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Electrochemistry, Recap Notes, Electrochemistry : It is the study of, production of electricity from energy, released during spontaneous chemical, , reactions and the use of electrical energy, to bring about non-spontaneous chemical, transformations., , Differences between electrochemical cell and electrolytic cell, Electrochemical cell, Electrolytic cell, (Galvanic or Voltaic cell), 1. It is a device which converts chemical energy 1. It is a device which converts electrical energy, into electrical energy., into chemical energy., 2. It is based upon the redox reaction which is 2. The redox reaction is non-spontaneous and, spontaneous. i.e., ∆G = –ve, takes place only when electrical energy is, supplied. i.e., ∆G = +ve, 3. Two electrodes are usually set up in two 3. B oth the electrodes are suspended in the, separate beakers., solution or melt of the electrolyte in the same, beaker., 4. The electrolytes taken in the two beakers are 4. Only one electrolyte is taken., different., 5. The electrodes taken are of different materials. 5. The electrodes taken may be of the same or, different materials., 6. The electrode on which oxidation takes place 6. The electrode which is connected to the –ve, is called the anode (or –ve pole) and the, terminal of the battery is called the cathode;, electrode on which reduction takes place is, the cations migrate to it which gain electrons, called the cathode (or +ve pole)., and hence, a reduction takes place, the other, electrode is called the anode., 7. To set up this cell, a salt bridge/porous pot 7. No salt bridge is used in this case., is used., Nernst equation : For a reduction reaction,, Mn+, (aq), , + ne, , –, , °, Ecell = Ecell, −, , M(s);, 2.303 RT, 1, log, +, nF, [ M(naq, )], , At 298 K,, °, Ecell = Ecell, −, , 0.0591, 1, log, +, n, [ M(naq, )], , For concentration cell, EMF at 298 K is, given by, 0.0591, C, Ecell =, log 2, n, C1, , where C2 > C1, X, , Applications of Nernst equation :, X To calculate electrode potential of a, cell :, ne−, , → xX + yY, aA + bB  

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x, y,  − 0.0591 log [ X ] [Y ] (At 298 K), Ecell = Ecell, n, [ A]a [ B]b, , X, ,  =, Ecell, , 0.0591, log K c at 298 K, n, , Relation between cell potential and, Gibbs energy change :, DG° = –nFE°cell ; DG° = –2.303 RT log Kc, , To calculate equilibrium constant :, At equilibrium, Ecell = 0, , Conductance in electrolytic solutions :, Property, Conductance (G), , Formula, 1 = a = κa, l, R ρl, , Units, Ohm, , –1, , –1, , (Ω )/Siemens (S), , Effect of dilution, Increases as larger number, of ions are produced., , Specific conductance, (k) or conductivity, , 1, l, or G, ρ, a, , Ohm–1 cm–1/S m–1, , Decreases as number of, ions per cm3 decreases., , Equivalent, conductivity (Λeq), , κ × V or, 1000, κ×, N, , Ω–1 cm2 eq–1/S m2 eq–1, , Increases with dilution due, to large increase in V., , Molar conductivity, (Λm), , κ × V or, 1000, κ×, M, , Ω–1 cm2 mol–1/S m2 mol–1 Increases with dilution due, to large increase in V., , Limiting molar conductivity : When, concentration approaches zero i.e., at infinite, dilution, the molar conductivity is known as, limiting molar conductivity (Λ°m)., Variation of molar conductivity with, concentration : For a strong electrolyte it, is shown by Debye–Huckel Onsager equation, as follows :, Λ m = Λ °m − A C, (KCl), Strong electrolyte, like KCl, m, , Weak electrolyte like, CH3COOH, C, , °, Here, Lm, = M olar conductivity at infinite, dilution ( L i m i t i n g m o l a r, conductivity), Lm = Molar conductivity at V-dilution, A = Constant which depends upon nature, of solvent and temperature, C = Concentration, Plot of Lm against C1/2 is a straight line with, °, intercept equal to Lm, and slope equal to ‘–A’., , Thus, Lm decreases linearly with C , when, C = 0, Lm = L°m and L°m can be determined, experimentally., , X, , X, , For weak electrolytes : There is a, very large increase in conductance with, dilution especially near infinite dilution, as no. of ions increases. Lm increases as, C decreases but does not reach a constant, value even at infinite dilution. Hence, their, L°m cannot be determined experimentally., For a strong electrolyte : There is, only a small increase in conductance, with dilution. This is because a strong, electrolyte is completely dissociated in, solution and so, the number of ions remain, constant and on dilution, interionic, attractions decreases as ions move far, apart., , Kohlrausch’s law of independent, migration of ions : It states that limiting, molar conductivity of an electrolyte can be, represented as the sum of the individual, contributions of the anion and cation of the, electrolyte., L°m = u+ l°+ + u–l°– ; where l°+ and l°– are the, limiting molar conductivities of the cation, and anion respectively and u + and u – are, stoichiometric number of cations and anions, respectively in one formula unit of the, electrolyte.

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CBSE Board Term-II Chemistry Class-12, , 4, , Practice Time, , OBJECTIVE TYPE QUESTIONS, , Multiple Choice Questions (MCQs), 1. Given below are the standard electrode, potentials of few half-cells. The correct order of, these metals in increasing reducing power will, be, K+|K = –2.93 V, Ag+|Ag = 0.80 V,, Mg2+|Mg = –2.37 V, Cr3+|Cr = –0.74 V., (a) K < Mg < Cr < Ag, (b) Ag < Cr < Mg < K, (c) Mg < K < Cr < Ag, (d) Cr < Ag < Mg < K, 2. DrG° for the cell with the cell reaction:, Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s), , + 2OH–(aq), [E°Ag2O/Ag = 0.344 V, E°Zn2+/Zn = –0.76 V], (a) 2.13 × 105 J mol–1, (c) 1.06 × 105 J mol–1, , (b) –2.13 × 105 J mol–1, (d) –1.06 × 105 J mol–1, , –, 3. For a cell reaction: Mn+, (aq) + ne → M(s), the, Nernst equation for electrode potential at any, concentration measured with respect to standard, hydrogen electrode is represented as, , (a) E, , ( M n+ / M ), , = E°, , ( M n+ / M ), , −, , RT, 1, ln, nF [ M n + ], , (b) E, = E°, −, (M / M n+ ), (M / M n+ ), , RT [ M n + ], ln, [M ], nF, , RT, 1, log, nF, [M ], RT, (d) E n +, = E° n+, −, ln[ M n + ], (M / M ), ( M / M ) nF, , (c) E n +, = E° n+, −, (M / M ), (M / M ), , 4. Limiting molar conductivity for some ions, is given below (in S cm2 mol–1) :, Na+ - 50.1, Cl– - 76. 3, H+ - 349.6, CH3COO– - 40.9,, Ca2+ - 119.0., What will be the limiting molar conductivities, (L°m) of CaCl2, CH3COONa and NaCl respectively?, (a) 97.65, 111.0 and 242.8 S cm2 mol–1, (b) 195.3, 182.0 and 26.2 S cm2 mol–1, (c) 271.6, 91.0 and 126.4 S cm2 mol–1, (d) 119.0, 1024.5 and 9.2 S cm2 mol–1, , 5. Electrical conductance through metals is, called metallic or electronic conductance and is, due to the movement of electrons. The electronic, conductance depends on, (a) the nature and structure of the metal, (b) the number of valence electrons per atom, (c) change in temperature, (d) all of these., 6. A galvanic cell has electrical potential of, 1.1 V. If an opposing potential of 1.1 V is applied, to this cell, what will happen to the cell reaction, and current flowing through the cell?, (a) The reaction stops and no current flows, through the cell., (b) The reaction continuous but current flows, in opposite direction., (c) The concentration of reactants becomes, unity and current flows from cathode to, anode., (d) The cell does not function as a galvanic cell, and zinc is deposited on zinc plate., 7. In a Daniell cell,, (a) the chemical energy liberated during the, redox reaction is converted to electrical, energy, (b) the electrical energy of the cell is converted, to chemical energy, (c) the energy of the cell is utilised in conduction, of the redox reaction, (d) the potential energy of the cell is converted, into electrical energy., 8. Mark the correct Nernst equation for the, given cell., Fe(s)|Fe2+(0.001 M)||H+ (1 M) | H2(g)(1 bar)|Pt(s), ° −, (a) Ecell = Ecell, , 0.591 [ Fe2+ ] [ H + ]2, log, 2, [ Fe][ H 2 ], , ° −, (b) Ecell = Ecell, , 0.591, [ Fe] [ H + ]2, log, 2, [ Fe2+ ][ H 2 ]

