38., , Adding powdered Pb and Fe to a solution containing 1 M each of Pb2+ and Fe2+ ions would result in the, 0, 0,  0.44V ),  0.126V and EFe, formation of – ( EPb, [NSEC-2010], 2, 2, / Fe, / Pb, (A) more of Pb and Fe2+ ions, (C) more of Pb and Fe, , (B) more of Fe and Pb2+ ions, (D) more of Pb2+ and Fe2+ ions, , The mass of copper deposited when a current of 10A is passed through a solution of copper(II) nitrate, for 30.6s is, [NSEC-2010], (A) 0.101 g, (B) 0.201 g, (C) 0.403 g, (D) 6.04 g, , 41., , In the conductometric titration of silver nitrate against KCI, the graph obtained is, , (B), , volume of KCl, 42., , volume of KCl, , (D), , volume of KCl, , volume of KCl, , Th emf of the cell (Zn | ZnSO4(0.1M) || CdSO4 (0.01M) | Cd) is, ( EoZn2 / Zn = – 0.76 V, EoCd2 / Cd = 0.40 V at 298 K), (A) +0.33 V, , 43., , (C), , [NSEC-2011], , Conductance, , (A), , Conductance, , 40., , Conductance, , The cell Al(s)|Al3+(aq) (0.001 M) | | Cu2+(aq) (0.10 M) | Cu(s) has a standard cell potential Eº = 2.00 V at, 25°C. The cell potential at the given concentration will be :, [NSEC-2010], (A) 2.07 V, (B) 2.03 V, (C) 1.97 V, (D) 1.94 V, , Conductance, , 39., , (B) +0.36 V, , (C) +1.13 V, , [NSEC-2011], (D) –0.36 V, , The conductivity of a metal decreases with increase in temperature because :, [NSEC-2012], (A) the kinetic energy of the electrons increases (B) the movement of electrons becomes haphazard, (C) the ions start vibrating, (D) the metal becomes hot and starts emiting radiation

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44., , The amount of electricity required to deposit 1.0 mole of aluminium from a solution of AlCl3 will be :, [NSEC-2012], (A) 1 faraday, (B) 3 faradays, (C) 0.33 faraday, (D) 1.33 faraday, , 45., , Which is the strongest oxidising agent among the species given below?, [NSEC-2013], (i) In3+, Eº = – 1.34V, (ii) Au3+, Eº =, 1.40V, (iii) Hg2+, Eº =, 0.867V, (iv) Cr3+, Eº =, – 0.786V, (A) Cr3+, (B) Au3+, (C) Hg2+, (D) In3+, , 46., , Which of the following aqueous solution has the lowest electrical conductance ?, [NSEC-2013], (A) 0.01M CaCl2, (B) 0.01M KNO2, (C) 0.01M CH3COOH (D) 0.01M CH3COCH3, , 47., , The value of the constant in Nernst equation, cons tant, E = Eº –, ln Q at 25ºC is :, n, (A) 0.592, (B) 0.0592, , [NSEC-2013], (C) 0.296, , (D) 0.0296, , 48., , When zinc rod is directly placed in copper sulphate solution, (A) the blue colour of the solution starts intensifying, (B) the solution remains electrically neutral, (C) the temperature of the solution falls, (D) the weight of zinc rod starts increasing, , [NSEC-2013], , 49., , For the following cell at 25ºC the E.M.F. is : [If Eº, , [NSEC-2014], , M2  /M, , (A) 0.089V, 50., , = 0.347 V ], , M(S) | M2+ (1M) || M2+ (0.01M) | M(S), (B) 0.598V, (C) 0.251V, , (D) 0.764V, , For a strong electrolyte, the change in the molar conductance with concentration is represented by :, [NSEC-2014], , (I), (A) I, , (II), (B) II, , (III), (C) III, , (IV), (D) IV, , 51., , The specific conductance of 0.01M solution of the weak monobasic acid is 0.20 × 10–3 Scm–1. The, dissociation constant of the acid is, [Given: ºHA = 400 Scm2mol–1], [NSEC-2014], (A) 5 × 10–2, (B) 2.5 × 10–5, (C) 5 × 10–4, (D) 2.5 × 10–11, , 52., , The reaction given below is the cell reaction in a galvanic cell., Cd(s) + Sn2+(aq)  Cd2+(aq) + Sn(s), Where, [Cd2+] = 0.1 M and [Sn2+] = 0.025 M, 0, 0, 1, Given: E Cd 2 / Cd  0.403V , E Sn2 / Sn  0.136V , F  96485Cmol, At 25ºC, the free energy change for this reaction is :, (A) — 48.05 KJ, (B) — 54.96 KJ, (C) — 100.58 KJ, , [NSEC-2014], (D) — 107.46 KJ, , 53., , A current of 5.0 A flows for 4.0 h through an electrolytic cell containing a molten salt of metal M. This, results in deposition of 0.25 mol of the metal M at the cathode. The oxidation state of M in the molten, salt is (1 Faraday = 96485 C mol–1), [NSEC-2015], (A) +1, (B) +2, (C) +3, (D) +4, , 54., , The limiting molar conductivities of KCl, KNO3, and AgNO3 are 149.9, 145.0 and 133.4 S cm2 mol–1,, respectively, at 25ºC. The limiting molar conductivity of AgCl at the same temperature in S cm 2 mol-1 is, [NSEC-2015], (A) 128.5, (B) 138.3, (C) 161.5, (D) 283.3

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55., , The emf of a cell corresponding to the following reaction is 0.199 V at 298 K., Zn (S) + 2 H+ (aq)  Zn2+ (0.1 M) + H2 (g), ( Eº Zn/Zn2  = 0.76V), , [NSEC-2015], , The approximate pH of the solution n at the electrode where hydrogen is being produced is (p H2 = 1, atm)., (A) 3, (B) 9, (C) 10, (D) 11, 56., , The standard electrode potentials, Eº of Fe3+/Fe2+and Fe2+/Fe at 300 K are +0.77 V and –0.44 V,, respectively. The Eº of Fe3+/Fe at the same temperature is, [NSEC-2015], (A) 1.21 V, (B) 0.33 V, (C) –0.036 V, (D) 0.036 V, , 57., , Three Faradays of electricity are passed through aqueous solutions of AgNO3, NiSO4 and CrCl3 kept in, three vessels using inert electrodes. The ratio (in moles) in which the metals Ag, Ni and Cr are, deposited is :, [NSEC-2016], (A) 1 : 2 : 3, (B) 3 : 2 : 1, (C) 6 : 3 : 2, (D) 2 : 3 : 6, , 58., , The standard potentials (E°) of MnO4– /Mn2+ and MnO2/Mn2+ half cells in acidic medium are 1.51 V and, 1.23 V respectively at 298 K. The standard potential of MnO4– /MnO2 half-cell in acidic medium at the, same temperature is :, [NSEC-2016], (A) 5.09 V, (B) 1.70 V, (C) 0.28 V, (D) 3.34 V, , 59., , Given the E0 values for the half reactions :, Sn4+ + 2e–  Sn2+, 0.15 V, 2Hg2+ + 2e–  Hg22+, 0.92 V, PbO2 + 4H+ + 2e–  Pb2+ + 2H2O, 1.45 V, Which of the following statements is true?, [NSEC-2016], (A) Sn2+ is a stronger oxidizing agent than Pb4+ (B) Sn2+ is a stronger reducing agent than Hg22+, (C) Hg2+ is a stronger oxidizing agent than Pb4+ (D) Pb2+ is a stronger reducing agent than Sn2+, , 60., , The conductivity of 0.10 M KCl solution at 298 K is 1.29 × 10–2 S cm–1. The resistance of this solution is, found to be 28.44 . Using the same cell, the resistance of 0.10 M NH4Cl solution is found to be 28.50, . The molar conductivity of NH4Cl solution in S cm 2 mol–1 is :, [NSEC-2016], (A) 0.130, (B) 13, (C) 130, (D) 1300, , 61., , Which of the following statements is not correct regarding the galvanic cells ?, (A) Oxidation occurs at the anode., (B) Ions carry current inside the cell., (C) Electrons flow in the external circuit from cathode to anode., (D) When the cell potential is positive, the cell reaction is spontaneous., , 62., , When a medal is electroplated with silver (Ag), (A) The medal is the anode, (C) The solution contains Ag+ ions, , [NSEC-2016], , [NSEC-2017], (B) Ag metal is the cathode, (D) The reaction at the anode is Ag+ + e– Ag, , Use the table given below to answer questions 63 and 64, Reaction, Ag Ag+ + e–, Cr3+ + 3e– 3Cr, Zn2++ 2e– Zn, 2(s) + 2e– 2 –, Co2+ + 2e– Co, Ni2+ + 2e–Ni, 63., 64., , E0/V, –0.80, –0.74, –0.76, 0.54, –0.28, –0.26, , The best reducing agent among the following is, (A) Ag+, (B) Zn2+, (C) Cr3+, Eº of the given cell is :, Ni | (Ni+2, 1.0 M) || (Co+2, 1.0 M) | Co, (A) +0.02V, (B) –0.02V, , [NSEC-2017], (D) –, [NSEC-2017], , (C) –0.54V, , (D) +0.54V

