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Electrochemistry

Question
CBSEENCH12006017

Calculate the cell emf at 25° C for the following cell:
Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s)
[Given E°Ni2+/Ni = – 0 25 V, E° Cu = + 0.34 V, 1 F = 96500]
Calculate the maximum work that can be accomplished by operation of this cell.

Solution

Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s)
At anode Ni(s) → Ni2+ (aq) + 2e
At cathode Cu2+ + 2e → Cu(s)
Net cell reaction
Nis) +Cu2+(aq)Ni2+(aq) +Cu(s)Ecell =Ecell0 -0.05912log [Ni2+][Cu2+]Ecell0 =ECu2+Cu0 -ENi2+Ni0+0.34V-(-0.25V) =0.59Vthus 0.59V -0.05912log 0.010.10.59V -0.05912log1100.59V -0.05912×-(1)=0.591V +0.295 =0.6195VG =-nE°F- 2×0.6195V×96500=-11956.500JG =-119.5635KJG=WMAX= 119.5635KJ