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° −, (c) Ecell = Ecell, , 0.0591 [ Fe2+ ] [ H 2 ], log, 2, [ Fe][ H + ]2, , ° − 0.0591 log [ Fe] [ H 2 ], (d) Ecell = Ecell, 2+, + 2, , 2, , [ Fe, , ][ H ], , 9. When an aqueous solution of AgNO 3 is, electrolysed between platinum electrodes, the, substances liberated at anode and cathode are, (a) silver is deposited at cathode and O 2 is, liberated at anode, (b) silver is deposited at cathode and H 2 is, liberated at anode, (c) hydrogen is liberated at cathode and O2 is, liberated at anode, (d) silver is deposited at cathode and Pt is, dissolved in electrolyte., 10. A standard hydrogen electrode has a zero, potential because, (a) hydrogen can be most easily oxidised, (b) hydrogen has only one electron, (c) the electrode potential is assumed to be zero, (d) hydrogen is the lightest element., 11. At 25°C, Nernst equation is, (a) Ecell = E °cell −, , 0.0591 [ion]RHS, log, n, [ion]LHS, , (b) Ecell = E °cell −, , 0.0591 [M ]RHS, log, n, [M ]LHS, , (c) Ecell = E °cell +, , 0.0591 [ion]RHS, log, n, [ion]LHS, , (d) Ecell = E °cell −, , 0.0591, [ion]LHS, log, n, [ion]RHS, , 12. Electrode potential data of few cells is given, below. Based on the data, arrange the ions in, increasing order of their reducing power., –, 2+, Fe3+, (aq) + e → Fe (aq) ; E° = +0.77 V, 3+, –, Al (aq) + 3e → Al(s) ; E° = –1.66 V, Br2(aq) + 2e– → 2Br–(aq) ; E° = +1.09 V, (a) Br– < Fe2+ < Al, (b) Fe2+ < Al < Br–, –, 2+, (c) Al < Br < Fe, (d) Al < Fe2+ < Br–, 13. Mark the correct relationship from the, following., (a) Equilibrium constant is related to emf as, log K =, , nFE, 2.303RT, , (b) EMF of a cell Zn | Zn2+(a1) || Cu2+(a2) | Cu is, E = E° −, , 0.591 [a2 ], log, n, [a1], , (c) Nernst equation is , Ecell = E °cell −, , 0.0591, [ Products], log, n, [ Reactants], , (d) For the electrode Mn+/M at 273 K, E = E° +, , 0.591, log[ M n + ], n, , 14. The specific conductivity of N/10 KCl solution, at 20°C is 0.0212 ohm–1 cm–1 and the resistance, of the cell containing this solution at 20°C is 55, ohm. The cell constant is, (a) 3.324 cm–1, (b) 1.166 cm–1, –1, (c) 2.372 cm, (d) 3.682 cm–1, 15. Following reactions are taking place in a, Galvanic cell,, Zn → Zn2+ + 2e– ; Ag+ + e– → Ag, Which of the given representations is the correct, method of depicting the cell?, +, (a) Zn(s) | Zn2+, (aq) || Ag (aq) | Ag(s), (b) Zn2+ | Zn || Ag | Ag+, +, (c) Zn(aq) | Zn2+, (s) || Ag (s) | Ag(aq), +, 2+, (d) Zn(s) | Ag (aq) || Zn (aq) | Ag(s), 16. What will be the molar conductivity of Al3+, ions at infinite dilution if molar conductivity of, Al2(SO4)3 is 858 S cm2 mol–1 and ionic conductance, of SO42– is 160 S cm2 mol–1 at infinite dilution?, (b) 698 S cm2 mol–1, (a) 189 S cm2 mol–1, (d) 429 S cm2 mol–1, (c) 1018 S cm2 mol–1, 17. E° value of Ni2+/ Ni is –0.25 V and Ag+ /Ag, is +0.80 V. If a cell is made by taking the two, electrodes what is the feasibility of the reaction?, (a) Since E° value for the cell will be positive,, redox reaction is feasible., (b) Since E° value for the cell will be negative,, redox reaction is not feasible., (c) Ni cannot reduce Ag+ to Ag hence reaction, is not feasible., (d) Ag can reduce Ni2+ to Ni hence reaction is, feasible., 18. A cell is set up as shown in the figure. It, is observed that EMF of the cell comes out to, be 2.36 V. Which of the given statements is not, correct about the cell?

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19. Limiting molar conductivity of NaBr is, (a) L°mNaBr = L°mNaCl + L°mKBr, (b) L°mNaBr = L°mNaCl + L°mKBr – L°mKCl, (c) L°mNaBr = L°mNaOH + L°mNaBr – L°mNaCl, (d) L°mNaBr = L°mNaCl – L°mNaBr, 20. Choose the option with correct words to fill, in the blanks., According to preferential discharge theory, out, of number of ions the one which requires _____, energy will be liberated ____ at a given electrode., (a) least, first, (b) least, last, (c) highest, first, (d) highest, last, 21. For the cell reaction :, 2Cu +(aq) → Cu (s) + Cu 2+, (aq) , the standard cell, potential is 0.36 V. The equilibrium constant, for the reaction is, (a) 1.2 × 106, (b) 7.4 × 1012, 6, (c) 2.4 × 10, (d) 5.5 × 108, 22. E° values of three metals are listed below., –, Zn2+, (aq) + 2e → Zn(s) ; E° = –0.76 V, –, Fe2+, (aq) + 2e → 2Fe(s) ; E° = –0.44 V, –, Sn2+, (aq) + 2e → Sn(s) ; E° = – 0.14 V, Which of the following statements are correct on, the basis of the above information?, (i) Zinc will be corroded in preference to iron if, zinc coating is broken on the surface., (ii) If iron is coated with tin and the coating, is broken on the surface then iron will be, corroded., (iii) Zinc is more reactive than iron but tin is less, reactive than iron., (a) (i) and (ii), (b) (ii) and (iii), (c) (i), (ii) and (iii), (d) (i) and (iii), 23. Which of the following is the correct order in, which metals displace each other from the salt, solution of their salts., (a) Zn, Al, Mg, Fe, Cu (b) Cu, Fe, Mg, Al, Zn, (c) Mg, Al, Zn, Fe, Cu (d) Al, Mg, Fe, Cu, Zn, 24. The reaction which is taking place in nickel, - cadmium battery can be represented by which, of the following equation?, , (a), (b), (c), (d), , Cd + NiO2 + 2H2O → Cd(OH)2 + Ni(OH)2, Cd + NiO2 + 2OH– → Ni + Cd(OH)2, Ni + Cd(OH)2 → Cd + Ni(OH)2, Ni(OH)2 + Cd(OH)2 → Ni + Cd + 2H2O, , 25. Molar conductivity of 0.15 M solution of KCl, at 298 K, if its conductivity is 0.0152 S cm–1 will, be, (a) 124 W–1 cm2 mol–1 (b) 204 W–1 cm2 mol–1, (c) 101 W–1 cm2 mol–1 (d) 300 W–1 cm2 mol–1, 26. Fluorine is the best oxidising agent because, it has, (a) highest electron affinity, (b) highest reduction potential, (c) highest oxidation potential, (d) lowest electron affinity., 27. During the electrolysis of dilute sulphuric, acid, the following process is possible at anode., (a) 2H2O(l) → O2(g) + 4H+(aq) + 4e–, 2–, –, (b) 2SO2–, 4(aq) → S2O 8(aq) + 2e, (c) H2O(l) → H+(aq) + OH–(aq), (d) H2O(l) + e– →, , 1, H, + OH–(aq), 2 2(g), , 28. Mark the correct choice of electrolytes, represented in the graph., Λm(S cm2 mol–1), , (a) Reduction takes place at magnesium, electrode and oxidation at SHE., (b) Oxidation takes place at magnesium, electrode and reduction at SHE., (c) Standard electrode potential for Mg2+ | Mg, will be –2.36 V., (d) Electrons flow from magnesium electrode to, hydrogen electrode., , A, B, C1/2 (mol L–1), , (a), (b), (c), (d), , A, A, A, A, , → NH4OH, B → NaCl, → NH4OH, B → NH4Cl, → CH3COOH, B → CH3COONa, → KCl, B → NH4OH, , 29. Molar conductivity of 0.025 mol L –1, methanoic acid is 46.1 S cm2 mol–1, the degree, of dissociation and dissociation constant, will be (Given : l°H+ = 349.6 S cm 2 mol –1 and, °, l°HCOO– = 54.6 S cm2 mol–1), (a) 11.4%, 3.67 × 10–4 mol L–1, (b) 22.8%, 1.83 × 10–4 mol L–1, (c) 52.2%, 4.25 × 10–4 mol L–1, (d) 1.14%, 3.67 × 10–6 mol L–1, 30. Electrode potential for Mg electrode, varies according to the equation,, E, , Mg 2+ |Mg, , = E°, , Mg 2+ |Mg, , −, , 0.059, 1, ., log, 2, [Μg 2+ ], , The graph of EMg2+|Mg vs log [Mg2+] is

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(a), , (c), , (b), , (d), , 31. E° values for the half cell reactions are given, below:, Cu2+ + e– → Cu+ ; E° = 0.15 V, Cu2+ + 2e– → Cu ; E° = 0.34 V, What will be the E° of the half-cell : Cu+ + e– → Cu?, (a) +0.49 V, (b) +0.19 V, (c) +0.53 V, (d) +0.30 V, 32. Given below are two figures of Daniell cell, (X) and (Y). Study the figures and mark the, incorrect statement from the following., , 33. Which of the following is/are an application, of electrochemical series?, (a) To compare the relative oxidising and, reducing power of substances., (b) To predict evolution of hydrogen gas on, reaction of metal with acid., (c) To predict spontaneity of a redox reaction., (d) All of these, 34. Two solutions of X and Y electrolytes are, taken in two beakers and diluted by adding, 500 mL of water. Lm of X increases by 1.5 times, while that of Y increases by 20 times, what, could be the electrolytes X and Y ?, (a) X → NaCl, Y → KCl, (b) X → NaCl, Y → CH3COOH, (c) X → KOH, Y → NaOH, (d) X → CH3COOH, Y → NaCl, 35. What would be the equivalent conductivity, of a cell in which 0.5 N salt solution offers, a resistance of 40 ohm whose electrodes are, 2 cm apart and 5 cm2 in area?, (a) 10 ohm–1 cm2 eq–1, (c) 30 ohm–1 cm2 eq–1, , (b) 20 ohm–1 cm2 eq–1, (d) 25 ohm–1 cm2 eq–1, , 36. The half-cell reactions with their appropriate, standard reduction potentials are, (i) Pb2+ + 2e– → Pb ; E° = –0.13 V, (ii) Ag+ + e– → Ag ; E° = +0.80 V, Based on the above data, which of the following, reactions will take place?, (a) Pb2+ + 2Ag → 2Ag+ + Pb, (b) 2Ag + Pb → 2Ag+ + Pb2+, (c) 2Ag+ + Pb → Pb2+ + 2Ag, (d) Pb2+ + 2Ag+ → Pb + Ag, , (a) In fig (X), electrons flow from Zn rod to Cu, rod hence current flows from Cu to Zn (Eext, < 1.1 V)., (b) In fig (Y), electrons flow from Cu to Zn and, current flows from Zn to Cu (Eext > 1.1 V)., (c) In fig (X), Zn dissolves at anode and Cu, deposits at cathode., (d) In fig (Y), Zn is deposited at Cu and Cu is, deposited at Zn., , 37. Units of the properties measured are given, below. Which of the properties has not been, matched correctly?, (a) Molar conductance = S m2 mol–1, (b) Cell constant = m–1, (c) Specific conductance = S m2, (d) Equivalent conductance = S m2 (g eq)–1, 38. When water is added to an aqueous solution, of an electrolyte, what is the change in specific, conductivity of the electrolyte?, (a) Conductivity decreases, (b) Conductivity increases, (c) Conductivity remains same, (d) Conductivity does not depend on number of, ions.

