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.
no. of faraday =
= 0.5Ag +
Cu++ → Ag+ + Cu2.00 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.Create a free account to view solution
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