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65., , The reduction of O2 to H2O in acidic solution has a standard reduction potential of 1.23 V. If the pH of, the acid solution is increased by one unit, half cell potential will, [NSEC-2017], O2(g) + 4H+(aq) + 4e–  2H2O(l), (A) decrease by 59 mV, (B) increase by 59 mV, (C) decrease by 236 mV, (D) increase by 236 mV, , 66., , From the given standard electrode potentials, [NSEC-2018], Sn4+(aq) + 2e–  Sn2+ (aq), Eº = 0.15V, Br2(I) + 2e–  2Br–(aq), Eº = 1.07V, The approximate free energy change of the process 2Br– (aq) + Sn4+ (aq)  Br2(l) + Sn2+(aq) is, (A) 117.6 kJ, (B) 355 kJ, (C) –177.6 kJ, (D) –355 kJ, , 67., , Concentration of K+ ions inside a biological cell was found to be 25 times higher than that outside. The, magnitude of the potential difference between the two sides of the cell is close to (2.303 RT/F-can be, taken as 59 mV; difference in concentrations of other ions can be taken as negligible), [NSEC-2018], (A) 4.2 mV, (B) 195 mV, (C) 82 mV, (D) –82 mV, , 68., , The standard redox potential for the reaction 2H2O  O2 + 4H+ + 4e– is –1.23V. If the same reaction is, carried out at 25ºC and at pH = 7, the potential will be, [NSEC-2018], (A) –0.82 V, (B) –3.28V, (C) 0.82V, (D) –1.18V, , 69., , The standard electrode potential (E°) of the Daniel cell is 1.1 V and the overall cell reaction can be, represented as Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s). Under which of the following conditions will the, cell potential be higher than 1.1 V ?, [NSEC-2018], (A) 1.0 M Zn2+, 1.0 M Cu2+, (B) 1.2 M Zn2+, 1.2 M Cu2+, (C) 0.1 M Zn2+, 1.0 M Cu2+, (D) 1.0 M Zn2+, 0.01 M Cu2+, , 70., , An electrochemical cell was constructed with Fe2+/Fe and Cd2+/Cd at 25ºC with initial concentrations of, [Fe2+] = 0.800 M and [Cd2+] = 0.250 M. The EMF of the cell when [Cd2+] becomes 0.100 M is, Half cell, Eº(V), Fe2+(aq)/Fe(s), – 0.44, Cd2+(aq)/Cd(s), – 0.40, [NSEC-2019], (A) 0.013 V, (B) 0.011 V, (C) 0. 051 V, (D) 0.022 V, , 71., , Molten NaCl is electrolysed for 35 minutes with a currect of 3.50 A at 40ºC and 1 bar pressure. Volume, of chlorine gas evolved in this electrolysis is, [NSEC-2019], (A) 0.016 L, (B) 0.98 L, (C) 9.8 L, (D) 1.96 L, , 72., , If the standard electrode potentials of Fe3+/Fe and Fe2+/Fe are –0.04 V and –0.44 V respectively then, that of Fe3+/Fe2+ is, [NSEC-2019], (A) 0.76 V, (B) –0.76 V, (C) 0.40 V, (D) –0.40 V, , PART - IV : HIGH LEVEL PROBLEMS (HLP), THEORY, Solubility product and EMF (Metal-Metal Insoluble Salt Electrode) :, , , , , A half cell containing metal M and its sparingly soluble salt MA in a saturated solution., i.e M(s) | MA (satd) or a metal, its sparingly soluble salt in contact with a solution of a soluble, salt NaA of the same anion, i.e. M(s) | MA(s) | NaA is set up., , The solubility product of a sparingly doubles salt is a kind of equilibrium constant., , MX (s), , M, , (s), , At Anode, M+ (aq) + e–, , M+(aq) + X– (aq), , Ksp = [M+] [X–], , At Cathode, + MX(s), , e–, , M(s) + X– (aq)

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When the switch is closed, mass of tin-electrode increase. If Eº (Sn2+ / Sn) = – 0.14 V and for, Eº (Xn+ / X) = – 0.78 V and initial emf of the cell is 0.65 V, determine n and indicate the direction of, electron flow in the external circuit., 6., , Equinormal Solutions of two weak acids, HA (pKa = 3) and HB (pKa = 5) are each placed in contact with, standard hydrogen electrode at 25°C (T = 298 K). When a cell is constructed by interconnecting them, through a salt bridge find the e.m.f. of the cell., , 7., , In two vessels each containing 500ml water, 0.5m mol of aniline (K b = 10–9) and 25 m mol of HCl are, added separately. Two hydrogen electrodes are constructed using these solutions. Calculate the emf of, cell made by connecting them appropriately., , 8., , Write cell reaction from given cell diagrams, (A) Cu | Cu2 || Cl– | Hg2Cl2 | Hg | Pt, (B) Ag (s) | AglO3 (s)|Ag+, HlO3 || Zn2+ | Zn (s), (C) Mn (s) | Mn (OH)2 (s) Mn2+, OH– || Cu2+|Cu (s), , 9., , For the galvanic cell : Ag | AgCl (s)|KCl (0.2M) || KBr (0.001M) |AgBr (s) |Ag,, 2.303RT, Calculate the EMF generated? (Take, = 0.06), F, [Ksp(AgCl)= 10–10 ; Ksp(AgBr)=10–13], , 10., , Given, E° = –0.27 V for the Cl– | PbCl2 |Pb couple and – 0.12 V for the Pb2+ | Pb couple, determine Ksp, 2.303RT, for PbCl2 at 25°C ? (Take, = 0.06), F, , 11., , The pKsp of Agl is 16. if the E° value for Ag+ | Ag is 0.8 V. Find the E° for the half cell reaction Agl(s) +, 2.303RT, e–  Ag + l– ? (Take, = 0.06 ), F, , 12., , The EMF of the standard weston cadmium cell Cd (12.5%) in Hg | 3CdSO 4, 8H2O (solid) | saturated, solution of CdSO4 || Hg2SO4(s) | Hg is 1.0180 volts at 25° C and the temperature coefficient of the cell,,  E , –5,  T  = – 4.0 × 10 V/degree. Calculate G, H and S for the reaction in the cell when n = 2., P, , 13., , H for the reaction Ag(s) +, , 14., , The standard electromotive force of the cell :, Fe | Fe2+ (aq) | | Cd2+ | Cd is 0.0372 V, The temperature coefficient of e.m.f. is –0.125 V K–1. Calculate the quantities Gº, Hº and Sº at 25ºC., , 15., , The voltage of a certain cell has standred potential at 25ºC and 20ºC are 0.3525 V and 0.3533 V, respectively. If the number of electrons involved in the overall reactions are two, calculate Gº, Sº and, Hº at 25ºC., , 16., , A metal is known to form fluoride MF2. When 10A of electricity is passed through a molten salt for 330, sec., 1.95g of metal is deposited. Find the atomic weight of M. What will be the quantity electricity, required to deposit the same mass of Cu from CuSO4 ?, , 17., , Find the volume of gases evolved by passing 0.965 A current for 1 hr through an aqueous solution of, CH3COONa at 250ºC and 1 atm., , 18., , One of the methods of preparation of per disulphuric acid, H2S2O8, involve electrolytic oxidation of H2SO4 at, anode (2H2SO4  H2S2O8 + 2H+ + 2e–) with oxygen and hydrogen as by-products. In such an electrolysis,, 9.722 L of H2 and 2.35L of O2 were generated at STP. What is the weight of H2S2O8 formed ?, , 1, Hg2 Cl2 (s)  AgCl(s) + Hg() is +1280 cal at 25°C. This reaction can, 2, be conducted in a cell for which the emf = 0.0455 volt at this temperature. Calculate the temperature, coefficient of the emf.

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19., , The standard reduction potential values, Eº (Bi 3+ / Bi) and Eº (Cu2+ /Cu) are 0.226V and 0.344V, respectively. A mixture of salts of bismuth and copper at unit concentration each is electrolysed at, 25°C. To what value can [Cu2+] be brought down before bismuth starts to deposit, in electrolysis., , 20., , Calculate the dissociation constant (kdissociation) of water at 25°C from the following data :, , , , Specific conductance of H2O = 5.8 × 10–8 mho cm–1, H = 350.0 and  OH¯, = 198.0 mho cm2 mol-1, , 21., , (a) Calculate Gº of the following reaction :, Ag+(aq) + Cl–(aq) AgCl(s), Given : Gº(AgCl) = – 109 kJ/mole, Gº(Cl–) = – 129 kJ/mole, Gº(Ag+) = 77 kJ/mole., Represent the above reaction in form of a cell., Calculate Eº of the cell. Find log10Ksp of AgCl at 25ºC., (b) 6.539 × 10–2 g of metallic Zn (atomic mass = 65.39 amu) was added to 100 mL of saturated solution, of AgCl., [Zn2 ], Calculate log10, at equilibrium at 25ºC, given that :, [Ag ]2, Ag+ + e–  Ag, Eº = 0.80 V, Zn2+ + 2e–  Zn, Eº = – 0.76 V, 114, 1.56, Also find how many moles of Ag will be formed. (Take, = 0.59,, = 26.44) [JEE 2005, 6/60], 193, 0.059, , ONLY ONE OPTION CORRECT TYPE, 22., , The standard potential of the reaction H2O + e– , (A) – 0.828 V, , (B) 0.828 V, , 1, H2 + OH– at 298 K by using Kw (H2O) = 10–14, is :, 2, (C) 0 V, (D) – 0.5 V, , 23., , Given : Hg22+ + 2 e 2 Hg , Eº = 0.789 V & Hg2+ + 2 e  Hg , Eº = 0.854 V,, calculate the equilibrium constant for Hg22+  Hg + Hg2+ ., (A) 3.13 × 103, (B) 3.13 × 104, (C) 6.26 × 103, (D) 6.26 × 104, , 24., , MnO4– + 8H+ + 5e–  Mn2+ + 4H2O,, If H+ concentration is decreased from 1 M to 10–4 M at 25ºC, where as concentration of Mn2+ and MnO4–, remain 1 M., (A) the potential decreases by 0.38 V with decrease in oxidising power, (B) the potential increases by 0.38 V with increase in oxidising power, (C) the potential decreases by 0.25 V with decrease in oxidising power, (D) the potential decreases by 0.38 V without affecting oxidising power, MnO4– + 8H+ + 5e–  Mn2+ + 4H2O,, , 25., , At equimolar concentrations of Fe2+ and Fe3+, what must [Ag+] be so that the voltage of the galvanic cell, made from the (Ag+ | Ag) and (Fe3+ | Fe2+) electrodes equals zero?, Fe2+ + Ag+, Fe3+ + Ag, Eº Ag |Ag = 0.7991 ; EºFe3  | Fe2  = 0.771, (A) 0.34, , (B) 0.44, , (C) 0.47, , (D) 0.61, , 26., , The cell Pt (H2) (1 atm) | H (pH = ?) || I (a = 1) | AgI(s), Ag(s) | Pt has emf, E298K = 0. The standard, electrode potential for the reaction AgI + e¯  Ag + I is – 0.151 volt. Calculate the pH value., (A) 3.37, (B) 5.26, (C) 2.56, (D) 4.62, , 27., , Using the information in the preceding problem, calculate the solubility product of AgI in water at 25°C, º, [ E(Ag, = + 0.799 volt], , , Ag), , +, , (A) 1.97 × 10–17, , (B) 8.43 × 10–17, , –, , (C) 1.79 × 10–17, , (D) 9.17 × 10–17