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39. The specific conductance of a saturated, solution of AgCl at 25°C is 1.821 × 10 –5 mho, cm–1. What is the solubility of AgCl in water, (in g L–1), if limiting molar conductivity of AgCl, is 130.26 mho cm2 mol–1?, (a) 1.89 × 10–3 g L–1, (b) 2.78 × 10–2 g L–1, –2, –1, (c) 2.004 × 10 g L, (d) 1.43 × 10–3 g L–1, 40. The standard reduction potential for the, half-cell reaction, Cl2 + 2e– → 2Cl– will be, (Pt2+ + 2Cl– → Pt + Cl2 , E°cell = –0.15 V ;, Pt2+ + 2e– → Pt, E° = 1.20 V), (a) –1.35 V, (b) +1.35 V, (c) –1.05 V, (d) +1.05 V, 41. Zn gives hydrogen with H2SO4 and HCl but, not with HNO3 because, (a) Zn acts as oxidising agent when reacts with, HNO3, (b) HNO3 is weaker acid than H2SO4 and HCl, (c) Zn is above the hydrogen in electrochemical, series, (d) NO3– is reduced in preference to H+ ion., 42. Given below are few reactions with some, expressions. Mark the expression which is not, correctly matched., (a) For concentration cell,, Ag | Ag+(C1) || Ag+(C2) | Ag ; Ecell = −, , 0.0591, C, log 1, 1, C2, , (b) For the cell, 2Ag+ + H2 (l atm) → 2Ag + 2H+ (1 M) ;, ° −, Ecell = Ecell, , 0.0591 [ Ag + ]2, log, 2, [ H + ]2, , (c) For a n el ectro chemi cal reacti o n, at, ne, , −, , , , equilibrium aA + bB , ,  cC + dD ;, ° =, Ecell, , 0.0591 [C ]c [ D ]d, log, n, [ A]a [ B ]b, , 44. In a cell reaction, Cu(s) + 2Ag+(aq) → Cu2+, (aq) , , + 2Ag(s), E°cell = +0.46 V. If the concentration of Cu2+ ions, is doubled then E°cell will be, (a) doubled, (b) halved, (c) increased by four times, (d) unchanged., 45. Molar conductivity of NH 4 OH can be, calculated by the equation,, (a) L°NH4OH = L°Ba(OH)2 + L°NH4Cl – L°BaCl2, (b) L°NH4OH = L°BaCl2 + L°NH4Cl – L°Ba(OH)2, (c) L°NH4OH =, (d) L°NH4OH =, , 2, , (a) 3.66%, (c) 2.12%, , (b) 3.9%, (d) 0.008%, , 47. Mark the incorrect statement., (a) The limiting equivalent conductance for, weak electrolytes can be computed with the, help of Kohlrausch’s law., (b) EMF of a cell is the difference in the, reduction potentials of cathode and anode., (c) For cell reaction to occur spontaneously, the, EMF of the cell should be negative., (d) Fluorine is the strongest oxidising agent as, its reducing potential is very high., 48. The process of chemical decomposition of the, electrolyte by the passage of electricity through, its melt or aqueous solution is called electrolysis., The following apparatus is used for the, electrolysis process:, DC source, , +, , –, , e–, , 0.0591, 1, log, n, [M n+ ], , 43. Which of the following is the cell reaction, that occurs when the following half-cells are, combined?, I2 + 2e– → 2I– (1 M) ; E° = +0.54 V, Br2 + 2e– → 2Br– (1 M) ; E° = +1.09 V, (a) 2Br– + I2 → Br2 + 2I–, (b) I2 + Br2 → 2I– + 2Br–, (c) 2I– + Br2 → I2 + 2Br–, (d) 2I– + 2Br– → I2 + Br2, , 2, Λ° NH 4Cl + Λ°Ba ( OH )2, , 46. The equivalent conductivity of N/10 solution, of acetic acid at 25°C is 14.3 ohm–1 cm2 equiv–1., What will be the degree of dissociation of acetic, acid (L∞CH3COOH = 390.71 ohm–1 cm2 equiv–1)?, , (d) For the cell, M n+ (aq) + ne – → M (s) ;, E = E° −, , Λ°Ba ( OH )2 + 2 Λ° NH 4Cl − Λ°BaCl 2, , Ammeter, , e–, , Anode, , Cathode, , –, –, –, –

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Nandini, a young scientist, tried different, electrolysis experiments using various, electrolytes., The incorrect observation of her experiment is, (a) cations which get reduced at cathode, preferentially are hydronium ions in, electrolysis of aqueous NaCl, (b) cations reaching to cathode are Cu2+ ions, during electrolysis of CuSO4 solution, (c) during electrolysis of conc. H2SO4, S2O82– is, formed at anode, (d) S2O82– is formed at anode during electrolysis, of CuSO4 solution., 49. Jiya, a class-12 student recorded Λm of various, electrolytes like acetic acid, sodium chloride and, AlPO4, etc., at various concentrations. Then she, plotted Λm versus C . Graphs obtained by her, are shown below:, , Λm, , Λm, , KCl, C, I, , CH3COOH, , Λm, , AlPO4, , C, , C, , II, , III, , Which of the given graph(s) is/are correct?, (a) I only, (b) I and II only, (c) I and III only, (d) I, II and III, 50. Which of the given Nernst equation, representation(s) is/are not correct for the given, cell?, Mg | Mg2+(0.130 M) || Ag+ (0.0001 M) | Ag, I., , , Ecell, , Ecell =, , (, , Mg 2+ , RT, , −, ln , 2, 2F,  Ag + , , , , ), , 2, ,  Ag + , , IV. Ecell =, , (, ,  Ag + , 0.059, , +, log , 2, Mg 2+ , , , , E +, Ag / Ag, , −, , (a) I only, (c) II and IV only, , E  2+, Mg / Mg, , ), , [ M 2+ ][H+ ], 0.059, log, 2, [ M 4+ ], , 2, , 2, , III. Ecell =, , 52. Arun, a class-12 student has a good habit of, practicing the topic at home whichever taught, in the class. After learning Nernst equation in, class, he tried writing few Nernst equations for, different cells. Next day when he shown the, work to his class teacher she said all are correct, except one., The incorrect Nernst equation is, 2+, (a) Pt(s) | H2(g), (1 bar) | H+(aq), 1 M || M4+, (aq), M (aq) | Pt(s),  −, Ecell = Ecell, , Mg 2+ , 0.059, II. Ecell = E  +, − E  2+, −, ln , 2, Ag / Ag, Mg / Mg, , , Ecell, , are very important and we can extract a lot of, useful informations from them. If the standard, electrode potential of an electrode is greater, than zero then its reduced form is more stable, compared to hydrogen gas. Similarly, if the, standard electrode potential is negative then, hydrogen gas is more stable than the reduced, form of the species., Based on the given data,, –, Fe(s); E° = –0.44 V, Fe2+, (aq) + 2e, 2+, Sn(s) ; E° = –0.14 V, Sn (aq) + 2e–, 2+, –, Zn(s) ; E° = –0.76 V, Zn (aq) + 2e, –, +, 3e, Cr, Cr3+, (aq), (s) ; E° = –0.74 V, He made following conclusions:, I. SnSO4 solution can be stored in Fe vessel., II. FeSO4 solution can be stored in Zn vessel., III. Cr2(SO4)3 solution can be stored in Sn vessel., IV. ZnSO 4 solution cannot be stored in iron, vessel., The correct conclusion(s) is/are, (a) I and II, (b) III and IV, (c) III only, (d) all of these., , 0.059, (0.0001)2, −, log, 2, (0.130), , (b) I and III only, (d) II, III and IV only, , 51. Shubh learnt during his electrochemistry, class that the standard electrode potentials, , (b) Pt | M|M3+ (0.001 mol L–1) || Ag+ (0.01 mol L–1) | Ag, , Ecell = Ecell, −, , [ M 3+ ], 0.059, log, 3, 3,  Ag + , , (c) Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s),  −, Ecell = Ecell, , [Cu 2+ ], 2.303 RT, log, 2F, [Zn2+ ], , +, (d) Ni(s) | Ni2+, (aq) || Ag (aq) | Ag(s), , Ecell = ( E, , +, , Ag /Ag, , − E, , Ni, , 2+, , /Ni, , )−, , [Ni2+ ], 0.059, log, 2, 2,  Ag + 

