Write Nernst equation and calculate the e.m.f. of the following cell at 298 K:
`Cu(s)|Cu^(2+)(0.130M)||Ag^(+)(1.0xx10^(-4)M)|Ag(s)`
Given that: `E_(Cu^(2+)//Cu)^(@)=+0.34V and E_(Ag^(+)//Ag)^(@)=+0.80V`
(log 0.130=-1.1139).

Explanation

`Cu(s)+2Ag^(+)(aq) to Cu^(2+)(aq)+2Ag(s)`
Oxidation : `Cu(s) to Cu^(2+)(aq)+2e^(-)`
Reduction : `Ag^(+)(aq)+e^(-) to Ag(s)xx2`
Overall : `Cu(s)+2Ag(aq) to Cu^(2+)(aq)+2Ag(s)`
Here n=2
Cell representation is,
`Cu(s)|Cu^(2+)(0.130M)||Ag^(+)(1.0xx10^(-4)M)|Ag(s)`
Nernst equator (at 298 K) is
`E=E^(@)(0.0591V)/(n)logQ_(C)`.
`=0.46V-(0.0591V)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))`
`=0.46V-0.0295V" log "(0.130)/((1xx10^(-4))^(2))`
`=0.46V-0.0295Vlog1.3xx10^(7)`
> `=0.46V-0.0295xx7.1139V`
`=(0.46-0.21)V=0.25V`