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28., , The solubility product of silver iodide is 8.3 × 10–17 and the standard reduction potential of Ag, Ag+, electrode is + 0.8 volts at 25° C. The standard reduction potential of Ag,Agl/l¯ electrode from these data, is, (A) – 0.30 V, (B) + 0.15 V, (C) + 0.10 V, (D) – 0.15 V, , 29., , The efficiency of an hypothetical cell is about 84% which involves the following reaction :, A (s) + B2+ (aq)  A2+ (aq) + B (s) : H = – 285 kJ, Then, the standard electrode potential of the cell will be, (A) 1.20, (B) 2.40 V, (C) 1.10 V, (D) 1.24 V, , 30., , The temperature coefficient, of the emf i.e., , dE, = – 0.00065 volt. deg–1 for the cell Cd | CdCl2 (1M) ||, dT, AgCl (s) | Ag at 25°C. Calculate the entropy changes S298K for the cell reaction, Cd + 2AgCl  Cd2+ +, , 2Cl + 2Ag, (A) – 105.5 JK–1, , (B) – 150.2 JK–1, , (C) – 75.7 JK–1, , (D) – 125.5 JK–1, , 31., , The standard emf of the cell, Cd(s) CdCl2(aq) (0.1 M) AgCl(s) Ag(s) in which the cell reaction is,, Cd(s) + 2AgCl(s)  2Ag(s) + Cd+2 (aq) + 2Cl–(aq) is 0.6915 V at 0°C and 0.6753 V at 25°C. The Hº, of the reaction at 25°C is :, (A) – 176 kJ, (B) – 234.7 kJ, (C) + 123.5 kJ, (D) – 167.26 kJ, , 32., , The potential of the Daniell cell, Zn, , 33., , Using the data in the preceding problem, calculate the equilibrium constant of the reaction at 25°C., , CuSO4, , Cu was reported by Buckbee, Surdzial and, (1 M), (1 M), Metz as E° = 1.1028 – 0.641 × 10–3 T + 0.72 × 10–5 T 2, where T is the temperature in degree celsius., Calculate S° for the cell reaction at 25°C :, (A) – 45.32 EU, (B) – 34.52 EU, (C) – 25.43 EU, (D) – 54.23 EU, , Zn + Cu2+, 24, , (A) 8.314 × 10, 34., , ZnSO4, , Zn2+ + Cu, K =, , [Zn2 ], [Cu2 ], , 31, , (B) 4.831 × 10, , (C) 8.314 × 1036, , (D) 4.831 × 1044, ,  dE ,  d( G) , G = H – TS and G = H + T , then  cell  is :, ,  dT ,  dT p, , (A), , S, nF, , (B), , nE, S, , (C) – nFEcell, , (D) + nEFcell, , 35., , One g equivalent of Na metal is formed from electrolysis of fused NaCl. No. of mole of Al from the fused, Na3AlF6 with the same current passed is :, (A) 1, (B) 3, (C) 1/3, (D) 2, , 36., , The specific conductivity of a saturated solution of AgCl is 3.40  10 6 ohm 1 cm 1 at 25 ºC. If,  Ag = 62.3 ohm 1 cm2 mol 1 &  Cl = 67.7 ohm 1 cm2 mol 1, the solubility of AgCl at 25 ºC is :, (A) 2.6  10 5 M, , 37., , (B) 4.5  10 3 M, , (C) 3.6  10 5 M, , List-1, (P) Conductivity does not change much then increases, (Q) Conductivity increases then does not change much, (R) Conductivity decreases then does not change much, (S) Conductivity decreases then increases, (T) Conductivity tends to zero at the end point, (P), (Q), (R), (S), (T), (A), 4, 2, 1, 5, 3, (B), (C), 5, 4, 3, 2, 1, (D), , (D) 3.6  10 3 M, , List-II, (1) NH3 is added in C6H5COOH, (2) CH3COOH is added in NaOH, (3) KOH is added in HCl, (4) Conc. KCl is added in dilute AgNO3, (5) MgSO4 is added in Ba(OH)2, (P), (Q), (R), (S), (T), 1, 2, 3, 4, 5, 4, 1, 2, 3, 5

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(A), , (B), , (C), end point, , end point, Volume of base added, , Volume of base added, , (D), , Conductance, , Conductance measurements can be used to detect the end point of acid-base titrations. Which of the, following plots correctly represent the end point of the titration of strong acid and a strong base ?, Conductance, , 49., , Conductance, , Pure water is saturated with pure solid AgCl, a silver rod is placed in the solution and the potential is, measured against normal calomel electrode at 25°C. This experiment is then repeated with a saturated, solution of AgI. If the difference in potential in the two cases is 0.177 V, what is the ratio of solubilty, product (Ksp) of AgCl and AgI at the temperature of the experiment ? (In both cases normal calomel, electrode is cathod), (A) 103, (B) 106, (C) 10–3, (D) 10–6, , Conductance, , 48., , end point, , end point, Volume of base added, , Volume of base added, , 50., , Which one of the following will increase the voltage of the cell ? (T = 298 K), Sn + 2Ag+  Sn2+ + 2Ag, (A) increase in the size of silver rod, (B) increase in the concentration of Sn+2 ions, +, (C) increase in the concentration of Ag ions, (D) none of the above, , 51., , In a H2 – O2 fuel cell, 6.72 L of hydrogen at NTP reacts in 15 minutes, the average current produced in, amperes is, (A) 64.3 amp, (B) 643.3 amp, (C) 6.43 amp, (D) 0.643 amp, , 52., , The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is, 0.79 V. The maximum amount of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution is, (A) 0.5 mmol, (B) 1.0 mmol, (C) 2.0 mmol, (D) 2.5 mmol, , 53., , A cell Cu | Cu++ || Ag+ | Ag initially contains 2M Ag+ and 2M Cu2+ ions in 1 L electrolyte. The change in, cell potential after the passage of 10 amp current for 4825 sec during usage of cell is:, 2.303RT, (Take, = 0.06), F, (A) – 0.009 V, (B) – 1.00738 V, (C) – 0.0038 V, (D) –1.2 V, , 54., , 55., ,  Eº , –3, –1, At 27ºC ,  = –1.45 × 10 V K and Eº = 1.36 V,  T  P, For the cell Pt | H2 (g)  HCl (aq) | Cl2 | Pt. Calculate entropy and enthalpy change in this standard state., (A) –962.48 JK–1, –346.435 KJ, (B) –279.85 JK–1, –346.453 KJ, –1, (C) –1326.23 JK , –346.435 KJ, (D) –280.24 KJK–1, –346.435 KJ., , If Ksp values of AgCl, AgBr & AgI at 298 K are 10–10, 10–13 & 10–17 respectively,, Compare Eo –, , Eo –, & Eo–, :, Cl / AgCl/ Ag, Br / AgBr / Ag, I / AgBr / Ag, o, o, (A) ECl– / AgCl/ Ag will have the least value and its value will be less than EAg / Ag, , o, o, (B) EI– / AgBr / Ag will have the least value and its value will be more than EAg / Ag, o, o, (C) ECl– / AgCl/ Ag will have the least value and its value will be more than EAg / Ag, , o, o, (D) EI– / AgBr / Ag will have the least value and its value will be less than EAg / Ag, , NUMERICAL VALUE QUESTIONS, 56., , Consider the cell Ag|AgBr(s)| Br– || Cl– | AgCl(s)| Ag at 25°C. The solubility product constants of AgBr &, AgCl are respectively 5 × 10–13 & 1 × 10–10 . For what ratio of the concentrations of Br– & Cl¯ ions would, the e.m.f. of the cell be zero ? Report as 1000  your answer.

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57., , A silver coulom meter is in series with a cell electrolyzing water. In a time of 1 minute at a constant, current, 1.08 g silver got deposited on the cathode of the coulometer. What total volume (in mL) of the, gases would have produced in other cell if in this cell the anodic and cathodic efficiencies were 90%, and 80% respectively. Assume STP conditions and the gases collected are dry. (Ag – 108) (Molar, volume of any ideal gas at STP = 22.4 L). Report as (your answer ÷ 10), , 58., , During electrolysis of CH3COONa(aq), the mole ratio of gases formed at anode and cathode is :, , 59., , Calculate the emf of the cell in mV, Ag (s), AgO3 (s)  Ag (xM), HO3 (1 M), If Ksp = 3 ×10–8 for AgO3 and Ka =, (log 3 = 0.48) (Take, , 60., , Zn2 (1 M) Zn(s), , 1,  2Ag+ + Zn is – 1.56 V., for HO3 and E0cell for 2Ag + Zn+2 , 6, , 2.303 RT, = 0.06) (Write magnitude of first two digits of your answer), F, , A saturated solution of MX is prepared KSP of MX is a × 10–b. If 10–7 mol of MNO3 are added in 1  of, this solution conductivity of this solution is 55×10–7 S m–1 :, º, º, º, –3, m,  = 6 × 10–3 S m2 mol –1 ;   = 8 × 10–3 ; , NO3– = 7 × 10, x, , Find the value of (a + b) ? Given that 10 < a < 100, 61., , Zn2 (aq)  4OH (aq) ,  Zn(OH)24 (aq), , Value of equilibrium constant (Kf) for above reaction is 10x then find x:, Given : Zn2 (aq)  2e  Zn(s); Eo  0.76V, o, Zn(OH)24 (aq)  2e  Zn(s)  4OH (aq); E  1.36V, , 2.303, 62., , RT,  0.06, F, , A cell reaction, Zn + 2Fe3+, , 2Fe2+ + Zn2+, works at 25ºC with the cell emf 1.2 volt and at 45ºC, , with the cell emf 1.718 volt. Assuming Sº to be constant in this temperature range, calculate Sº in, kJ/K. (Give your answer in the nearest integer)., , ONE OR MORE THAN ONE OPTIONS CORRECT TYPE, 63., , Which of the following statements is wrong about galvanic cells ?, (A) Cathode is the positive electrode, (B) Cathode is the negative electrode, (C) Electrons flow from cathode to anode in the external circuit, (D) Reduction occures at cathode, , 64., , When a cleaned strip of zinc metal is placed in a solution of CuSO 4, a spontaneous reaction occurs., Which of the following observation(s) is/are made ?, (A) the mass of zinc metal decreases gradually, (B) the copper metal starts depositing on either zinc plate or settles down to the vessel, (C) the solution remains electrically neutral, (D) the temperature of the solution decreases as it is an endothermic reaction., , 65., , Mark out the correct statement(s), (A) Copper metal cannot reduce iron (II) ions in acidic solutions., (B) Sodium can be obtained by the electrolysis of aqueous solution of NaCl using Pt electrodes., (C) The current carrying ions in an electrolytic cell are not necessarily discharged at the electrodes., (D) Cations having more negative oxidation potential than –0.828 V are reduced in preference to water.

