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Electrochemistry

Question
CBSEENCH12005986

Calculate the emf of the cell Zn/Zn2+ (0.1 M) || Cd2+ (0.01 M) | Cd at 298 k. (given)
Zn2+/Zn = – 0.76 V and E°Cd2+/Cd = – 0.40 V).

Solution

Zn/Zn2+(0.1 M) || Cd2+ (0.01) | Cd

we have the reaction Zn(s) Zn2+ +2e-Cd2+ +2e- Cd(s)thus overall reaction is Zn(s) +Cd2+(aq)Zn2+(aq) + Cd(s)Ecell =Ecell0 - 0.05912log [Zn2+][Cd2+]= ECd2+Cd0 - EZn2+Zn0   - 0.05912log 0.10.01=[-0.40-(-0.76V)] -0.0592log 10= +0.36V -0.0295 =0.3305V