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SUNSHINE COACHING TUNDLA, M.P. ROAD, KANAK PALACE TUNDLA, SUN SHINE, MOB. 9458022469, 8923508689, Coaching Tundla, Detailed Solutions, 8., (i), The cell reactions are :, The device which converts the chemical energy, liberated during the, electrical energy is called electrochemical cell., If external potential applied becomes greater than, E al, 1., chemical reaction, Zng, Zn + 2e (Anode), to, Curag + 2e, Net reaction :, > Cu (Cathode), of electrochemical cell then the cell behaves, as an electrolytic cell and the direction of flow of, current is reversed., Representation of the galvanic cell for the given, Zn + Cuag), (ii) E°cell - E°right -Elef = 0.34 V-(-0.76 V)= 1.10 V, (iii) Copper electrode will be positive on which, reduction takes place., Zn) + Cu, 2., reaction is :, 9., The reaction is, Zn | Zn || Cug)| Cu, Zn + 2Agian) → Zn) + 2Ag), Cell can be represented as, Zn | Znap || Agian) Ag, Anode, Salt, Cathode, bridge, 3., 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, Flow of, Electrons, Current, Zn electrode, (-), the electrical neutrality of the solutions in the two, Ag electrod, (Calvode), (Anode),, KCI salt bridge, half cells., Because standard electrode potential of, Tit' /Ti" is less than that of Fe*/Fe* so, it cannot, oxidise Fe" to Fe"., At anode : Sn+, 4., Zn, Ag, solution, NO,, solution, Oxidation, Reduction, 7.n Zn + 2e, (i) The zinc clectrode 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 → Zn) + 2e", At silver electrode (cathode): Agiaa) + e > Ags, 5., (ag), Snag) + 2e ] x 5, Agan e- Ag, At cathode : MnOtag) + 8H',, + 5e →, (aq), Mnug) + 4H,Ol x 2, (aq), Net cell reaction :, 2MnOa, + 16Hag), 2Mn (ag), E° cell = E°cathode- Eanode = 1.51 V - 0.15 V = 1.36 V., Since, cell potential is positive therefore the reaction, + 5Snag) + 8H,On, 10. (a) Given : E°cell = 2.7I V, For the reaction,, is product favoured., Mg) + Cuag), n = 2, A,G° = ?, Mg, (aq) + Cu, The reducing power increases with decreasing, value of electrode potential. Hence, the order is, Ag < Cu < Fe < Cr < Mg < K., 6., Using formula, A,G = - nFE, cell, A,G =-2 x 96500 Cmolx 2.71 V, A,G° = 523.03 kJ mol", 11. Ag + =ーA+ B), Here, n = 2, 7., 'The cell reaction is, 2+, Zna + 2Agag) Zna) + 2Ags), The cell is represented as, Zn | Zn) || Agag) | Ag, (i) Anode i.e., zinc electrode will be negatively, charged., (ii) At anode :, Zn > Zn2+, At cathode:, Ag aa) + e Aga (Reduction), or, using formula,, 0.059, log K., Eell =, (ag) + 2e" (oxidation), 0.059, cell =, -log 10, 2, ..-00245 V, (iii) lons are the carriers of current within the cell, SUN SHINE COACHING CENTER, TUNDLA, 9458022469, 1