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66., , 67., , When a lead storage battery is recharged, (A) PbSO4 is formed, (B) Pb is formed, , (C) SO2 is consumed, , (D) H2SO4 is formed, , Which of the following statements is / are correct ?, (A) The conductance of one cm3 (or 1 unit3) of a solution is called conductivity., (B) Specific conductance increases while molar conductivity decreases on progressive dilution., (C) The limiting equivalent conductivity of weak electrolyte cannot be determine exactly by, extraplotation of the plot of eq against, , c., , (D) The conductance of metals is due to the movement of free electrons., 68., , Peroxodisulphate salts (Na2S2O8) are strong oxidizing agents used as bleaching agents for fats, oil etc., Given, O2(g) + 4H(aq) + 4e–  2H2O(), Eº = 1.23 V, S2O8–2 + 2e–  2SO4–2 (aq), Eº = 2.01 V, Which of the following statements is (are) correct ?, (A) Oxygen gas can oxidize sulphate ion to per-oxo disulphate ion (S2O8–2) in acidic solution., (B) O2(g) is reduced to water, (C) Water is oxidised to O2, (D) S2O8–2 ions are reduced to SO4–2 ions., , 69., , 0.1 molar solution of NaBr solution is electrolysed by passing 965 column charge. After electrolysis, which statement is correct for resulting solution., (A) Specific conductance increases, (B) molar conductance increases, (C) No change in molar conductance., (D) Specific resistance increases., , 70., , A beaker contains a small amount of iron Fe(s). Which of the following aqueous solution, when added, to the beaker, would dissolve the iron i.e. convert Fe(s) to Fe2+ (aq) ?, Half cells, Eº at 25ºC, –0.76, Zn2+ + 2e–  Zn, –0.41, Fe2+ + 2e–  Fe, –1.66, Al3+ + 3e–  Al, 0.70, O2 + 2H+ + 2e–  H2O2, 1.23, Cr2 O2– + 6e– + H+  2Cr3+, 7, , O2 + 4H+ + 4e–  2H2O, (A) Cr2O72– (acidic solution), , 1.30, (B) H2O2 (acidic solution), , (C) Al3+, , (D) Zn2+, , COMPREHENSION, Comprehension # 1, , Given, , EoZn2 |Zn = – 0.76 V, , Kf [Cu(NH3)4]+2 = 4 × 1011, , EoCu2 |Cu, , = 0.34 V, , Answer the following.

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71., , The emf of cell at 200 k is [Given :, on temperature.], (A) 1.7 V, , (B) 1.08 V, , 2.303  R, F, , = 2  10-4 and assume that Eº values are independent, (C) 1.09 V, , (D) 1.10 V, , 72., , When 1 mole NH3 added to cathode compartment than emf of cell is (at 298K), (A) 0.81 V, (B) 1.91 V, (C) 1.1 V, (D) 0.72 V, , 73., , At what conc of Cu+2 emf of the cell will be zero (at 298K) and conc. of Zn+2 is remain same, (A) 1.19 × 10–37, (B) 1. 19 × 10–20, (C) 3.78 × 10–4, (D) 0.0068, , Comprehension # 2, Strong Acid Versus Strong Base, The principle of conductometric titrations is based on the fact that during the titration, one of the ions is, replaced by the other and invariably these two ions differ in the ionic conductivity with the result that the, conductivity of the solution varies during the course of the titration. Take, for example, the titration, between a strong acid, say HCl, and a strong base, say NaOH. Before NaOH is added, the, conductance of HCl solution has a high value due to the presence of highly mobile hydrogen ions. As, NaOH is added, H+ ions are replaced by relatively slower moving Na+ ions. Consequently, the, conductance of the solution decreases and this continues right upto the equivalence point where the, solution contains only NaCl. Beyond the equivalence point, if more of NaOH is added, then the solution, contains an excess of the fast moving OH– ions with the result that its conductance is increased and it, continues to increase as more and more of NaOH is added., If we plot the conductance value versus the amount of NaOH added, we get a curve of the type shown, in Fig., , The descending portion AB represents the conductances before the euivalence point (solution contains, a mixture of acid HCl and the salt NaCl) and the ascending portion CD represents the conductances, after the equivalence point (solution contains the salt NaCl and the excess of NaOH). The point E which, represents the minium conductance is due to the solution containing only NaCl with no free acid or, alkali and thus represents the equivalence point. This point can, however, be obtained by the, extrapolation of the lines AB and DC, and therefore, one is not very particular in locating this point, experimentally as it is in the case of ordinary acid-base titrations involving the acid-base indicators., Weak Acid versus Strong Base, Let us take the specific example of acetic acid being titrated against NaOH. Before the addition of alkali,, the solution shows poor conductance due to feeble ionization of acetic acid. Initially the addition of alkali, casuse not only the replacement of H+ by Na+ but also suppresses the dissociation of acetic acid due to, the common ion Ac– and thus the conductance of the solution decreases in the beginning. But very, soon the conductance starts increasing as addition of NaOH neutralizes the undissociated HAc to, Na+Ac– thus causing the replacement of non-conducting HAc with strong-conducting electrolyte Na+ Ac–, . The increase in conductance continunes right up to the equivalence point. Beyond this point, conductance increases more rapidly with the addition of NaOH due to the highly conducting OH– ions., The graph near the equivalence point is curved due to the hydrolysis of the salt NaAc. The actual, equivalence point can, as usual, be obtained by the extrapolation method., In all these graphs it has been assumed that the volume change due addition of solution from burrette is, negligible, hence volume change of the solution in beaker the conductance of which is measured is almost, constant throughout the measurement.

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(B), , (C), , volume of NaOH, , volume of NaOH, , volume of NaOH, , (A), , (B), , Conductance, , The most appropriate titration curve obtained when a mixture of a strong acid (say HCl) and a weak, acid (say CH3COOH) is titrated with a strong base(say NaOH) will be, , Conductance, , 75., , (D), , Conductance, , (A), , Conductance, , The nature of curve obtained for the titration between weak acid versus strong base as described in, the above passage will be :, Conductance, , 74., , V1, , V1 + V2, , (D), , V1, , V2, , volume of NaOH, , Conductance, , Conductance, , (C), , V2, , volume of NaOH, , volume of NaOH, , V1, , V2, , volume of NaOH, , PART - V : PRACTICE TEST-2 (IIT-JEE (ADVANCED Pattern)), Max. Time : 1 Hr., , Max. Marks : 66, , Important Instructions, A. General %, 1. The test is of 1 hour duration., 2. The Test Booklet consists of 22 questions. The maximum marks are 66., B. Question Paper Format, 3., Each part consists of five sections., 4., Section 1 contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out, of which ONE is correct., 5., Section 2 contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out, of which ONE OR MORE THAN ONE are correct., 6., Section 3 contains 6 questions. The answer to each of the questions is a numerical value, ranging from 0, to 9 (both inclusive)., 7., Section 4 contains 1 paragraphs each describing theory, experiment and data etc. 3 questions relate to, paragraph. Each question pertaining to a partcular passage should have only one correct answer among, the four given choices (A), (B), (C) and (D)., 8., Section 5 contains 1 multiple choice questions. Question has two lists (list-1 : P, Q, R and S; List-2 : 1, 2,, 3 and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY ONE, is correct., C. Marking Scheme, 9., For each question in Section 1, 4 and 5 you will be awarded 3 marks if you darken the bubble, corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one, (–1) mark will be awarded.

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10., , 11., , For each question in Section 2, you will be awarded 3 marks. If you darken all the bubble(s), corresponding to the correct answer(s) and zero mark. If no bubbles are darkened. No negative marks will, be answered for incorrect answer in this section., For each question in Section 3, you will be awarded 3 marks if you darken only the bubble corresponding, to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for, incorrect answer in this section., SECTION-1 : (Only One option correct Type), This section contains 10 multiple choice questions. Each questions has four choices (A), (B),, (C) and (D) out of which Only ONE option is correct., , 1., , An initial solution of x M, 1L Fe+2 was reduced to Fe(s) on passage of 1 A current for 965 seconds. If, after electrolysis 0.1M, 10 ml acidified KMnO4 solution was required to oxidize remaining Fe+2 solution, then the value of ‘x’ is (A) 10–2, (B) 10–3, (C) 5 × 10–3, (D) 5 × 10–2, , 2., , A solution of 100 mL, 0.2 M CH3COOH is mixed with 100 mL, 0.2 M NaOH solution. The molar, conductance for 0.1 M CH3COOH at infinite dilution is 200 S cm 2 mol–1 and at any concentration is 2.0, S cm2 mol–1. Then calculate pH of the solution?, (A) 7, (B) 8, (C) 5, (D) 9, , 3., , The specific conductance of saturated solution of silver bromide is K (–1 cm–1). The limiting ionic, conduction of Ag+ and Br– ions are a & b respectively. The solubility of AgBr in g lit –1 is:, (Molar mass of AgBr = 188g mol–1), 1000, K, K  1000  188, a  b 1000, (A) K ×, (B), × 188, (C), (D), ×, ab, ab, ab, K, 188, , 4., , The conductance ratio, , , = 0.936 given this for a certain solution of KCl and  = 122 –1cm2 eq–1 and, º, , º, , 0.98, =, . Calculate the limiting values of Ionic conductance of K and Cl– ions in –1cm2 eq–1., º, 1.98, (A) 64.51, 65.83, (B) 74.60, 26.40, (C) 30.31, 69.69, (D) 70.12, 29.88, 5., , Osmotic pressure of 0.1 M weak acid HA is 3 atm. If molar conductance of 0.1 M HA is 30–1cm2mol–1., than molar conductance at infinite dilution is :, (A) 150 –1cm2 mole–1, (B) 300 –1cm2 mole–1, –1, 2, –1, (C) 100  cm mole, (D) 200 –1cm2 mole–1, , 6., , The molar conductivity of 0.05 M solution of MgCl2 in a cell with electrodes of 1.5 cm2 surface area and, 0.5cm apart and 0.15 amphere current flow when a potential difference of 5 volt is applied between two, electrodes –, (A) 200–1 cm2 mol–1, (B) 195.6 –1 cm2 mol–1, –1, 2, –1, (C) 149.8 cm mol, (D) 169.5 –1 cm2 mol–1, , 7., , 5 litre solution of 0.4 M CuSO4(aq) is electrolyzed using Pt electrode. A current of 482.5 ampere is, passed for 4 minutes. The concentration of CuSO4 left in solution is (Assume volume of solution to be, remained unchanged):, (A) 0.16 M, (B) 0.28 M, (C) 0.34 M, (D) 0.40 M, , 8., , For a concentration cell :, Pt | Ag (s) | Ag+ (aq., C1) || Ag+ (aq., C2) | Ag (s) | Pt, C2, EMF of the cell is X volt then calculate the ratio of, ?, C1,  x , (A) anti log  0.059 , , , , x, , ,  4x , (B) anti log , (C) anti log , , ,  2  0.059 ,  0.059 , , (D) None of these