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Case Based MCQs, Case I : Read the passage given below and, answer the following questions from 53 to 57., The study of the conductivity of electrolyte, solutions is important for the development of, electrochemical devices, for the characterisation, of the dissociation equilibrium of weak electrolytes, and for the fundamental understanding of charge, transport by ions. The conductivity of electrolyte, is measured for electrolyte solution with, concentrations in the range of 10–3 to 10–1 mol L–1,, as solution in this range of concentrations can, be easily prepared. The molar conductivity (Λm), of strong electrolyte solutions can be nicely fit, by Kohlrausch equation., Λm = L°m – K C , …(i), Where, Λ°m is the molar conductivity at infinite, dilution and C is the concentration of the, solution. K is an empirical proportionality, constant to be obtained from the experiment., The molar conductivity of weak electrolytes,, on the other hand, is dependent on the degree, of dissociation of the electrolyte. At the limit of, very dilute solution, the Ostwald dilution law, is expected to be followed,, Λm CA, 1, 1, , …(ii), =, +, , Λm Λ, ( Λ  )2 K d, m, , m, , where, CA is the analytical concentration of the, electrolyte and Kd is dissociation constant. The, molar conductivity at infinite dilution can be, decomposed into the contributions of each ion., Λ°m = n+l°+ + n– l°–, …(iii), Where, l+ and l– are the ionic conductivities of, positive and negative ions, respectively and n+, and n– are their stoichiometric coefficients in, the salt molecular formula., 53. Which statement about the term infinite, dilution is correct?, (a) Infinite dilution refers to hypothetical, situation when the ions are infinitely far, apart., (b) The molar conductivity at infinite dilution of, NaCl can be measured directly in solution., (c) Infinite dilution is applicable only to strong, electrolytes., (d) Infinite dilution refers to a real situation, when the ions are infinitely far apart., 54. Which of the following is a strong electrolyte, in aqueous solution?, , (a) HNO2, (c) NH3, , (b) HCN, (d) HCl, , 55. Which of the following is a weak electrolyte, in aqueous solution?, (a) K2SO4, (b) Na3PO4, (c) NaOH, , (d) H2SO3, , 56. If the molar conductivities at infinite dilution, for NaI, CH 3 COONa and (CH 3 COO) 2 Mg are, 12.69, 9.10 and 18.78 S cm2 mol–1 respectively, at 25°C, then the molar conductivity of MgI2 at, infinite dilution is, (a) 25.96 S cm2, mol–1 (b) 390.5 S cm2 mol–1, (c) 189.0 S cm2 mol–1, (d) 3.89 × 10–2 S cm2 mol–1, 57. Which of the following is the correct order of, molar ionic conductivities of the following ions, in aqueous solutions?, (a) Li+ < Na+ < K+ < Rb+, (b) Li+ > Na+ > K+ > Rb+, (c) Rb+ < Na+ < Li+ < K+, (d) Li+ < Rb+< Na+ < K+, Case II : Read the passage given below and, answer the following questions from 58 to 62., The electrochemical cell shown below is, 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, concentrations of M2+ ions at the two electrodes., The emf of the cell at 298 K is 0.059 V., 58. The solubility product (K sp , mol 3 dm –9 ), of MX 2 at 298 K based on the information, available for the given concentration cell is, (take 2.303 × R × 298/F = 0.059), (b) 4 × 10–15, (a) 2 × 10–15, –12, (c) 3 × 10, (d) 1 × 10–12, 59. The value of DG (in kJ mol–1) for the given, cell is (take 1 F = 96500 C mol–1), (a) 3.7, (b) –3.7, (c) 10.5, (d) –11.4, 60. The equilibrium constant for the following, reaction is, Fe2+ + Ce4+  Ce3+ + Fe3+, (Given: E°Ce4+/Ce3+ = 1.44 V and E°Fe3+/Fe2+ = 0.68 V)

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(a) 7.6 × 1012, (c) 5.2 × 109, , (b) 6.5 × 1010, (d) 3.4 × 1012, , 61. The solubility product of a saturated solution, of Ag2CrO4 in water at 298 K if the emf of the cell, Ag | Ag+ (satd. Ag2CrO4 soln) || Ag+ (0.1 M) | Ag is, 0.164 V at 298 K, is, (a) 3.359 × 10–12 mol3 L–3, (b) 2.287 × 10–12 mol3 L–3, (c) 1.158 × 10–12 mol3 L–3, (d) 4.135 × 10–12 mol3 L–3, 62. To calculate the standard emf of the cell,, which of the following options is correct if E° is, reduction potential values?, (a) emf = E°cathode – E°anode, (b) emf = E°anode – E°cathode, (c) emf = E°anode + E°cathode, (d) None of these, Case III : Read the passage given below and, answer the following questions., Nernst equation relates the reduction potential, of an electrochemical reaction to the standard, potential and activities of the chemical species, undergoing oxidation and reduction., nM(s), Let us consider the reaction, Mn+(aq), For this reaction, the electrode potential, measured with respect to standard hydrogen, electrode can be given as, , E, , ( M n+ / M ), , = E, , ( M n+ / M ), , −, , RT, [M ], ln n+, nF [M ], , In the following questions (Q. No. 63-67), a, statement of assertion followed by a statement, of reason is given. Choose the correct answer, out of the following choices on the basis of the, above passage., (a) Assertion and reason both are correct, statements and reason is correct explanation, for assertion., (b) Assertion and reason both are correct, statements but reason is not correct, explanation for assertion., (c) Assertion is correct statement but reason is, wrong statement., (d) Assertion is wrong statement but reason is, correct statement., 63. Assertion : For concentration cell,, Zn(s) | Zn2+(aq) || Zn2+(aq) | Zn, C1, , C2, , For spontaneous cell reaction, C1 < C2, Reason : For concentration cell, Ecell =, , C, RT, log 2, nF, C1, , For spontaneous reaction, Ecell = +ve so, C2 > C1., 64. Assertion : For the cell reaction,, Zn(s) + Cu2+, Zn2+(aq) + Cu(s), (aq), voltmeter gives zero reading at equilibrium., Reason : At the equilibrium, there is no change, in concentration of Cu2+ and Zn2+ ions., 65. Assertion : The Nernst equation gives the, concentration dependence of emf of the cell., Reason : In a cell, current flows from cathode, to anode., 66. Assertion : Increase in the concentration of, copper half cell in a cell, increases the emf of the, cell., , Reason : Ecell = Ecell, +, , 0.059, [Cu2+ ], log, 2, [Zn2+ ], , 67. Assertion : Electrode potential for the, electrode Mn+/Mn with concentration is given, by the expression under STP conditions., 0.059, E = E° +, log[Mn + ], n, Reason : STP conditions require the temperature, to be 273 K., Case IV : Read the passage given below and, answer the following questions from 68 to 72., 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), 68. For the above cell,, (a) Ecell = 0 ; DG > 0, (c) Ecell < 0 ; DG° > 0, , (b) Ecell > 0 ; DG < 0, (d) Ecell > 0 ; DG° = 0, , 69. 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, (a) 130 mV, (b) 185 mV, (c) 154 mV, (d) 600 mV

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70. The value of equilibrium constant for a, feasible cell reaction is, (a) < 1, (b) = 1, (c) > 1, (d) zero, 71. What is the emf of the cell when the cell, reaction attains equilibrium?, (a) 1, (b) 0, (c) > 1, (d) < 1, 72. The potential of an electrode change with, change in, (a) concentration of ions in solution, (b) position of electrodes, (c) voltage of the cell, (d) all of these., Case V : Read the passage given below and, answer the following questions from 73 to 75., All chemical reactions involve interaction of, atoms and molecules. A large number of atoms/, molecules are present in a few gram of any, chemical compound varying with their atomic/, molecular masses. To handle such large number, conveniently, the mole concept was introduced., All electrochemical cell reactions are also based, , on mole concept. For example, 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 electrode. The amount of products formed, can be calculated by using mole concept., 73. The total number of moles of chlorine gas, evolved is, (a) 0.5, (b) 1.0, (c) 1.5, (d) 1.9, 74. If cathode is a Hg electrode, then the, maximum weight of amalgam formed from this, solution is, (Given : Atomic mass of Na = 23u and, Hg = 200.59 u), (a) 300 g, (b) 446 g, (c) 396 g, (d) 296 g, 75. In electrolysis of aqueous NaCl solution, when Pt electrode is taken, then which gas is, liberated at cathode?, (a) H2 gas, , (b) Cl2 gas, , (c) O2 gas, , (d) None of these, , Assertion & Reasoning Based MCQs, For question numbers 76-90, a statement of assertion followed by a statement of reason is given. Choose, the correct answer out of the following choices., (a) Assertion and reason both are correct statements and reason is correct explanation for assertion., (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion., (c) Assertion is correct statement but reason is wrong statement., (d) Assertion is wrong statement but reason is correct statement., 76. Assertion : The conductivity depends on, the charge and size of the ions in which they, dissociate, the concentration of ions or ease with, which the ions move under potential gradient., , chloride ions respectively, then the limiting, molar conductivity for sodium chloride is given, , Reason : The conductivity of solutions of, different electrolytes in the same solvent and, at a given temperature is same., , Reason : This is according to Kohlrausch law, of independent migration of ions., , 77. Assertion : If standard reduction potential, for the reaction,, Ag+ + e– → Ag is 0.80 volt, then for the reaction,, 2Ag+ + 2e– → 2Ag, it will be 1.60 volt., Reason : If concentration of Ag+ ions is doubled,, the standard electrode potential remains same., 78. Assertion : If λ o + and λ o − are molar, Na, Cl, limiting conductivities of the sodium and, , by the equation, Λ °NaCl = λ°, , Na +, , + λ°, , Cl −, , ., , 79. Assertion : The conductivity of solution is, greater than pure solvent., Reason : Conductivity depends upon number of, the ions present in solution., 80. Assertion : At the end of electrolysis using, platinum electrodes, an aqueous solution of, copper sulphate turns colourless., Reason : Copper in CuSO 4 is converted to, Cu(OH)2 during the electrolysis.