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12. Here n = 2, E°., F = 96500 C mol"!, A,G° = -nFE°ell, A,G° = - 2 x 1.1 x 96500 = - 212300 J mol, = - 212.3 kJ mol-, 13. Refer to answer 12., cell = 1.1 V, 1.98 V = Ecell -, 6., 0.0591 1og10-, 0.0591, x2, 6, 1.98 V = E'cell, 0.0591, E°cell = 1.98 +, = 1.99 V, 6, 14. 'Ihe electrode reaction written as reduction, reaction is, 19. The cell reaction is, Zn2* + 2e Zn (n = 2), Applying Nernst equation, we get,, 0.0591, Fe + 2Ha) – Fe, + He, (ag), E° cell = 0.00 - (-0.44) = 0.44 V, (Fe2+], log, [H*]?, EIZn Zn /IZn, = E°., 0.0591, log, Ecell = Ecell-, 2, As 0.1 M ZnSO, solution is 95% dissociated, this, 0.0591, 0.001, means that in the solution,, = 0.44 –, log, 2, (0.01)?, (Zn"] =, 95, x 0.1 = 0.095 M, = 0.44 - 0.02955 = 0.41045 V, 100, 0.0591, -log-, 2, 1, 20. The cell reaction is, Zn) + 2H'ag) -> Znag) + H2e), - 0.00 - (- 0.76) = 0.76 V, Ezn?+/Zn = - 0.76 -, 0.095, - - 0.76 - 0.02955 (log 1000 - log 95), cell, = - 0.76 - 0.0295 (3 - 1.9777) = - 0.79021 V, 15. Nia + 2Aga → Nia + 2Ag,s» E = 1.05 V, [Zn2*], log, 0.0591, E = E"cel, (ag), Here, n - 2, nE cell, 0.0591, (0.001), -log, (0.01)2, Using formula, log K., = 0.76 -, 0.059, 2 x1.05, = 0.76 - 0.02955 = 0.730 V, or log K. =, = 35.5932, 0.059, K. = antilog 35.5932 or K, = 3.92 x 10*, Again, AG = -nFE"edl, AG° = -2 x 96500 x 1.05 = - 202650 J, 21. At anode : Ni > Ni? + 2e, At cathode: [Ag' +e Ag] x 2, Cell reaction : Ni + 2Ag" > Ni* + 2Ag, E cell - E cathode - Eanode, AG° = -202.65 kJ, E, Ag*1Ag", E'N?, = 0.80 V - (-0.25), Ni, 16. E= E rishi - Elef = - 0.44 (-0.74) 0.30 V, E°ell, 1.05 V, 0.0591, log-, 6, (0.01)3, (0. 1)?, Ecel, 0.0591, log K., Esd = 0.30 –, = 0.30 - 0.0394 = 0.26 V, Ecell Xn, 1.05 x 2, log K., %3D, 17. Refer to answer 16., 0.0591, 0.0591, logk, = 35.53, K, = antilog 35.53 3.38 x 105, 18. Given cell,, 2Al + 3Cu*" (0.01 M) 2Al"(0.01 M) + 3Cu,a, Eell = 1.98 V, Fe = ?, Using Nernst equation at 298 K, 22. E ell = +0.80 V -0.77 V = +0.03 V, A,G° = -nFE",cell = -1 x 96500 x 0.03, 0.0591,, (AI P, = -2895 J mol- = -2.895 kJ mol, AG = - 2.303 RT log K., - 2895 = -2.303 x 8.314 x 298 x log K, or log K. = 0.5074, Eel = E" cel -, log, (Cu**, [10-2)², 0.0591, log, 6, (10-23, 1.98 V = Ecdi -, K. = Antilog (0.5074) = 3.22, SUN SHINE COACHING CENTER, TUNDLA, 9458022469, 2

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23. The cell reaction can be represented as:, Mg + Cu) → Mga) + Cu), Given: E = +2.71 V, T = 298 K, According to the Nernst equation :, (103)2, Esel = 0.46 V - 0.0295 log, 10-1, = 0.46 - 0.0295, log 10-5= 0.46 - 0.0295 (-5), = 0.46 + 0.0295 x 5 = 0.6075 V, 0.0591, log, 27. The cell reaction in button cell :, Zn + Ag,O + H;O0 Zn) +2Ag +20Hag), (i) Calculation of E° cel, Ecel = Ecell, (Cuap), 2+, 0.0591, 0.1, = 2.71-, log-, 0.01, = 2.6805 V, 2, Reactions:, Anode : Zn, Cathode :, Ag,O9 + H,Oy + 2e > 2Ag +20Hn) + 2e, n = 2, > Znao) + 2e, 3, 24. Al,O, (2Al+ 30-)- 2Al +, O, n = 6e, 2, 4, → Al + O,, x 6e = 4e, n =, E° e = E°,, = E'AR,O/Ag - E°Zn2+rZn, E anode, cel, cathode, AG = 960 x 1000 = 960000, Now, AG = -nFEcel, = + 0.80 - (- 0.76) V = 1.56 V, (ii) Calculation of A,G°, AG, E° = --, -960000, A,G° = -nFE°cl, = - 2 x 96500 Ç mol- x 1.56 V, nF, 4x96500, Eel = -2.487 v, Minimum potential difference needed to reduce, Al,O, is -2.487 V., 25. Fe| Fe"(0.001 M) || H*(1 M) |H2(1 bar) | Pt), = - 301080 C V mol, =- 301080 J mol =- 301 kJ mol, 28. At anode : Al), Al'ag) + 3e"] x 2, At cathode : Ni* + 2e"Ni] x 3, Cell reaction : 2Al) + 3Ni) → 2Ala) + 3Nia), Applying Nernst equation to the above cell reaction,, 0.0591, Reactions :, Anode : Fe- Fe?, Cathode : 2H,, + 2e, (aq), + 2e > H2g}, LA, Cell reaction : Fe, + 2Ha7), » Fe) + H2g), Ecel = Ecell -, log, 2x3, n - 2., Now, Ecell = EN?