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9., , A saturated solution of Fe(OH)3 is present in a solution of pH = 12, what is the reduction potential of, 0, 2.303  RT, Fe3+/Fe in solution ( EFe3  /Fe = –0.036V, Ksp of Fe(OH)3 = 10–26), [, = 0.06], F, (A) –0.436V, (B) 0.39V, (C) + 0.36V, (D) – 1.2 V, , 10., , Under which of the following condition direction of flow of current will be opposite i.e. from Zn electrode, 2.303  RT, to Cu electrode at 298 K : [Given :, = 0.06] ; Eºcell for Zn|Zn2+||Cu2+|Cu = 1.1 V, F, (A) [Zn2+ ] > e84.4 [Cu2+], (B) [Zn2+] < e84.4 [Cu2+], 2+, 84.4, 2+, (C) [Zn ] = e, [Cu ], (D) [Cu2+] = e84.4 [Zn2+], Section-2 : (One or More than one options correct Type), This section contains 6 multipole choice questions. Each questions has four choices (A), (B),, (C) and (D) out of which ONE or MORE THAN ONE are correct., , 11., , Two test tubes  &  contain solutions of sodium salts of halide in water. When Br 2 was added to both, the solutions then following observations were noted., Test Tube, Observation, Violet vapous emerged, , No reaction occurred, , If halides in the tubes I & II are X– and Y– (and their molecular forms being X2 & Y2 respectively) then, the true options would be :, (A) SRP of Br2 is more than the SRP of X2, (B) SRP of Br2 is more than the SRP of Y2, (C) Y2 can oxidize X– into X2, (D) Y2 can oxidize Br– into Br2., , 12., , In the concentration cell, HA, HA, Pt H2 (g), H2(g) Pt, NaA NaA, Value of cell potential will depend on –, (A) Value of pKa of HA, (C) Concentration of HA in two electrodes, , (B) Temperature, (D) Concentration of NaA in two electrodes, , 13., , 20 millimolar solution of aq. CuSO4 (500 ml) is electrolysed with sufficient amount and a total of 0.04, faraday of electricity is supplied. Then :, (A) Total volume of gases evolved at STP = 224 ml, (B) Total volume of gases evolved at STP = 448 ml, (C) Total volume of gases evolved at STP = 672 ml, (D) Resulting solution after electrolysis becomes acidic, , 14., , Emf of cell Ag|Ag+ (saturated solution of Ag2CrO4) || Ag+(0.1 M) | Ag is 0.164 volt at 298 K. Then, (A) Ksp of Ag2CrO4 in water is nearly 2.3 × 10–12, (B) Given cell is a concentration cell, (C) Ksp of Ag2CrO4 can’t be determined by given data., (D) Concentration of Ag+ ion in anode compartment when EMF is 0.164 volt is nearly 1.66 × 10–4 M, , 15., , EºMg2  / Mg  2.4V , EºSn4  / Sn2  0.1 V , EºMnO, , –, 4, , , H / Mn2 , ,  1.5, , V , Eº , , 2, , / –, ,  0.5, , V, , Here,, (A) MnO4– is the strongest Oxidizing Agent and Mg is the strongest Reducing Agent., (B) Sn4+ + 2–  Sn2+ + 2 is a nonspontaneous reaction., (C) Mg2+ + Sn2+  Mg + Sn4+ is a spontaneous reaction., (D) Here, Weakest oxidizing agent is Sn4+ and weakest reducing agent is Mn2+

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16., , The following diagram shows an electrochemical cell in which the respective half cells contain aqueous, 1.0 M solutions of the salts XCl 2 and YCl3. Given that:, 3X(s) + 2Y3+(aq)  3X2+(aq) + 2Y(s) Ecell > 0, , Which of the following statements is correct?, (A) The electrode made from metal X has positive polarity., (B) Electrode Y is the anode, (C) The flow of electrons is from Y to X, (D) The reaction at electrode X is an oxidation, Section-3 : (Numerical Value Questions), This section contains 6 questions. Each question, when worked out will result in a numerical, value from 0 to 9 (both inclusive), 17., , By how many of the following actions, can the Ecell be increased (S = + ve) for the cell reaction, A | A+ (aq) || Cl– | Cl2(g) | Pt, (a) By dilution of anodic solution., (b) By dilution of cathodic soltuion., (c) By decreasing temperature., (d) By increasing pressure of Cl2 in cathodic compartment., (e) By increasing the mass of anode (A(s)), (f) By increasing temperature, , 18., , At infinite dilution the molar conductance for CH3COONa is 150 S cm 2 mol–1, for HCl is 200 S cm 2 mol–1, and for NaCl is 125 S cm2 mol–1. Then calculate pH of 0.001 M CH3COOH ?, (Given : Molar conductance of CH3COOH at 0.001 M concentration is 2.25 S cm 2 mol–1)., , 19., , The conductivity of an aqueous solution of a weak monoprotic acid is 0.000032 ohm –1cm–1 at a, concentration, 0.2 M. If at this concentration the degree of dissociation is 0.02, calculate the value of 0, (ohm–1 cm2 /eqt)., , 20., , Pt, H2(g) | 2 M CH3COONH4(aq) || 2 M NaCl(aq) | H2(g), Pt, 20 atm, 0.2 atm, Given pKa(CH3COOH) = 4.74 pKb = (NH4OH) = 4.74, If E is emf of the cell in volt, calculate 1000 E., , 21., , [Take :, , 2.303 RT, = 0.059], F, , EMF of the following cell is 0.634 volt at 298 K Pt | H2 (1 atm) | H+ (aq) || Hg22+(aq.,1N) | Hg(). The pH, of anode compartment is :, 0, Given EHg, = 0.28 V and, 2, |Hg, 2, , 2.303 RT, = 0.059, F

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SECTION-5 : Matching List Type (Only One options correct), This section contains 1 questions, each having two matching lists. Choices for the correct, combination of elements from List-I and List-II are given as options (A), (B), (C) and (D) out of, which one is correct, 28., List-I, (P), , Molar conductivity, , (1), , List-II, Conductivity, Molarity, , (Q), , Conductivity, , (2), , Conductivity, Limiting molar conductivity, , (R), , Degree of dissociation, , (3), , (S), , Solubility of sparingly soluble salt, , (4), , Molar conductivity, Limiting molar conductivity, Decreases with dilution, , Codes :, P, 4, 1, , (A), (C), , Q, 1, 2, , R, 2, 3, , S, 3, 4, , (B), (D), , P, 2, 1, , Q, 3, 4, , R, 4, 3, , S, 1, 2, , Practice Test-2 (IIT-JEE (ADVANCED Pattern)), OBJECTIVE RESPONSE SHEET (ORS), Que., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , 25, , 26, , 27, , 28, , Ans., Que., Ans., Que., Ans.

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PART - I, 1., , (1), , 2., , (4), , 3., , (3), , 4., , (2), , 5., , (4), , 6., , (3), , 7., , (3), , 8., , (3), , 9., , (2), , 10., , (3), , 11., , (1), , 12., , (4), , 13., , (4), , 14., , (2), , 15., , (1), , 16., , (2), , 17., , (1), , 18., , (3), , 19., , (2), , 20., , (3), , 21., , 6, , 22., , 3 (I, II, III), , 23., , 2, , 24., , 13, , 25., , 50, , PART - II, 1., , (2), , 2., , (1), , 3., , (3), , 4., , (4), , 5., , (1), , 6., , (3), , 7., , (1), , 8., , (3), , 9., , (1), , 10., , (1), , 11., , (4), , 12., , (4), , 13., , (1), , 14., , (3), , 15., , (1), , 16., , (4), , 17., , (2), , 18., , (3), , 19., , (3), , 20., , (4), , 21., , (4), , 22., , (1), , 23., , (3), , 24., , (2), , 25., , (1), , 26., , (2), , 27., , (3), , 28., , (4), , 29., , (1), , PART - III, 1., , (B), , 2., , (B), , 3., , (C), , 4., , (C), , 5., , (A), , 6., , (C), , 7., , (B), , 8., , (C), , 9., , (C), , 10., , (D), , 11., , (B), , 12., , (A), , 13., , (C), , 14., , (B), , 15., , (A), , 16., , (C), , 17., , (B), , 18., , (D), , 19., , (B), , 20., , (D), , 21., , (A), , 22., , (B), , 23., , (D), , 24., , (A), , 25., , (C), , 26., , (B), , 27., , (D), , 28., , (A), , 29., , (C), , 30., , (C), , 31., , (A), , 32., , (C), , 33., , (B), , 34., , (B), , 35., , (B), , 36., , (B), , 37., , (A), , 38., , (A), , 39., , (B), , 40., , (A), , 41., , (B), , 42., , (C), , 43., , (B), , 44., , (B), , 45., , (B), , 46., , (D), , 47., , (D), , 48., , (B), , 49., , (), , 50., , (B), , 51., , (B), , 52., , (A), , 53., , (C), , 54., , (B), , 55., , (C), , 56., , (C), , 57., , (C), , 58., , (B), , 59., , (B), , 60., , (C), , 61., , (C), , 62., , (C), , 63., , (D), , 64., , (B), , 65., , (A), , 66., , (A), , 67., , (C), , 68., , (A), , 69., , (C), , 70., , (B), , 71., , (B), , 72., , (A)