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81. Assertion : The electrical resistance of any, object decreases with increase in its length., , electrolyte at infinite dilution is equal to the, sum of molar conductance of cations and anions., , Reason : The electrical resistance of any object, decreases with increase in its area of crosssection., , Reason : Kohlrausch’s law is applicable for, strong electrolytes., , 82. Assertion : Substances like glass, ceramics,, etc. having very low conductivity are known as, insulators., Reason : They do not allow the passage of, electric current through them., 83. Assertion : Molar conductivity of a weak, electrolyte at infinite dilution cannot be, determined experimentally., Reason : Kohlrausch law helps to find the molar, conductivity of a weak electrolyte at infinite, dilution., 84. Assertion : The observed conductance, depends upon the nature of the electrolyte and, the concentration of the solution., Reason : The cell constant of a cell depends, upon the nature of the material of the electrodes., 85. Assertion : The molar conductivity of, strong electrolyte decreases with increase in, concentration., Reason : At high concentration, migration of, ions is slow., 86. Assertion : The molar conductance of weak, , 87. Assertion : Equivalent conductance of, all electrolytes decreases with increasing, concentration., Reason : More number of ions are available per, gram equivalent at higher concentration., 88. Assertion : Specific conductance decreases, with dilution whereas equivalent conductance, increases., Reason : On dilution, number of ions per, millilitre decreases but total number of ions, increases considerably., 89. Assertion : The ratio of specific conductivity, to the observed conductance does not depend, upon the concentration of the solution taken in, the conductivity cell., Reason : Specific conductivity decreases with, dilution whereas observed conductance increases, with dilution., 90. Assertion : Kohlrausch law helps to find, the molar conductivity of weak electrolyte at, infinite dilution., Reason : Molar conductivity of a weak electrolyte, at infinite dilution cannot be determined, experimentally., , SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), 1. Express the relation between conductivity, and molar conductivity of a solution held in a, cell?, 2. Limiting molar conductivity of an electrolyte, cannot be determined experimentally. Why?, 3. Following reactions occur at cathode during, the electrolysis of aqueous silver chloride, solution :, Ag+(aq) + e– → Ag(s), E° = +0.80 V, 1, H+(aq) + e– → H2( g) , E° = 0.00 V, 2, On the basis of their standard reduction, electrode potential (E°) values, which reaction, is feasible at the cathode and why?, , 4. Give reason :, Molar conductivity of CH3COOH increases on, dilution., 5., , Give reason :, , On the basis of E° values, O2 gas should be liberated, at anode but it is Cl2 gas which is liberated in the, electrolysis of aqueous NaCl., , 6. What is the necessity to use a salt bridge in, a Galvanic cell?, 7. What is the use of platinum foil in the, hydrogen electrode?, 8. Out of HCl and NaCl, which do you expect, will have greater value for Lm and why?

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9. State Kohlrausch’s law of independent, migration of ions. Write its one application., 10. Following reactions occur at cathode during, the electrolysis of aqueous copper (II) chloride, solution :, –, Cu2+, (aq) + 2e → Cu(s) ; E° = +0.34 V, , H+(aq) + e– →, , 1, H2( g) ; E° = 0.00 V, 2, , On the basis of their standard reduction, electrode potential (E°) values, which reaction, is feasible at the cathode and why?, , Short Answer Type Questions (SA-I), 11. Define the term degree of dissociation. Write, an expression that relates the molar conductivity, of a weak electrolyte to its degree of dissociation., 12. (i) Explain why fluorine is the strongest, oxidising agent?, (ii) Lithium metal is the strongest reducing, agent. Why?, 13. The standard electrode potential (E°) for, Daniell cell is +1.1 V. Calculate the DrG° for the, reaction., 2+, Zn(s) + Cu2+, (aq) → Zn (aq) + Cu(s), (1 F = 96500 C mol–1), 14. What is the difference between electronic, and electrolytic conductors?, 15. Define electrochemical cell. What happens if, external potential applied becomes greater than, E°cell of electrochemical cell?, 16. Why a galvanic cell stops working after, sometime?, 17. In the plot of molar conductivity (Lm) vs square, root of concentration (c1/2),following curves are, obtained for two electrolytes A and B., , Answer the following :, (i) Predict the nature of electrolytes A and B., (ii) What happens on extrapolation of L m, to concentration approaching zero for, electrolytes A and B?, 18. Two half-reactions of an electrochemical cell, are given below :, MnO–4(aq) + 8H+(aq) + 5e– → Mn2+, (aq) + 4H2O(l),, , E° = + 1.51V, 4+, –, →, Sn, +, 2e, ,, E°, =, +, 0.15, V, Sn2+, (aq), (aq), Construct the redox equation from the standard, potential of the cell and predict if the reaction is, reactant favoured or product favoured., 19. The conductivity of 0.001 M acetic acid is, 4 × 10–5 S/cm. Calculate the dissociation constant, of acetic acid, if molar conductivity at infinite, dilution for acetic acid is 390 S cm2/mol., 20. Equilibrium constant (Kc) for the given cell, reaction is 10. Calculate E°cell., A2+, A(s) + B2+, (aq), (aq) + B(s), 21. Given that the standard electrode potential, (E°) of metals are :, K+/K = –2.93 V, Ag+/Ag = 0.80 V,, Cu2+/Cu = 0.34 V,, Mg2+/Mg = –2.37 V, Cr3+/Cr = –0.74 V,, Fe2+ /Fe = –0.44 V., Arrange these metals in an increasing order of, their reducing power., , Short Answer Type Questions (SA-II), 22. Mention few applications of electrochemical, series., 23. A voltaic cell is set up at 25°C with the, following half cells :, Al/Al3+ (0.001 M) and Ni/Ni2+ (0.50 M), Write an equation for the reaction that occurs, , when the cell generates an electric current and, determine the cell potential., E°Ni2+/Ni = – 0.25 V and E°Al3+/Al = – 1.66 V., (log 8 × 10–6 = – 5.09), 24. A cell is prepared by dipping copper rod in, 1 M copper sulphate solution and zinc rod in 1 M

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zinc sulphate solution. The standard reduction, potential of copper and zinc are 0.34 V and –0.76 V, respectively., (i) What will be the cell reaction?, (ii) What will be the standard electromotive, force of the cell?, (iii) Which electrode will be positive?, 25. Resistance of a conductivity cell filled with, 0.1 mol L–1 KCl solution is 100 W. If the resistance, of the same cell when filled with 0.02 mol L–1 KCl, solution is 520 W, calculate the conductivity and, molar conductivity of 0.02 mol L–1 KCl solution., The conductivity of 0.1 mol L–1 KCl solution is, 1.29 × 10–2 W–1 cm–1., 26. Calculate the potential for half-cell, containing, 0.10 M K2Cr2O7(aq) , 0.20 M Cr3+, (aq) and, 1.0 × 10–4 M H+(aq). The half cell reaction is :, +, –, 3+, Cr 2O 2–, 7(aq) + 14H (aq) + 6e → 2Cr (aq) + 7H 2 O (l), and the standard electrode potential is given as, E° = 1.33 V., 27. Estimate the minimum potential difference, needed to reduce Al 2O 3 at 500°C. The Gibbs, energy change for the decomposition reaction,, 2, 4, Al2O3 → Al + O2 is 960 kJ., 3, 3, (F = 96500 C mol–1), 28. For the cell reaction,, +, Ni(s) | Ni2+, (aq) || Ag (aq) | Ag(s), Calculate the equilibrium constant at 25°C., How much maximum work would be obtained, by operation of this cell?, E, , (Ni2+ /Ni), , = − 0.25 V and E, , Ag + /Ag, , = 0.80 V, , 29. Calculate DrG° and logKc for the following, reaction., 2+, Cd2+, (aq) + Zn(s) → Zn (aq) + Cd(s), Given : E°Cd2+/Cd = –0.403 V ; E°Zn2+/Zn = –0.763 V, , 30. The equivalent conductivity of 0.05 N, solution of a monobasic acid is 15.8 mho cm2 eq–1., If equivalent conductivity of the acid at infinite, dilution is 350 mho cm 2 eq –1 , calculate the, (a) degree of dissociation of acid (b) dissociation, constant of acid., 31. The electrical resistance of a column of, 0.05 M NaOH solution of diameter 1 cm and, length 50 cm is 5.5 × 10 3 ohm. Calculate its, resistivity, conductivity and molar conductivity., 32. Depict the galvanic cell in which the reaction, Zn (s) + 2Ag +(aq) → Zn 2+, (aq) + 2Ag (s) takes place., Further show :, (i) Which of the electrode is negatively charged?, (ii) The carriers of the current in the cell., (iii) Individual reaction at each electrode., 33. What is the difference between a chemical, and a concentration cell?, 34. A copper-silver cell is set up. The copper, ion concentration is 0.10 M. The concentration, of silver ion is not known. The cell potential, when measured was 0.422 V. Determine the, concentration of silver ions in the cell., Given : E, , Ag + /Ag, , = + 0.80 V , E, , Cu 2+ /Cu, , = + 0.34 V, , 3, , 35. The resistance of 100 cm aqueous solution of, 0.025 M CuSO4 is 520 ohm at 298 K. Calculate, the molar conductivity if the cell constant of the, conductivity cell is 153.7 m–1., 36. When a certain conductance cell was filled, with 0.1 M KCl, it has a resistance of 85 ohms, at 25°C. When the same cell was filled with an, aqueous solution of 0.052 M unknown electrolyte,, the resistance was 96 ohms. Calculate the, molar conductance of the electrolyte at this, concentration., [Specific conductance of 0.1 M KCl, , = 1.29 × 10–2 ohm–1 cm–1], , Long Answer Type Questions (LA), 37. E°cell for the given redox reaction is 2.71 V., Mg(s) + Cu2+ (0.01 M) → Mg2+ (0.001 M) + Cu(s), Calculate E cell for the reaction. Write the, direction of flow of current when an external, opposite potential applied is, (i) less than 2.71 V and (ii) greater than 2.71 V, 38. (a) Calculate standard emf of the cell in, which following reaction takes place at 25°C., , Cu2+ + 2Cl–, Cu(s) + Cl2(g), E°Cl2/Cl– = +1.36 V, E°Cu2+/Cu = + 0.34 V, Also calculate standard free energy change and, equilibrium constant of the reaction., (b) The emf of a galvanic cell composed of two, hydrogen electrode is 0.16 volt at 25°C. Calculate, pH of the anode solution if the cathode is in a, solution with pH = 1.

