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Electrochemistry

Question
CBSEENCH12006138

Calculate the cell e.m.f. and ΔG for the cell reaction at 298 K for the cell
Zn(s) | Zn2+ (0.0004 M) || Cd2+(0.2 M) | Cd(s)
(Given E°Zn2+/Zn = 0.763 V, E°cd 2+ / cd = 0.403 Vat 298 K, F = 96500 C mol–1)

Solution

(i)    According to nernst equation:
          E = E°-0.0591nlogZn2+(aq)Cd2+ (aq)E°cell = E°Cd2+/Cd-E°Zn2+/Zn          = (0.403) - (-0.763) = 0.36 V
          [Zn+(aq)] = 0.0004 M,  [Cd2+(aq)] = 0.2 M,  n = 2
                        E=(0.36)-(0.0591 V)2log0.00040.2  = 0.36  - (0.0591 V)2×(-2.69990) = 0.36 V + 0.08 = 0.44 V
(ii)     
          G = -nFEcellE°cell = 0.44 V, n = 2 mol, F = 96500 c mol-1G = -2 (mol) × (96500 C mol-1) × (0.44 V)        = - 84920 CV = -84920 J.