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. `(-0.0266V)`
C. `(-0.04V)`
D. None of these

Answer

B

Explanation

Givnen Cell : `Cu|Cu^2+ (2M)||Ag^(+) (2M)|Ag`
Number of faradays supplied = `(It)/(96500) = (1xx96500)/(96500) = 1`
After cell has supplied 1.0 F electricity :
At anode : `Cu rightarrow Cu^(2+) + 2e^(-) `
`Rightarrow` 2F electricity `equiv 1 mol e Cu^(2+)` Product
1.0 F electricity `equiv` 0.5 mole `Cu^(2+)` product `Rightarrow ([Cu^(2+)]) _(n ew) = (2xx1+ 0.50)/(1) = 2.5M`
At cathode : `Ag^(+) + e^(-) rightarrow Ag`
`Rightarrow1F` electricity `equiv` mole `Ag^(+)` used `Rightarrow([Ag^(+)])_("new") = (2xx1- 1.0)/(1) = 1.0 M`
usnig : `E_(cell) = C_(cell)^(@) =- (0.059)/(2) log _(10)([Cu^(2+)])/([Ag^(+)])^(2)`
`E_(cell) = E_(cell)^(@)-(0.059)/(2) log _(10) (2)/(2)^(2)` and `E_(cell) _ (2)= E_(Cell)^(@)-(0.059)/(2) log_(10) (2.5)/(2)` `Rightarrow` Change in `E_(cell) = E_(cell)_(2) - (E_cell)_(1)= (0.059)/(2) log _(10)2.5 - (0.059)/(2) log_(10)^(2)`
`=- (0.059)/(2) log_(10) 5 =- 0.0266V`