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39. (a) Calculate the cell emf and DG° for the, cell reaction at 25°C for the cell :, Zn(s) | Zn2+ (0.0004 M) || Cd2+ (0.2 M) | Cd(s), E° values at 25°C : Zn2+/ Zn = – 0.763 V;, Cd2+/Cd = – 0.403 V; F = 96500 C mol–1;, R = 8.314 J K–1 mol–1., (b) If E° for copper electrode is 0.34 V, how will, you calculate its emf value when the solution in, contact with it is 0.1 M in copper ions? How does emf, for copper electrode change when concentration, of Cu2+ ions in the solution is decreased?, , OBJECTIVE TYPE QUESTIONS, , 1. (b) : Higher the oxidation potential, more easily it is, oxidised and hence greater is the reducing power. Hence,, increasing order of reducing power is Ag < Cr < Mg < K., 2., , (b) : E°cell = E°Ag2O/Ag – E°Zn2+/Zn, = 0.344 – (–0.76) = 1.104 V, , DG° = –nFE°cell = –2 × 96500 × 1.104, = –2.13 × 105 J mol–1, 3., , (a) : E, , (M n + / M ), , = E°, , (M n + / M ), , −, , RT, [M ], ln, nF [M n + ], , Since concentration of solid is taken as unity,, RT, 1, E n+, = E ° n+ −, ln, (M / M ), (M /M ) nF [M n + ], 4. (c) : Lm° CaCl2 = l°Ca2+ + 2l°Cl–, = 119.0 + 2 × 76.3 = 271.6 S cm2 mol–1, Lm° CH3COONa = l°CH3COO– + l°Na+, = 40.9 + 50.1 = 91 S cm2 mol–1, Lm° NaCl = l°Na+ + l°Cl–, = 50.1 + 76.3 = 126.4 S cm2 mol–1, 5. (d) : The electronic conductance depends on all these, factors., 6. (a) : If an external potential of 1.1 V is applied to the, cell, the reaction stops and no current flows through the cell., Any further increase in external potential again starts the, reaction but in opposite direction and the cell functions as, an electrolytic cell., 7. (a) : Daniell cell converts the chemical energy liberated, during the redox reaction to electrical energy and has an, electrode potential of 1.1 V., 8., , (c) : At anode : Fe → Fe2+(0.001 M) + 2e–, , 40. (a) Equivalent conductance of a 0.0128 N, solution of acetic acid is 1.4 mho cm 2 eq –1, and conductance at infinite dilution is, 391 mho cm2 eq–1. Calculate degree of dissociation, and dissociation constant of acetic acid., (b) The equivalent conductances of sodium, acetate, sodium chloride and hydrochloric, acid are 83, 127 and 426 mho cm 2 eq –1 at, 250°C respectively. Calculate the equivalent, conductance of acetic acid solution., , At cathode : 2H+ (1 M) + 2e– → H2 (1 bar), Net reaction : Fe + 2H+ → Fe2+ + H2, Nernst equation for the given cell,, ° −, E cell = E cell, , 9., , 0.0591 [Fe2+ ][H2 ], log, 2, [Fe][H+ ]2, , (a) : At cathode : Ag+(aq) + e– → Ag(s), , 1, O2(g) + H2O(l) + 2e–, 2, 10. (c) : According to convention, the standard hydrogen, electrode is assigned a zero potential at all temperatures., –, , At anode : 2OH (aq) →, , 11. (a) : E cell = E °cell − 0.0591log [Ion]RHS, n, [Ion]LHS, 12. (a) : Lower the reduction potential, more is the reducing, power. Thus, the order is, Br– < Fe2+ < Al., nFE °cell, 2.303RT, ° − 0.0591log [a1], (b) E cell = E cell, 2, [a2 ], (d) Expression is valid at 298 K, not at 273 K., 13. (c) : (a) log K =, , 14. (b) : κ = G ×, , l, A, , 1, l, = κ × = κ × R = 0.0212 × 55 = 1.166 cm−1, A, G, 15. (a) : Zn + 2Ag+ → Zn2+ + 2Ag can be represented as, 2+, , +, , Zn(s) | Zn (aq) | | Ag (aq) | Ag(s), 16. (a) : L°Al2(SO4)3 = 2l°Al3+ + 3l°SO42–, λ Al3+, , =, , Λ Al2 (SO4 )3 − 3λ SO2−, 4, , 2, 858 − (3 × 160), =, = 189 S cm2 mol−1, 2

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28. (d) : For strong electrolytes, the plot between Lm and, C1/2 is a straight line., For weak electrolytes, Lm increases steeply on dilution,, especially near low concentrations., , 17. (a) : The cell reaction will be, Ni(s) + 2Ag, , +, , (aq), , →, , Ni2+(aq), , + 2Ag(s), , E °cell = E °cathode – E °anode, = 0.80 – (–0.25) = +1.05 V, , 29. (a) : l°HCOOH = l°H+ + l°HCOO–, , DG° = –nFE°cell, , = 349.6 + 54.6 = 404.2 S cm2 mol–1, , As E°cell = +ve,, , Λm, , DG° = –ve, hence reaction is feasible., , α=, , 18. (a) : Mg2+(aq) + 2e– → Mg(s); E° = –2.36 V, 2H+ + 2e– → H2(g); E° = 0.00 V, Thus, oxidation takes place at magnesium electrode and, reduction at hydrogen electrode., , Ka =, , 19. (b) : L°m NaBr = L°m NaCl + L°m KBr – L°m KCl, 20. (a) : The ion which requires less energy is liberated first., °, 21. (a) : log K c = nE cell, 0.0591, For the given reaction, n = 1, 1 × 0.36, log K c =, = 6.09, 0.0591, Kc = antilog 6.09 = 1.2 × 106, , 22. (c) : Iron coated with zinc does not get rusted even if, cracks appear on the surface because Zn will take part in, redox reaction not Fe as Zn is more reactive than Fe. If iron is, coated with tin and cracks appear on the surface, Fe will take, part in redox reaction because Sn is less reactive than Fe., 23. (c) : In reactivity series,, Mg > Al > Zn > Fe > Cu, , =, , 46.1, = 0.114 × 100 = 11.4%, 404.2, , C α2 0.025 × (0.114)2, =, 1− α, 1 − 0.114, , 0.025 × 0.114 × 0.114, = 3.67 × 10–4 mol L–1, 0.886, 0.059, log[Mg2+]., 30. (b) : E = E° +, 2, Hence, plot of E vs log [Mg2+] will be linear with positive, slope and intercept = E°., =, , 31. (c) : Cu2+ + e– → Cu+ ; E°1 = 0.15 V, DG°1, n1 = 1, E°2 = 0.34 V, DG°2, n2 = 2, Cu2+ + 2e– → Cu ;, +, –, E°3 = ?, DG°3, n3 = 1, Cu + e → Cu ;, °, °, °, DG 3 = DG 2 – DG 1, –n3 FE°3 = –n2FE°2 + n1FE°1, –E°3 = –2 × 0.34 + 1 × 0.15, E°3 = 0.68 – 0.15 = +0.53 V, 32. (d) : In fig. (Y), zinc is deposited at the zinc electrode, and copper dissolves at copper electrode., 33. (d), , Reactivity decreases, , Hence, Mg can displace Al, Al can displace Zn and so on., 24. (a) : Nickel-Cadmium battery, Anode - Cd ; Cathode - NiO2 ; Electrolyte - KOH, –, , At anode : Cd(s) + 2OH(aq) → Cd(OH)2(s) + 2e–, –, , At cathode : NiO2(s) + 2H2O(l) + 2e– → Ni(OH)2(s) + 2OH(aq), Cd(s) + NiO2(s) + 2H2O(l) → Cd(OH)2(s) + Ni(OH)2(s), 25. (c) : Λ m =, , Λ °m, , κ × 1000 1.52 × 10 −2 × 1000, =, M, 0.15, , = 101 W–1 cm2 mol–1, 26. (b) : Higher the reduction potential, stronger is the, oxidising agent., 27. (a) : During the electrolysis of dilute sulphuric acid, the, following process is possible at anode:, +, 2H2O(l) → O2(g) + 4H (aq) + 4e–, , 34. (b) : Electrolyte X is strong electrolyte as on dilution, the number of ions remain same, only interionic attraction, decreases and hence not much increase in Lm as seen. While, Lm for a weak electrolyte increases significantly., 1 l, 1 2, 35. (b) : κ = × = ×, R A 40 5, 1000 1 2 1000, Λ eq = κ ×, = × ×, = 20 ohm−1 cm2 eq −1, N, 40 5 0.5, 36. (c) : At cathode : Ag+ + e– → Ag ; E° = +0.80 V, At anode : Pb → Pb2+ + 2e–; E° = +0.13 V, E°cell = E°cathode – E°anode = 0.80 – 0.13 = 0.67 V, Hence, the reaction will be, 2Ag+ + Pb → Pb2+ + 2Ag, 37. (c) : Specific conductance = S m–1, 38. (a) : Conductivity decreases because number of ions per, unit volume decreases., κ × 1000, 39. (c) : Solubility =, Λ °m