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PART - IV, 1., , (a) 0.34 < E° < 0.8 ; (b) –0.44 < E° < 0, , 2., , –0.8825 volt, , 3., , 0.836 volt, 1.1937 volt, , 4., , 0.936V, , 5., , n = 3, Since mass of Sn increasing, Sn - electrode is working as cathode and X - metal electrode anode, and electrons are flowing from X-electrode to Sn-electrode in the external circuit., 7., , E = 0.395 V, , 9., , –0.042 V, , 10., , KSP = 10–5, , 13., , 3.389 × 10–4 volt deg–1, , 11. –0.16V, , 6., , E = 0.059 V, , 12., , G = – 196.5 kJ ; H = 198.8 kJ ; S = – 7.72 J deg–1, , 14., , Sº = – 24.125 kJ K–1 ;, , 15., , Sº = – 30.88 JK–1, , 16., , A = 114, Q = 5926.8C., , 17., , V = 1.763 L, , 19., , [Cu2+] = 10–4 M., , 20., , 2 × 10–16 mole / litre., , 21., , (a) Eº = 0.59 V, log10Ksp = – 10 ; (b) 52.88, 10–6 mole., , 22., , (A), , 23., , (C), , 24., , (A), , 25., , (A), , 26., , (C), , 27., , (B), , 28., , (D), , 29., , (D), , 30., , (D), , 31., , (D), , 32., , (D), , 33., , (C), , 34., , (A), , 35., , (C), , 36., , (A), , 37., , (D), , 38., , (B), , 39., , (B), , 40., , (B), , 41., , (A), , 42., , (A), , 43., , (C), , 44., , (A), , 45., , (B), , 46., , (A), , 47., , (B), , 48., , (B), , 49., , (A), , 50., , (C), , 51., , (A), , 52., , (B), , 53., , (A), , 54., , (B), , 55., , (D), , 56., , 5, , 57., , 14, , 58., , 3, , 59., , 11, , 60., , 26, , 61., , 20, , 62., , 5, , 63., , (BC), , 64., , (ABC), , 65., , (ACD), , 66., , (BD), , 71., , (B), , Gº = – 7179.6 J ;, Hº = – 77.23 kJ, , Hº = – 7196.43 kJ, , Gº = – 68.03 kJ, 18., , 43.456 g, , 67., , (ACD), , 68., , (CD), , 69., , (AB), , 70., , (AB), , 72., , (A), , 73., , (A), , 74., , (A), , 75., , (C), , PART - V, 1., , (A), , 2., , (D), , 3., , (C), , 4., , (A), , 5., , (A), , 6., , (A), , 7., , (B), , 8., , (A), , 9., , (A), , 10., , (A), , 11., , (ACD), , 12., , (BCD), , 13., , (BD), , 14., , (AD), , 15., , (AB), , 16., , (D), , 17., , 4 (a, b, d, f), , 18., , 5, , 19., , 8, , 20., , 59, , 21., , 6, , 22., , (B), , 23., , (C), , 24., , (A), , 25., , (B), , 26., , (C), , 27., , (A), , 28., , (D)

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PART - I, 1., , 0, Ecell  E0Sn4  / Sn2  + EFe, 2, / Fe3 , , 2., , 2Cu+1  Cu + Cu+2, 2Cu+1 + 2e 2Cu, Cu – 2e Cu+2, ––––––––––––––––––––––, 2Cu+1 Cu+2 + Cu, ––––––––––––––––––––––, 2  0.521  2( 0.337), , E° =, = 0.184, 2, WAg, W1 W2, 4, =, ;, ; W Ag = 36, , E1 E2, 12 108, , 3., 4., , E10 = 0.15 v G10 = - n E10 F, , Cu+ + 1e–  Cu, , E02 = 0.50 v G02 = – n E02 F, , 1, , 2, , Cu + 2e  Cu, , Gº = Gº + Gº, , –, , 1, , (–1) n Eº F = (–1) n, , 1, ,  n2E02, , E10, , 2, , F+ (–1) n E02 F, 2, , 0.15  1  0.50  1, = 0.325V, 2, 1, 1, .0591, H+ + e–  H2., E=0–, log10  = + 0.0591 log10[H+]., 2, 1, [H ], Eº =, , 6., , 0.15 — 0.77 = – 0.62 V, , Cu2+ + 1e–  Cu+, , 2+, , 5., , , , n1E10, , n, , =, , E1 = 0 {pH = 0}., E2 = + 0.0591log10[10–7] = – .0591 × 7 {at pH = 7} = – 0.41 V., 0.0591, 0.0591, Ecell = Eºcell –, log10[H+] [Cl–], and, E’cell = Eºcell –, log10 100[H+] [Cl–]., 1, 1, , E'cell – Ecell = – 2 × 0.0591 = – 0.1182., 8., 9., , Cell notation is anode || cathode., 0.06, 0.34 =, log Keq, 2, log Keq = 11.3 or Keq = 2 × 1011, , 10., , Number of moles of Cu2+ produced from anode = number of moles of Cu2+ deposited at cathode., , 11., , For same charge passed mole of H2 produced = 2 × moles of O2 produced., , 13., , W, it, , , E 96500, , , , 3 9.95  10  60,  E = 48.5, , E, 96500, , 1 , 1  ,  0.0112 =, , R  a , 55  a , , 14., , K=, , 15., , 1, , –3,   G *   10, eq =  R, , N, , 16., , m (NaCl),  m Na , m(Cl–) keep on increasing as concentration decreases but  keeps on, ,  , , decreasing with dilution., , a, , = 0.616, ,  1, ,  G *   10 –3, ,  10–2 =  50, , 1/10, ,  G* = 50 m–1

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1, ×, , = 1.4 × 50 m–1., R, A, A, Now, new soltuion has M = 0.5, R = 280 , K=, , 1, K, 1, 1, 1, 1, 4, , ,  5  104, K=, ×, =,  M =,  1.4  50 , 1000  M, 1000 0.5, 2000, 4, R, A 280, , 23., , c     B C, , 24., , Reason: Higher the position of element in the electrochemical series, more difficult is the reduction of, its cations., If Ca2+ (aq) is electrolysed, water is reduced in preference to it. Hence it cannot be reduced, electrolytically from an aqueous solutions., , 25., , (Debye Huckel onsagn equation), , E0 1.51V, , E0 1.18V, , 1, 2, Mn2 ,  Mn2+ , ,  Mn, 2+, 0,  for Mn disproportionation, E = – 1.51 V –1.18 V, = – 2.69 V < 0, Reaction is non-spontaneous., , 27., ,  Cu, Cu2+ + 2e– , 2F, 1 mole, = 63.5 g., Galvanization is applying a coating of Zn., , 28., , For strongest reducing agent EOP should be maximum., , 26., , EOP Cr / Cr 3 = 0.74 V, Whereas,, , EOP Mn2 /MnO  = –1.51 V, , , , 4, , 29., , EOP Cr3  / Cr O2 = –1.33 V, 2, , , , 7, , B2H6 + 3O2 B2O3 + 3H2O, 1 mol 3 mol, 3 mol O2 is required for Burning 1 mol B2H6, 1, Electroly sis, H2O , (V.F. of O2 = 4),   H2  O2, 2, Equivalent of O2,  mol of O2 =, V.F.of O2, ,  (100A)  t sec.  1 = 3, ,  4, 96500, , , , 3, , , t = 3  96500  4 hr. = 3.22hrs., 100  3600, , PART - IV, 1., , (a) Metal should below hydrogen and Cu2+ but should above Ag+ in series., (b) Metal should above hydrogen but should below from Zn2+ and Fe2+ both., , 2., , TiO2+ + 2H+ + e–  Ti3+ + H2O , 0.1 V G1º = – 2 × F × 0.1, Ti3+ + 3e–  Ti – 1.21 V, G2º = – 3 × (– 1.21) × F, TiO2+ + 2H+ + 4e–  Ti + H2O, – 4 × Eº × F = – 1 × 0.1 × F + – 3 × (– 1.21) × F, 0.1– 3.63, Eº =, = – 0.8825 volt., 4, , EOP Cl– / Cl = –1.36 V, 2

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3., , 4., , 0.0591, log 92, 2, Ereduction = –Eoxidation = 0.836 volt, In neutral medium,, 0.0591, EOxidation = – 0.78 –, log (10–7)2, 2, Eoxidation = – 0.78 –, , 0.0591, × 2 × log 9 = –0.836 volt, 2, , = –1.1937 volt, , Cr2O72– + 14H+ + 6e–  2 Cr3+ + 7 H2O, 1.33 volt, E = 1.33 –, = 1.33 –, , 5., , = – 0.78 –, , 0.0591, 0.0591, (0.01)2, log, = 1.33 –, log 10–2 × 1042, 3 14, 6, 6, (0.01)  (10. ), 0.0591, 0.0591, × log 1040 = 1.33 –, × 40 = 0.936 volt, 6, 6, , 0.0591, 0.0591, 1 , , , , log (0.1) + 0  0.14 –, log, 0.65 = Eoxid + Ered = 0.78 –, , n, 2, 0.5 , , , , , 0.01 = –, , 0.0591, 0.0591,  1 0.301, × (– 1) –, × 0.301= 0.0591  –, n=3, n, 2 , n, 2, , Electrons flow from X electrode to Zn electrode., 6., , Pt / H2O / H+ (HA) // H+ (HB) / H2 / Pt, H2  2H+ + 2e–, 2H+ +2e–  H2, 2H+(HB)  2H+HA, Eº = 0, E=0–, , [H ]2, 0.0591, log  HA, 2, 2, [H ]HB, , But Ka =, , [H ]2, C, , 10 3  C, 0.0591, log, = – 0.0591, 2, 10 –5  C, the cell is constructed in reversed direction., Ecell = 0.0591 volt., , =–, , 7., , Pt / H2 / H+ (C6H5NH2(C) // H+(HCI) / H2 / Pt, H2  2H+(10–8M) + 2e– ,, C6H5NH2 + H2O  C6H5NH3+ + OH–, 2H+ + 2e–  H2, , Kb =, , (OH– )2, 5 10 4, 0.5, , 2H+(5×10–2)  2H+(10–8), E=0–, , 0.0591, 0.0591, 0.0591, (108 )2, log, =–, log 10–14 × 4 =, . [log4 – 14] = 0.396 volt, 2, 2, 2, (5  102 )2, , 8., , (A) Hg2Cl2(s) + Cu(s) Cu2+(aq) + 2Cl– (aq) + 2Hg(l), (B) 2Ag(s) + 2IO3– + Zn2+ 2AgIO3(g) + Zn(s), (C) Mn(s) + 2OH– + Cu2+ Mn(OH)2(s) + Cu(s), , 9., , E=0–, , 10., , E0cell =, , 11., , E0Cell = 0.06(–log KSP), , 10 –10 / 0.2, .06, log, = –0.042 V, 1, 10 –13 /10 –3, , 1, .06, log, K, 2, SP, , , , – 0.12 + 0.27 = .03 log, , , , 0.8 – EI0– / AgI / Ag = 0.96, , 1, K SP, , , , KSP = 10–5, , , , EI0– / AgI/ Ag = 0.16V