+ /Ni, - E", Al3+ /Al, Using Nernst equation at 298 K, 0.0591, log, 2, [Fe* ]× PH,, = - 0.25 V - (-1.66) = 1.41 V, Ecell = Eel, (H", (10-3)2, log, 6, 0.0591, .. Ecell = 1.41 V-, (0.5), For the given cell,, E° cell = E°cathode - E'anode = E, 0.0591, log (8 x10-6), = 1.41 V -, = 0 - (-0.44) = + 0.44 V, Given (Fe*] = 0.001 M; (H*] = 1 M; PH, =1 bar, Putting in Nernst equation, 0.0591, = 1.41 V-, (5.09), = 1.41V + 0.050 V = 1.46 V, 0.001x1, Ecel = 0.44 - 0.0295 log, 12, - 0.44 - 0.0295 log 103, = 0.44 - ((0.0295) x (-3)], 29. 2Fe + 2e > 2Fe" and 21" 1, + 2e, Hence, for the given cell reaction, n 2, A,G° =-nFE=-2x96500x0.236=-45.55kJ mol, A,G° = - 2.303 RT log K., = 0.44 + 0.0885 - 0.53 V, A,G°, 2.303RT, 26. The cell may be represented as, log K. =, or, Ag, JAg' (10 M)||Cu (10- M)| Cu, -45.55 kJ mol-1, = 7,983, [Ag*, log, 0.0591, 2.303x 8.314 x 10kJ K-'mol"x 298K, K. = Antilog (7.983) = 9.616 x 10, Using formula Ecell = E° cell -, (Cu2+], SUN SHINE COACHING CENTER, TUNDLA, 9458022469

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30. For half cell reaction,, Cr,Oag) + 14Ha) + 6e" → 2Crag), 0.0591, log, 2, q) + 7H,O, Ecell = E"cell-, [Ag*, 0.0591, log, Ecell = E°cell -, 0.1, log, 0.0591., or, 0.422 V = 0.46 V -, [Ag*, Given, E°celt = 1.33 V, n = 6, (Cr*] = 0.2 M, 0.1, [Cr.O]= 0.1 M, [H*] = 1 x 10M, - 0.038 V = - 0.0295 log, [Ag*?, (0.20)?, log, 0.0591, Ecel, = 1.33 V, -0.038, 0.1, or, log, [Ag*j?, (0.1) (10-4)14, = 1.288, -0.0295, 0.0591, = 1.33 V-, 6, log (4 x 105), 0.1, or,, antilog 1.288 = 19.41, [Ag*P, 0.0591, [log 4 + log 10551, 6., 0.1, = 5.1519 x 10, - 1.33 V -, |Ag J =, 19.41, 0.0591, = 1.33 V -, [log 4 + 55 log 10], 6., [Ag"] = 7.1 x 10-2 M, 34. Refer to answer 26., 0.0591, = 1.33 V -, [0.602 + 55], 35. Refer to answer 28., 36. E°cel=Eathode-Enude-0.40 V-(-0.44V)%3D0.04V, = 1.33 V - 0.548 V= 0.782 V, 31. (a) Oxidation half reaction :, Using formula, log K, =, at 298 K, 0.0591, Zn, >Znfre + 2e, (ac), 2x0.04 V, Reduction half reaction :, or, K. antilog, 0.0591 V, Cu o) + 2e -> Cu, or, K. = antilog 1.356 = 22.38, (b) Ee=0.34 - (-0.76) = 1.10 V, 37. The cell may be represented as, 0.0591, E= E - log, 2, Zng Znan(1 M) | AgiaplAgto), Zn2+, 2+, [Cu²*], E°cel = Ecathode - Einode =0.80 V - (-0.76 V) = 1.56 V, Using formula,, 0.0591., 2, log, 0.5, El = 1.10 -, [Zn2*], log, 2, 0.0591, Ecell = E° cell, 0.0591, 2, log, 2, [Ag f, = 1.10 -, 0.5, 0.0591., 1, log, 2, 0.059, or, 1.48 = 1.56-, - 1.10 -, 2, x 0.6021, [Ag*P, or, log [Ag"] = - 1.354, or, [Ag'] antilog (- 1.354), or, (Ag'] = 4.426 x 10 2 M, 38. T= 273 + 25 C= 298 K and n 6, = 1.10 - 0.0177 = 1.0823 V, 32. AG"= -RT In K = - 2.303 RT log K., -- 212300 = - 2.303 x 8.314 x 298 x log K,, 212300, L° cell = Esathude - E°anode = -0.40 (-0.74) = 0.34 V, A,G° =- nFE",, Again A,G = - 2.303 RT log K, - 196860 = - 2.303 x 8.314 x 298 x log K, or log K, =, 37.2074, 2.303x8.314 x 298, °cell=-6x96500 x0.34 -196860Jmol, K, = Antilog 37.2074 1.6 x 10"., 33. The given cell may be represented as, Cu, Cu (0.10 M)|| Ag (C)| Ag6), log K = 34.5014, K = antilog 34.5014 = 3.172 x 1034, Ecel = Ethode -ode =0.80 V - 0.34 V = 0.46 v, SUN SHINE COACHING CENTER, TUNDLA, 9458022469

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42. Eell = E cathode, 39. (i) The cell can be represented as, Cra| Cr || Feag Fea, (ii) Ees = Ecathode - Eanode, -- 0.44 - (-0.74) = -0.44 + 0.74 0.30 V, (iii) Refer to answer 16., anode, = - 0.403 - (- 0.763) = -0.403 +0.763 = 0.360 V, 2.303RT, Znagl, * Eel - E, log, nF, 40. Mg, | Mg*(0.001 M) || Cu*(0.0001 M) | Cu, (ag), Reactions :, Mg) > Mgaa) + 2e, 0.059, -log, Anode :, Ecel = E"cell, Cathode :, Cu*,, (ag), + 2e, → Cu s), Net cell reaction : Mg + Cua), Mgag) + Cu,, 2.303RT, n= 2, = 0,059, F, Using Nernst equation:, [Mg+], 0.059, -log, 0.1, = 0.36 -, 0.01, 0.059, -log 10, 2, 2.303RT, = 0.36 -, log, nF, [Cu*1, E = E°, sell -, = 0.36 - 0.0295 0.3305 0.33 V, A,G° = - 1FE° cell, For the given cell, E°,dl =, = 0.34 V - (-2.37 V) = 2.71 V, Given (Mg'] = 0.001 M, [Cu'] = 0.0001 M, Putting in Nernst equation at 298 K, E°, anode = E'Cu2+/Cu - EMg?+/Mg, cathode, = - 2 x 96500 x 0.36 = - 69480 | mol, = - 69.48 kJ mol, 43. The limiting molar conductivity of, an, 0.059, -log, 2, 0.001, E- - 2.71 V -, electrolyte is defined as its molar conductivity when, 0.0001, the concentration of the electrolyte in the solution, E = 2.71 – 0.0295 log 10 = 271 - 003 = 2.68 V, A,G = - nFE el, = - 2 x 96500 C molx 2.68, -- 517,240 J mol=- 517.24 kJ mol, approaches zero., Conductivity of an electrolyte decreases with dilution, beçause the number of current carryi, L.e., ions present per cm' of the solution becomes less, and less on dilution., -1, particles, 41. We have, Ni2+, 44. Kohlrausch's law of independent migration, of ions : It states that limiting molar conductivity of, 0.059, (ag), Ecel = E° cell-, log, 2+, an electrolyte can be represented as the sum of the, individual contributions of the anion and cation of, 0.01, 0.059, log, 2, 0.059 E, [Here n = 2], celi, 0.1, the electrolyte., If A°Na' and 2°cr are limiting molar conductivities, of the sodium and chloride ions respectively then, 0.059, -log, 0,059 = Eell, 0.059, 0.059 = E"cell, (- log 10), 2, the limiting molar conductivity for sodium chloride, is given by, 0.059, 0.059 = E'cell+, 2, 0.059, a E° et = 0.059-, 2, 0.059, E ell-, Kohlrausch's law helps in the calculation of degree of, dissociation of weak electrolyte like acetic acid., The degree of dissociation o is given by, = 0.0295 = 0.03, Am, Now Ecell = E calhode - E"anode, 0.03 = 0.34 - E°,, E Am, where A, is the molar conductivity and A°, is the, anode, Einode = 0.34 - 0.03 = 0.31 V, Hence, EN2N = + 0.31 V, limiting molar conductivity., SUN SHINE COACHING CENTER, TUNDLA, 9458022469