A cell `Cu|Cu^(2+)||Ag^(o+)|Ag` inintially contains `2 M Ag^(o+)` and `2 M Cu^(2+)` ion in `1L` solution each . The change in cell potential after it has supplied `1A` current for `96500 s` is

Options

A. `-0.003V`
B. `-0.02V`
C. `-0.04V`
D. None of these

Answer

b

Explanation

`Cu|Cu^(2+)(2M)||Ag^(o+)(2M)|Ag`
Number of faradays supplied `=(It)/(96500)=(1xx96500)/(96500)=1`
After cell has supplied `1.0F` electricity` :`
At anode `:Cu rarr Cu^(2+)+2e^(-)`
`implies 22F` electricity `-=1 mol Cu^(2+)` produced .
`implies 1.0F` electricity `-=0.50 mol Cu^(2+)` produced
`implies [Cu^(2+)]_(new)=(2xx1+0.50)/(1)=2.5M`
At cathode `:Ag^(o+)+e^(-) rarr Ag`
`implies 1F` electricity `-=mol e Ag ^(o+) ` used
`implies [Ag^(o+)]_(new)=(2xx1-1.0)/(1)=1.0M`
`E_(cell)=E^(c-)._(cell)-(0.059)/(2) log .([Cu^(2+)])/([Ag^(o+)]^(2))`
`implies(E_(cell))_(1)=E^(c-)._(cell)-(0.059)/(2)log.(2)/(2^(2))`
and `(E_(cell))_(2)=E^(c-)._(cell)-().059)/(2)log .(2.5)/(1^(2))`
`implies` Change in `E_(cell)=(E_(cell))_(2)-(E_(cell))_(1)`
`=-(0.059)/(2)log 2.5-(0.059)/(2)log 2`
`=-(0.059)/(2) log 5=-0.02V`