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24., , 25., , MnO4– + 8H+ + 5e–  Mn2+ + 4H2O, , E1 = E° –, , [Mn2 ], 0.0591, log, 5, [MnO4– ]  18, , E2 = E° –, , [Mn2 ], 0.0591, log, 5, [MnO4– ]  10 –4, , 0 = (–0.771 + 0.7991) –, log x =–, , 26., , , , 0.0281, 0.0591, , 0.0591, 1, log, 1, x, , , 0 = (– 0.151 – 0) –, , E0, , Ag | Ag I |I –, , = E0Ag |Ag –, , – 0.151 = 0.799 –, , =–, , 0.0591, × 32 = –0.37824, 5, , , , E1–E2 = 0.38 Volt., , 0 = 0.0281 +0.0591 log x, , x = 0.335 M, , 0.0591, log [H+]., 1, , 0.0591 × log[H+] = – 0.151, , 27., , , , 8, , ;, , pH =, , 0.151, = 2.56, 0.0591, , 1, 0.0591, log, K SP, 1, , 1, 0.0591, log, K sp, 1, , 0.0591 log Ksp = –0.151 –0.799, log Ksp= – 16.074, Ksp = 8.43 × 10–17, 28., , EI– / AgI / Ag = 0.8 –, , 1, 0.0591, log, K sp, 1, , = 0.8 + 0.0591 × log 8.3 × 10–17 = – 0.15 Volt, or, 1, 8.31  298  2.303, Eº Ag / Ag /  –  Eº, = – 0.8 –, log ksp, RT . ln, Ag / Ag , k, 96500, sp, F, = – 0.8 – 0.0591. log 8.3 × 10–17 = – 0.8 + 0.095 = 0.15 V., 29., , 2, 2, A(s) + Baq.,  A aq., + B(s) , H° = – 285 KJ, , Assuming s to negligible , G° = H° = –285×103×0.84, E° = 1.24 Volt, 30., , 31., , = –2×E°×96500, , d, = –0.00065 Vol deg–1, dt, dE, S298 = n.F., = 2×96500× (–0.00065) = – 125.5 J/K., dT, , dE (0.6753 – 0.6915), =, = –6.48×10–4 V deg–1, (25 – 0), dT, dE, = – 2 × 0.6753 × 96500 + 2 × 96500 × 298 × (–6.48×10–4), dT, = 2 × 96500 (– 0.6753 – 0.1931) = – 167.6 KJ., H298 = – neF + nFT

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32., , E° = 1.1028 – 0.641 × 10–3 T + 0.72 × 10–5 T2,  dEº , –3, –5,  dT  2S = –0.641 × 10 + 2 × 0.72 × 10 T, , S° = nF, 33., , = (–0.641 + 0.36) × 10–3 = –0.281 × 10–3, , dEº, = 2 × 96500 × (–281 × 10–3) = – 54.23 EU, dT, , E298 = 1.1028 – 0.641 × 10–3 × 25 + 0.72 × 10–5 × (25)2 = 1.091275 Volt, , = –8.314 × 298 ln K= – 2 × 1.091275 × 96500, K = 1036.91 = 8.128 × 1036., 34., ,  d, ,  dE , S = –  ( G) = + nF ,  dT  p,  dT, p, , S,  dE ,  dT  = nF, p, 35., , 36., , Na+ + e–  Na(s), 1mole 1 Faraday, Al3+ + 3e– Al(s), 1 Faraday, 1, No. of mole of Al = mole., 3,  Ag = 62.3 Scm2 mol–1 ,  Cl = 67.7 Scm2 mole–1, , KAgcl = 3.4 × 10–6 Scm–1,  AgCl = (62.3 + 67.5) =, , S=, , 1000  3.4  10 –6, S, , 3.4  10 –3, = 2.6 × 10–5 M, (62.3  67.5), , 38., , Higher the std. reduction potential, higher is the oxidising power., , 39., , Ecell = 0.29 –, , 40., , Eºcell = 1.89 ; EºCe4+/Ce3+ + EºCo/Co2+ = E + 0.277  E = 1.62 V, , 41., , EºMnO / MnO =, , 42., , S =, , 43., , – 0.413 = 0 – 0 .059 log, , 44., , Z > Y > X (Non metals like F2 > Cl2 > Br2), So, Y will oxidise X– but not Z–, Z will oxidise both X– and Y–, X can't oxidise Y– or Z– ., , 45., , Eºcell = 0.8 – (– 0.76) = 1.56 V, , 4, , 2, , 0.059, 0.01  (0.01)2, log, 2, (0.01)2  1, , or, , Ecell = 0.35 volt, , 5  1.5 – 2  1.23, = 1.7 volt, 3, , nFd Ecell, d t, , d Ecell,  S, =, d t, n F, , or, 1, , , [H ], , or, , 0.414, = – log H+ = pH, 0.059, , or, , pH = 7

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0.059, 1.5, log, = 0.652 V, 1, 0.015, , 46., , Ecell = 0.77 –, , 47., , 1.1591 = 1.1 –, , 48., , Ag, Ag+ + e–, E1 = Eoxid + Ecalomel, 0.0591, = E' –, log Ksp1 + Ecalomel, 1, E2 = E' –, , K sp1, K sp2, , 0.059, [Zn2 ], log, 2, [Cu2 ], , 0.0591, log, 1, , Ksp2 + Ecalomel, , or, , , , [Zn2 ], [Cu2 ], , = 10–2 = 0.01, , E2 – E1 = 0.177 = 0.0591 log, , K sp1, K sp2, , = 106., , 49., , H+ + CI– + NaOH  Na+ + CI– + H2O to conductance Ist decreases since no. of ions decreases after, end point it inceases., , 50., , Ecell = Eºcell –, , 51., , i  15  60, 6.72, =, ×2 , 96500, 22.4, , 52., , 0.059, [Sn2 ], log, 2, [Ag ]2, , Ag+ increase, Ecell, , i = 64.3 amp., , AgCl + e–  Ag + Cl–, E° = 0.2 V, Ag  Ag+ + e–, E° = – 0.79 V, ––––––––––––––––––––––––––––––––––––––––, e, , , , , Ag+ + Cl–, , E° = – 0.59 V, 0.059, 0.059, E° =, log K, , – 0.59 =, log KSP, n, 1, Now solubility of AgCl in 0.1 M AgNO3, S (S + 0.1) = 10–10, , S = 10–9 mol/L, 9, Hence 1 mole dissolves in 10 L solution, hence in 106 L amount that dissolves in 1 m mol., AgCl, , 53., , increase., , Q = 10 × 4825, , = 48250 C, 48250, no. of faraday =, = 0.5, 96500, 1, Ag, +, Cu++ , 2, 2.00, 2–0.25, Ecell = EºCell –, , Ag+, , +, , 2.00, 2 + 0.50, , 0.0591, [Ag ], log, 1, [Cu ]1/ 2, , E1 = EºCell –, , 2.00, 0.0591, log, 1, (2.00)1/ 2, , E2 = EºCell –, , 2.50, 0.0591, log, 1, (1.75)1/ 2, , 1, Cu, 2, , , , KSP = 10–10

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E = E2 – E1 =, , 54., , 0.0591 , 2.50  0.0591, [log 1.41 – log 1.88], log 2  log, =, 1, 1 , 1.75 , 0.0591, 0.0591, =, [0.1492 – 0.2742] = –, × 0.125 = – 0.00738 V., 1, 1, ,  Eº , –3, –1, nF ,  = Sº = –2 × 96500 × 1.45 × 10 = –279.85 JK,  T , , Gº = –nFEº = –2 × 1.36 × 96500 = –262.48 KJ., Hº = Gº + TSº, = –262.48 × 103 – 300 × 279.85, = –262480 – 83955 = –346.435 KJ, 55., , EoX– / AgX / Ag = EoAg / Ag + 0.059 log10 Ksp (AgX), 1, o, Thus, salt having least value of Ksp will have least value of EX– / AgX / Ag and all values will be less than, , EoAg / Ag (since 2nd term will always have a negative values)., 56., , If cell is taken to be conc cell, Eºcell = 0, Anode :, Ag Ag+a + e–, Cathode :, Ag+c + e–  Ag, –––––––––––––––––––, Agc+, Aga+, From Nernst eq,, Ecell = Eºcell –, , [Ag ]a, 0.059, log, 1, [Ag ]c, ,  [Ag+]a = [Ag+]c , 57., , K sp of AgBr, , , [Br ], , , , =, , K sp of AgCI, , , [CI ], , 0=0, or ,, , –, , 5  10 13, 10, , 10, , [Ag ]a, 0.059, log, 1, [Ag ]c, , =, , [Br  ], , , [CI ], , =, , [Br  ], , , [CI ], , =, , 1, ., 200, , Charge passed = 0.01 Faraday, 1, , , At the anode  H2O  O2  2H  2e   with 90 % efficientcy 0.01 × 0.9 F have been used and will, , , 2, , 1, × 0.01 × 0.9 mole of O2 i.e. 0.00225 mol O2 ., 4, 2e,  H2 + 2OH–, At the cathode 2H2O , 0.01  0.8, moles of H2 produced =, mol = 0.004 mol, 2, Total moles produced of gases = 0.004 + 0.00225 = 0.00625 mol, vol. at STP = 0.00625 × 22400 mL = 140 mL, produce, , 59., , Ka =, , , , C 2, 1 , , , , 2, 1, =, 1 , 6, , 1  (1)2  4  6  1, 1  1  24 1, =, =, 12, 3, 12, 1, 1, [O3–] = 1 ×, =, M, 3, 3, =

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, , [Ag+] =, , 3  10 8, = 9 × 10–8 M, 1, 3, , , 2e, Now 2Ag + Zn+2 ,  2Ag+ + Zn, , E = – 1.56 +, , Gives:, , 1, 0.06, log, = – 1.1376 V, 2, (9  10 8 )2, , Ans. 11, 60., , MX, , M, , +, , + X–, , a + 10–7, a, KSP = (a + 10–7) a, 55  10 7 = (6 × 10–3 (a + 10–7) + 8 × 10–3 a + 7 × 10–3 × 10–7), 1000, 55 × 10–10 = 6 × 10–3 a + 6 × 10–10 + 8 × 10–3 a + 7 × 10–10, 42 × 10–10 = 14 × 10–3 a, a = 3 x 10–7, KSP = 12 × 10–14, , 61., , o, o, Eocell  ERP(RHS),  ERP(LHS), , = – 0.76 – (– 1.36) = 0.6, r G  RT ln K eq ;, o, , or log K eq , , nFEo, 2  0.6, =, = 20, RT  2.303, 0.06, , , , 2  0.6,  20; K f  1020, 0.06, , dE, 1.718  1.2, = 2 × 96500 ×, = 5000 Joule/K., dt, 20, , 62., , Sº = nF, , 63., , Reduction and electronation take place at cathode elctrode, so it become positive electrode., , 64., , (A, B, C) Reduction Potential of Ce is higher than that of Zn., , 65., , º, º, (A) because ECu, > EFe, ., 2, 2, / Cu, / Fe, , 66., , Recharging reaction:, , 67., , On dilution specific conductance decreases while molar conductivity increases., , 68., , Create a cell with required cell reaction, ,  Pb(s) + PbO2(s) + 2H2SO4 (aq), 2PbSO4 (s) + 2H2O. , , O2 + SO 4–2  S2O8–2 + H2O, , Eocell = 1.23 – 2.01 < 0, ,  Nonspontaneous cell reaction, 69., , Electrolysis of NaBr Solution, At Anode, 2Br–  Br2 + 2e–, –, , –, 2e, 2H2O ,  H2 + 2OH, It is clear that Br– ion are replaced by OH–., Hence molar conductance & specific conductance increases., , at cathode

