ElectrochemistryHard

Question

A cell Cu | Cu++ || Ag+ | Ag initially contains 2M Ag+ and 2M Cu++ ions in 1L electrolyte. The change in cell potential after the passage of 10 amp current for 4825 sec is :

Options

A.- 0.00738 V
B.- 1.00738 V
C.- 1.00738 V - 0.0038 V
D.none

Solution

Q = 10 × 4825 = 48250 C
no. of faraday = = 0.5
Ag     + Cu++ →  Ag+      +    Cu
              2.00             2.00
              2-0.25         2 + 0.50
Ecell = EoCell -
E1 = EoCell -
E2 = EoCell -
ᐃE = E2 - E1 = [log 1.41 - log 1.88] 
= [0.1492 - 0.2742]  = - × 0.125  = - 0.00738 V.

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