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0.164 =, , 0.059, 0.1, log +, 1, [Ag ]Satd. Ag CrO, 2, , +, , 4, , –4, , [Ag ]Satd. Ag2CrO4 = 1.66 × 10 M, So,, , 1.66 × 10, [CrO24− ] =, , −4, , 2, , Ksp (Ag2CrO4) = [Ag+]2 [CrO42–],  1.66 × 10 −4 , = (1.66 × 10 −4 )2 , , 2, , , = 2.287 × 10, , –12, , 3 –3, , mol L, , 62. (a), , C , 63. (a) : log  1  < 0 for spontaneity., C, \ C1 < C2  2 , 64. (a) , , 65. (b), , 66. (a), 67. (d) : Nernst equation is measured at 298 K. At STP, conditions, temperature to be 273 K., 68. (b) : M(s) → M +(aq) (0.05 M) + e–, M +(aq) (1.0 M) + e– → M(s), M +(aq) (1.0 M) → M +(aq) (0.05 M), 0.059 0.05, For concentration cell, E cell = −, log, 1, 1, 0.059, E cell = −, log(5 × 10 −2 ), 1, 0.059, E cell = −, [( −2) + log5] = –0.059(–2 + 0.698), 1, = –0.059(–1.302) = 0.0768, DG = –nFEcell, If Ecell is positive, DG is negative., E, log 0.05, 69. (c) : 1 =, E 2 log 0.0025, E1 log5 × 10 −2, =, E 2 log25 × 10 −4, E1 = 0.0768, 0.0168 −1.3 1, =, or E 2 = 154 mV, =, −2.6 2, E2,  nE ° , 70. (c) : K = antilog ,  0.0591, For feasible cell, E° is positive, hence from the above equation,, K > 1 for a feasible cell reaction., 71. (b) , , 72. (a), 4 × 500, 73. (b) : nNaCl =, = 2 mol, 1000, \ nCl2 = 1 mol, , 74. (b) : nNa deposited = 2 mol, \ nNa – Hg formed = 2 mol, \ Mass of amalgam formed = 2 × 223 = 446 g, 75. (a) : H2 gas at cathode., 76. (c) : The conductivity of solutions of different electrolytes, in the same solvent and at a given temperature is different., Effect of concentration on electrode potential is found by, Nernst equation., 77. (d) : Standard reduction potential of an electrode has a, fixed value., 78. (a) : According to Kohlrausch law, “limiting molar, conductivity of an electrolyte can be represented as the sum, of the individual contributions of the anion and cation of the, electrolyte.”, 79. (a) : When electrolytes are dissolved in solvent they, furnish their own ions in the solution hence, its conductivity, increases., 80. (c) : Cu2+ ions are deposited as Cu., 81. (d) : The electrical resistance of any object is directly, proportional to its length, l, and inversely proportional to, its area of cross-section, A. So, it increases with increase in, length of object and decreases with increase in area of crosssection of object., 82. (a) : The substances which do not allow the flow of, electric current through them are termed as insulators., 83. (b) : In the plot of molar conductivity versus, concentration, the extrapolation to zero concentration is not, possible. Since the graph is not linear., 84. (c) : The cell constant depends upon the distance, between the electrodes and their area of cross section., 85. (a), ∞, , ∞, , 86. (c) : Λ ∞, AB = λ A + + λ B −, Kohlrausch’s law is applicable for weak electrolytes., 87. (c) : At higher concentration, mobility of ions decreases., Hence, conductance decreases., 88. (c) : Total number of ions will increase slightly on, dilution (not considerably)., 89. (b) , , 90. (a), SUBJECTIVE TYPE QUESTIONS, , κ × 1000, in CGS units, M, κ × 10 −3, in SI units, Lm =, M, where, k is the conductivity, M is the molar concentration and, Lm is molar conductivity., Λm =, 1. 

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2. In weak electrolyte, the conductivity of the solution, increases very slowly with dilution of solution and goes on, increasing up to infinity. Therefore, it cannot be measured, experimentally., Λm, 1/2, , C, , 3. The species that get reduced at cathode is the one, having higher value of standard reduction potential. Hence,, the reaction that will occur at cathode is, Ag+(aq) + e– → Ag(s)., 4. Molar conductivity increases with decrease in, concentration. This is because the total volume, V, of solution, containing one mole of electrolyte also increases. It has been, found that decrease in K on dilution of a solution is more than, compensated by increase in its volume., 5. The reaction at anode with lower value of E° is preferred, i.e., O 2 gas should be liberated but on account of over, potential of oxygen reaction at anode, preferred reaction is, 1, Cl(−aq ) → Cl2( g ) + e −, 2, i.e., Cl 2 gas is liberated at anode in the electrolysis of, aq. NaCl., 6. The salt bridge allows the movement of ions from one, solution to the other without mixing of the two solutions., Moreover, it helps to maintain the electrical neutrality of the, solutions in the two half cells., 7., , It is used for the inflow and outflow of electrons., , 8. HCl will have greater value of Lm because H+ ions are, smaller than Na+ ions and hence H+ ions have greater ionic, mobility than Na+ ions., 9. Kohlrausch’s law of independent migration, of ions : It states that limiting molar conductivity of an, electrolyte can be represented as the sum of the individual, contributions of the anion and cation of the electrolyte., Kohlrausch’s law helps in the calculation of degree of, dissociation of weak electrolytes like acetic acid., 10. The species that get reduced at cathode is the one which, have higher value of standard reduction potential. Hence, the, reaction that will occur at cathode is, –, Cu2+, (aq) + 2e → Cu(s), 11. The fraction of the total number of molecules present in, solution as ions is known as degree of dissociation., Molar conductivity (lm) = al°m, where l°m is the molar conductivity at infinite dilution., , 12. (i) Because fluorine has highest reduction potential., (ii) Lithium metal is strongest reducing agent because Li, has lowest reduction potential i.e., E°Li+/Li = –3.05 V, 13. Here n = 2, E°cell = 1.1 V, F = 96500 C mol–1, DrG° = –nFE°cell, DrG° = – 2 × 1.1 × 96500 = – 212300 J mol–1, = – 212.3 kJ mol–1, 14. The substance which conducts electricity by ions present, in solution is called electrolytic conductor e.g., NaCl solution., Substances which conduct electricity in solid state are called, electronic conductors. These are made up of metals. e.g.,, Cu, Zn, Al. (Electrolytes are electrolytic conductors while, electrodes are electronic conductors)., 15. The device which converts the chemical energy liberated, during the chemical reaction to electrical energy is called, electrochemical cell., If external potential applied becomes greater than E°cell of, electrochemical cell then the cell behaves as an electrolytic, cell and the direction of flow of current is reversed., 16. With time, concentrations of the electrolytic solutions, change. Hence, their electrode potentials change when the, electrode potentials of the two half-cells become equal, the, cell stops working., 17. (i) Electrolyte A is a strong electrolyte while electrolyte, B is a weak electrolyte., (ii) For electrolyte A, the plot becomes linear near high, dilution and thus can be extrapolated to zero concentration, to get the molar conductivity at infinite dilution., For weak electrolyte B, Lm increases steeply on dilution and, extrapolation to zero concentration is not possible. Hence,, molar conductivity at infinite dilution cannot be determined., 4+, –, 18. At anode : Sn2+, (aq) → Sn (aq) + 2e ] × 5, –, +, –, At cathode : MnO4(aq) + 8H (aq) + 5e →, Mn2+, (aq) + 4H2O(l)] × 2, Net cell reaction :, +, 2+, 4+, 2MnO–4(aq) + 5Sn2+, (aq) + 16H (aq) → 2Mn (aq) + 5Sn (aq) , , + 8H2O(l), E°cell = E°cathode – E°anode = 1.51 V – 0.15 V = 1.36 V, Since, cell potential is positive therefore the reaction is, product favoured., , 19. C = 0.001 M, k = 4 × 10–5 S cm–1,, L∞m = 390 S cm2/mol, κ ×1000, Lmc =, C, Substituting the values,, Lmc =, , 4 × 10 −5 × 1000, = 40 S cm2/mol, 0.001

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a=, , Λ cm, , ∞, Λm, , =, , 40, = 0.10256 ≈ 0.103, 390, CH3COO– + H+, , CH3COOH, c, c (1 – a), , 0, ca, , 0, ca, , Ka =, , [CH3COO− ] [H+ ] cα ⋅ cα cα2, =, =, c (1 − α ) 1 − α, [CH3COOH], , Ka =, , 0.001(0.103)2 1.061 × 10 −5, –5, =, (1 − 0.103), 0.897 = 1.18 × 10, , 20. A(s) + B2+, A2+, (aq), (aq) + B(s), Here, n = 2, using formula,, 0.059, log K c, E°cell =, n, , 21. The reducing power increases with decreasing, value of electrode potential. Hence, the order is, Ag < Cu < Fe < Cr < Mg < K., 22. (i) Ions with higher reduction potentials are strong, oxidising agents while lower reduction potentials are strong, reducing agents., (ii) The electrode with higher electrode potential (E°) acts, as cathode while with lower electrode potential will act as, anode., (iii) Predicting the feasibility of redox reaction., (iv) Predicting the capability of metal to evolve H2 gas from, acid., –, , + 3e ] × 2, 23. At anode, At cathode : Ni + 2e → Ni(s)] × 3, 3+, Cell reaction : 2Al(s) + 3Ni2+, (aq) → 2Al (aq) + 3Ni(s), Applying Nernst equation to the above cell reaction,, 3+ 2, , 0.0591 [Al ], log 2+ 3, 2×3, [Ni ], , , , , Now, E cell, = E Ni, − E Al, 2+, 3+, /Ni, /Al, , = – 0.25 – (–1.66) = 1.41 V, \, , 0.0591 (10 −3 )2, E cell = 1.41 −, log, 6, (0.5)3, 0.0591, log (8 × 10 −6 ), 6, 0.0591, = 1.41−, ( −5.09), 6, = 1.41 + 0.050 = 1.46 V, , = 1.41 −, , −1, , cell constant 129 m, = 0.248 S m–1, =, 520 Ω, R, Concentration, C = 0.02 mol L–1, = 1000 × 0.02 mol m–3 = 20 mol m–3, , 0.059, log10, 2, E°cell = 0.0295 V, , , E cell = E cell, −, , 25. Resistance of 0.1 M KCl solution R = 100 W, Conductivity k = 1.29 S m–1, Cell constant G* = k × R = 1.29 × 100 = 129 m–1, Resistance of 0.02 M KCl solution, R = 520 W, Conductivity, k =, , E°cell =, , : Al(s) → Al3+, (aq), 2+, –, , 24. (i) The cell reactions are :, –, Zn(s) → Zn2+, (aq) + 2e (Anode), –, Cu2+, (aq) + 2e → Cu(s) (Cathode), Net reaction :, 2+, Zn(s) + Cu2+, (aq) → Zn (aq) + Cu(s), (ii) E°cell = E°cathode – E°anode = 0.34 V – (– 0.76 V), = 1.10 V, (iii) Copper electrode will be positive on which reduction, takes place., , −1, κ 0.248 S m, =, C 20 mol m−3, = 0.0124 S m2 mol–1, , Molar conductivity, Lm =, , 26. For half cell reaction,, +, –, 3+, Cr2O2–, 7(aq) + 14H (aq) + 6e → 2Cr (aq) + 7H2O(l), , −, E cell = E cell, , 0.0591, [Cr3+ ]2, log, n, [Cr2O27− ][H+ ]14, , Given, E°cell = 1.33 V, n = 6, [Cr3+] = 0.2 M, +, –4, [Cr2O2–, 7 ] = 0.1 M, [H ] = 1 × 10 M, E cell = 1.33 −, = 1.33 −, , 0.0591, (0.20)2, log, 6, (0.1) (10 −4 )14, 0.0591, log (4 × 1055 ), 6, , = 1.33 −, , 0.0591, [log 4 + log 1055 ], 6, , = 1.33 −, , 0.0591, [log 4 + 55 log 10], 6, , 0.0591, [0.602 + 55], 6, = 1.33 – 0.548 = 0.782 V, = 1.33 −, , 3, 27. Al2O3 (2Al3++ 3O2–) → 2Al + O2, n = 6e–, 2, 2, 4, 2, \, Al2O3 → Al + O2, n = × 6e– = 4e–, 3, 3, 3, DG = 960 × 1000 = 960000 J, Now, DG = –nFEcell, ∆G −960000, E°cell = –, = –2.487 V, =, nF 4 × 96500