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70., , (A) For Cr2O72– (acidic solution), o, Eº = 1.23 which is greater than E(Fe2 /Fe) hence it can oxidize Fe, , (B) H2O2  2H+ + O2 + 2e–, O2 + 4H+ + 4e–  2H2O, , Eº = –0.70 V, Eº = 1.30 V, (–0.70  2)  (4  1.30), Eº =, = 1.9, 2, , H2O2 + 2H+  2H2O, , o, Here Eº is grater than E(Fe2 /Fe) hence H2O2 in acidic medium can oxides Fe., , 71., , Ecell = Eºcell –, , 72., , Cu+2 + 4 NH3, 0.2, 1, x, 1–0.8, , 2.303 RT, 2.303  8.31  200, 2, (Zn2 ), log, = 0.76 + 0.34 –, log, = 1.08 volt., , 2, nF, 2  96500, 0.2, (Cu ), , [Cu (CH3)4]+2, 0, 0.2, 0.2, 1, kf = 4.0  1011 =, =, 4, x  (0.2), x  (0.2)3, x=, , 1011, (0.2)3  4, , Ecell = 0.75 + 0.34 –, = 1.1 –, , 73., , [2], 2, , [Cu ], , x = 3.125 × 10–10, , [Cu+2] = 3.125 × 10–10, , 2, 0.0591, log, 2, 3.125  10–10, , 0.0591, (10 – 0.194) = 1.1 – 0.29 = 0.81 volt, 2, , [2], 0.0591, log, =0, 2, [Cu 2 ], , Ecell = 1.1 –, log, , , , =, , 1.1  2, = 37.23, 0.0591, , 2, , , , [Cu 2 ], , = 1.68 × 1037, , , , [Cu+2] = 1.19 × 10–37, , 74., , First conductance decreases due to nutralisation of free H+ ions of weak acid, then it increases due to, formation of salt and after equivalence point it increases more fastly due to increasing of OH– ions., , 75., , First conductance decreases due to nutralisation of strong acid H+ ion then after it increases due to, nutralisation of weak acid and after equivalence point it increases more fastly., , PART - V, 1., , Fe+2 + 2e–  Fe, Number of milimoles of e– passed =, , (965)(1), × 1000 = 10, 96500, ,  Milimoles of Fe+2 reduced = 5,  Milimoles of Fe+2 left = 1000 x – 5,  By equating mili equivalent = (1000x – 5) × 1 = (0.1)(10)(5), 2., milli moles, Then,, , CH3COOH + NaOH, 100×0.2, 100×0.2, 0, 0, , CH3COONa + H2O, –, –, 100×0.2, , , , x = 10–2

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[CH3COONa] =, Then,, , , 100  0.2  10 3, 2.0, × 1000 = 0.1 D.O.D () for CH3COOH = m, =, = 10–2, 0, 200, 200, m, , Ka of CH3COOH = C2 = 0.1 × (10–2)2 = 10–5, , , pKa = 5 for CH3COOH., So, pH of CH3COONa salt is :, 1, 1, pH = 7 + pKa + logC., 2, 2, 3., , =7+, , 1, 1, × 5 + log0.1 = 9., 2, 2, , , , m =  Ag +  Br –, , =a+b, m, , 4., , 1000, S, , K, × 1000, m, , , , S=, , , , º, 0.98, =, 130.34, 1.98, , º–, º, =, 1, –, º, º, , , , º, 0.98, º =1 –, , 1.98, , º–, 1, 130.34 = 1.98, , , , 130.34, º– = 1.98, , =K×, , , , S (g/lit) =, , 122, , = º = 0.936, , º, , º = 130.34 –1cm2 eq–1, º, 0.98, =, º, 1.98, , ºK = º = 64.51 –1cm2 eq–1, And, , 5., , = 65.83 –1cm2 eq–1, ,  = iCRT, 3 = i × 0.1 ×, , 1, × 300, 12, , i = 1.2, i = 1 + (n–1), 1.2 = 1 + (2–1) = 0.2, 30, 0.2 = , m, , , , , m, = 150 –1cm2 mol–1, , 100, V, 5, =, =, , 3, i, 0.15, 1, 3, 0.5, K= , =, ×, = 10–2, R a, 100, 1.5, M = K  1000 = 10–2 × 1000 = 200–1 cm2 mol–1, M, 0.05, , 6., , V = iR, , R=, , 7., , +, 2e– , , Cu, 482.5  4  60, 2mol, mol, 96500, = 1.2 mol electrons, 2 mole electrons then Cu+2 reacted = 1 mole, Cu2+, , K, × 1000 × 188, ab

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1, × 1.2 = 0.6, 2, so, remaining Cu+2 moles = 2 – 0.6 = 1.4 moles, 1.4, Remaining [CuSO4] =, = 0.28 M., 5, When 1.2 mole electrons then Cu+2 reacted =, , 8., , For concentration cell Eºcell = 0, ,  Ag+ (C1) + e–, Ag (s) ,  Ag (s), Ag+ (C2) + e– , ____________________________, ,  Ag+ (C1), Ag+ (C2) , Then,, , C , Ecell = X = 0 – 0.059 log  1 ,  C2 , , C , X,  log  2  =,  C1  0.059, , , 9., ,  x , C2, = anti log  0.059 , , , C1, , Fe+3 + 3OH– ;, , Fe(OH)3, , 0, EFe3 /Fe = EFe, 3, /Fe –, , 10., , [Fe+3] =, , K sp, [OH– ]3, , =, , 1.1 –, , = 10–20, , 1, 0.06, 0.06, log, = –0.036 –, × 20 = –0.036 – 0.4 = – 0.436, [Fe3 ], 3, 3, , [Zn2 ], 0.06, log, <, [Cu2 ], 2, , [Zn2 ], 0.06, log, <, [Cu2 ], 2, , [Zn2 ], 0.06, log, [Cu2 ], 2, log, , n, , [Zn2 ], [Cu2 ], [Zn2 ], [Cu2 ], , 0, 0, , >, , 1.1, , >, , 36.67, , >, , 84.4, , [Zn2+] > [Cu2+]e84.4, , 12., , (10–2 )3, , For cell reaction to take place in opposite direction. Ecell must be negative., Ecell, <, 0, Eºcell –, , 11., , 10–26, , X¯ is I¯, Y¯ is Cl¯, SRP Cl2 > Br2 > 2, , Ecell = –, , [H ]anode, , RT, n , [H ]cathode, nF, , K a [HA]anode, [NaA]anode, RT, =–, n, K, a [HA]cathode, nF, [NaA]cathode

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13., , 2H2O  O2 + 4OH¯ + 4e–, , Anode :, , 0.01 mol, 2+, , Cathode :, , Cu, , 0.04 F, –, , +, , 2e, , –3, , , , 20×10 ×0.5, = 10–2 mol, , 0.04 faraday, –0.02, , = 0.01 mol, , = 0.02 F, , Cu(s), , 2H2O + 2e–  H2(g) + 2OH¯, 0.02F, , 0.01 mol, , Total volume of gases evolved at STP = (0.01 + 0.01) × 22.4 = 448 ml, 14., , 0.164 = 0 +, , 0.1, 0.0591, log10 [Ag ], anode, 1, ,  [Ag+]anode = 1.66 × 10–4 M.,  1.66  10 –4 , –4, Ksp = [Ag+]2 × [CrO2–, =, 1.66, ×, 10, ×, ], , , 2, 4, , , , 15., , Hint: Reverse of (B) & (C) is spontaneous; weakest Oxidizing Agent here is Mg2+], , 17., , (a, b, d, f) E = Eº –, , 18., , ºCH3COOH = ºCH3COONa + ºHCl – ºNaCl, , [A  ][Cl ], 0.6, log, PCl2, n, , = 150 +200 – 125 = 225 S cm 2 mol–1., , , c, , CH3COOH =, , 2.25 S cm2 mol–1., , cCH3COOH, , =, , 0CH3COOH, , =, , 2.25, = 10–2, 225, , Then [H+] for CH3COOH = C = 0.001 × 10–2 = 10–5, , 19., , pH = – log[H+] = – log(10–5) = 5, , K = 3.2 × 10–5 –1.cm–1, =, , 103 K, C, , , , =, , 3.2  10 2, = 16 × 10–2, 0.2, , , , =, , , , , , ,  =, , 16  10 2, , =, =8, 0.02, 

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20., , E = Eºcell –, Q=, , 21., , 0.059, . log Q, 2, , 0.2, (10–7 )2, 1, ×, =, –7 2, 20, 100, (10 ), , E=0–, , 0.059, 1, 0.059, . log, =, × 2 = 0.059, 2, 100, 2, , , , 1000E = 1000 × 0.059 = 59, , 1,  Hg(l), Hg22+ + e– , 2, , At cathode :, , 1,  H+(aq) + e–, H2(g) , 2, ______________________________, , At anode :, , 1, 1,  Hg(l) + H+(aq), Hg22+ + H2(g) , 2, 2, 0.059, log [H+], 1, 0.634 = (0.28 – 0) + 0.059 pH, Ecell = E0cell –, or, , 0.634  0.28, =6, 0.059, Overall reaction should be the one which is written in term of species present in the given electrode/cell., or, , 22., , 23., , pH =, , Hence at anode, , H2 + 2OH¯  2H2O + 2e–, , At cathode, , 2AgCl + 2e–  2Ag + Cl¯, , We can assume the given cell to be:, Pt | H2 | H+ (aq), Cl¯ (aq) | AgCl (s) | Ag, o, o, With this assumption, Eocell = EAgCl/ Ag– – ESHE, , = 0.22 V, And cell reaction is :, , 1, H2(g) + AgCl(s)  H+(aq) + Ag(s) + Cl¯(aq), 2,  Ecell = Eocell –, , 0.06, log (H+) (Cl–), 1, , 1.05 = 0.22 – 0.06 log, , Kw, (Cl¯), (OH¯), , , (Cl– ) , – logK w – log, , , 0.83 = 0.06 , (OH– ) , , , 0.012, 83, = pKw – log, 0.01, 6, 83, = pKw – log(1.2), 6,  pKw =, , 83, + log(1.2) = 13.91, 6