A cell, `Ag|Ag^(o+)||Cu^(2+)|Cu` , initially contains `1 M Ag^(o+)` and `1M Cu^(2+)` ions. Calculate the change in the cell the potential after the passage of `9.65A` of current for `1h.`

Answer

`-0.010V`

Explanation

For a cell `Ag|Ag^(o+)||Cu^(2+)|Cu`
Cell reaction is `:`
`Ag(s)+(1)/(2) Cu^(2+)(aq)rarr Ag^(o+)(aq)+(1)/(2) Cu(s)(n=1)`
As we know that
`m=ZxxIxxt=(Ew)/(96500)xxIxxt`
`(W)/(Ew)` or gram equivalent of `Cu^(2+)=(1)/(96500)xxIxxt`
`=(1)/(96500)xx9.65xx60xx60=0.36`
Decrease in the concentration of copper ion is `(0.36)/(2)M`.
Remaining concentration of coper `=1-0.18=0.82M`
So, increase in the concentration of silver ion is `0.36M`
Now concentration of silver ion `=1+0.36M`
On applying Nernst equation
`E_(cell)=E^(c-)._(cell)-(0.0591)/(n)log.([Ag^(o+)])/(sqrt([Cu^(2+)]))`
or `E^(c-)._(cell)-E_(cell)=DeltaE=(0.0591)/(n)log.([Ag^(o+)])/(sqrt([Cu^(2+)]))`
`:.` Change in cell potential,
`DeltaE=(0.0591)/(1)log.(1.36)/(sqrt(0.82))=-0.010V`