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Minimum potential difference needed to reduce Al 2O 3 is, –2.487 V., 28. At anode : Ni → Ni2+ + 2e–, At cathode : [Ag+ + e– → Ag] × 2, Cell reaction : Ni + 2Ag+ → Ni2+ + 2Ag, E°cell = E°cathode – E°anode, = E°Ag+/Ag – E°Ni2+/Ni = 0.80 V – (– 0.25) V, E°cell = 1.05 V, 0.0591, , E cell, =, log K c, n, E  × n 1.05 × 2, log K c = cell, =, 0.0591 0.0591, logKc = 35.53, Kc = antilog 35.53 = 3.38 × 1035, 29. E°cell = E°cathode – E°anode = –0.403 – (–0.763) = 0.36 V, DrG° = –nFE°cell = –2 × 96500 × 0.36, = –69480 J = –69.48 kJ, Using formula, log K c =, , , nE cell, 2 × 0.36, =, = 12.20, 0.059, 0.059, , Kc = antilog 12.20 = 1.58 × 1012, Λ eq, 30. (a) Degree of dissociation, a = ∞, Λ eq, 15.8, \ a=, = 0.04514, 350, (b) For monobasic acid, HA, H+ + A–, Cα2, = Cα2, (1 − α ), As a < < < 1 hence (1 – a) ≈ 1, \ K = 0.05 × (0.04514)2 ⇒ K = 1.019 × 10–4, K=, , 31. Given : Diameter = 1 cm, length = 50 cm, R = 5.5 × 103 ohm, M = 0.05 M, r = ? k = ? Lm = ?, Area of the column,, , 32. The reaction is, Zn(s) + 2Ag+(aq) → Zn2+, (aq) + 2Ag(s), Cell can be represented as, +, Zn | Zn2+, (aq) || Ag (aq) | Ag, , (i) The zinc electrode is negatively charged (anode) as it, pushes the electrons into the external circuit., (ii) Ions are the current carriers within the cell., (iii) The reactions occurring at two electrodes are :, –, At zinc electrode (anode) : Zn(s) → Zn2+, (aq) + 2e, At silver electrode (cathode) : Ag+(aq) + e– → Ag(s), 33. A chemical cell is a galvanic cell in which electrical, energy produced is due to chemical changes occurring within, the cell and no transfer of matter takes place. It involves the, use of two different electrode dipped in solutions of different, electrolytes., A concentration cell is a galvanic cell in which electrical, energy is produced due to physical change involving transfer, of matter from one part of the cell to the other. It involves, the use of the same electrodes dipped in solutions of the, same electrolyte with different concentrations (or electrodes, of different concentration dipped in the same solution of the, electrolyte)., 34. The given cell may be represented as, Cu(s) |Cu2+ (0.10 M)|| Ag+ (C)| Ag(s), E°cell = E°cathode – E°anode = 0.80 V – 0.34 V = 0.46 V, Ecell = E°cell –, , 2, , 1, 3.14 2, a = πr 2 = 3.14 ×  cm =, cm, 2 , 4, Resistivity,, a, 3.14 cm2, ρ = R ⋅ = 5.5 × 103 ohm ×, = 86.35 ohm cm, l, 4 × 50 cm, 1, Again, conductivity, κ =, ρ, 1, =, = 1.158 × 10 −2 ohm−1 cm−1, 86.35, 103, and molar conductivity, Λ m = κ ⋅, M, 103, = 1.158 × 10 ohm cm ×, 5 × 10 −2, –1, 2, –1, = 231.6 ohm cm mol, −2, , −1, , −1, , or, , 0.0591 [Cu2+ ], log + 2, 2, [Ag ], , 0.422 V = 0.46 V –, , 0.0591, 0.1, log + 2, 2, [Ag ], , – 0.038 V = – 0.0295 log, or, or, \, , 0.1, [Ag+ ]2, , − 0.038, 0.1, =, = 1.288, + 2, −, 0.0295, [Ag ], 0.1, = antilog 1.288 = 19.41, [Ag+ ]2, 0.1, [Ag+]2 =, = 5.1519 × 10–3, 19.41, log, , [Ag+] = 7.1 × 10–2 M, 35. Given : V = 100 cm3, M = 0.025 M, R = 520 ohm, G° = 153.7 m–1 = 1.537 cm–1, Lm = ?

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0.16 = 0.0591 [log [H+]c – log [H+]a], 0.16 = 0.0591 [pHa – pHc], 0.16 = 0.0591 [pHa – 1], 0.16, = 2.70, or pHa − 1 =, 0.0591, or pHa = 2.70 + 1 = 3.70, 39. (a) E°cell = E°cathode – E°anode = – 0.403 – (– 0.763), = 0.36 V, The net cell reaction is, 2+, Zn(s) + Cd2+, (aq) → Zn (aq) + Cd(s), Here, value of n = 2, , or, or, , 1, 1, = 1.537 cm−1 ×, R, 520 ohm, = 2.95 × 10–3 ohm–1cm–1, , κ =G×, , Again, Λ m =, =, , κ × 103, M, 2.95 × 10 −3 ohm−1 cm−1 × 103, 0.025 mol cm−3, , Lm = 118.0 ohm–1 cm2 mol–1, 36. k = 1.29 × 10–2 ohm–1 cm–1, 1, k=, × Cell constant, R, ⇒ Cell constant = k × R, = 1.29 S m–1 × 85 W = 109.65 m–1, For second solution,, 1, k = 1 × Cell constant =, × 109.65 m–1, 96 Ω, R, = 1.142 W–1m–1, 1.142 Ω −1m−1 × 1000 cm3, Lm = κ × 1000 =, M, 0.052, Λm =, , 1.142 Ω −1cm−1 × 10 −2 × 1000 cm3, 0.052 mol, , = 219.62 S cm2 mol–1, 2+, 0.0591 [Mg ], log 2+, n, [Cu ], 0.0591 0.001, = 2.71 −, log, = 2.73955 V, 2, 0.01, (i) If external opposing potential is less than 2.71 V, then current will flow from Cu to Mg., (ii) If external opposing potential is greater than 2.71 V, then current will flow in opposite direction i.e. from Mg, to Cu., , 37. E cell = E °cell −, , 38. (a) The given cell may be represented as, Cu(s) | Cu2+ || Cl2 | Cl–, (i), , 0.0591 [Zn2+ ], log 2+, 2, [Cd ], 0.0591 0.0004, = 0.36 −, log, 2, 0.2, 0.0591, = 0.36 −, ( −2.69) = 0.36 + 0.08 = 0.44 V, 2, \ DG = – nFEcell = – 2 × 96500 × 0.44, = – 84920 J/mol, 2+, –, (b) Cu (aq) + 2e → Cu(s), 0.059, [Cu], , E Cu2+ /Cu = E Cu, −, log 2+, 2+, /Cu, 2, [Cu ], 0.059, 0.059, 1, = 0.34 −, log, = 0.34 −, log 10, 2, 0.1, 2, 0.059, = 0.34 −, × (1) = 0.34 – 0.0295 = 0.3105 V, 2, When the concentration of Cu 2+ ions is decreased, the, electrode potential for copper decreases., , −, E cell = E cell, , 40. (a) Given : Leq = 1.4 mho cm2 eq–1,, L∞eq = 391 mho cm2 eq–1, a = ?, Ka = ?, Using formula, α =, , Ka =, , ∆r G ° = − nFE ° = − 2 × 96500 C × 1.02 V = 196.86 kJ, 0.0591, log K, (iii) E°Cell =, n, 2 × 1.02, K = antilog, = antilog (34.51), 0.0591, K = 3.236 × 1034, (b) The given cell may be represented as, Pt, H2 (1 atm) | H+ (pH = ?) || H+ (pH = 1) | H2 (1 atm), [H+ ]c, Using formula, E cell = 0.0591 log, 1, [H+ ]a, , =, , 1.4 mho cm2 eq−1, , ∞, Λ eq, 391 mho cm2 eq−1,           = 0.00358, , , E cell, = E c − E a = ( +1.36 V) − ( + 0.34 V) = 1.02 V, , (ii), , Λ eq, , (0.00358)2 × 0.0128, α2C, =, 1− α, 1 − 0.00358, , 1.64 × 10 −7, = 1.64 × 10–7, 0.99642, (b) Given : Λ°eq (CH3COONa) = 83 mho cm2 eq–1, Λ°eq (NaCl) = 127 mho cm2 eq–1, Λ°eq (HCl) = 426 mho cm2 eq–1, Λ°eq (CH3COOH) = ?, Using Kohlrausch law of independent migration of ions, Λ°eq (CH3COOH) = Λ°eq (CH3COONa) + Λ°eq (HCl) – Λ°eq (NaCl), or Λ°eq (CH3COOH) = 83 + 426 – 127, = 382 mho cm2 eq–1, , , , =