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//Cu^(2+) (0.130M)//Ag^(+) (1 xx 10^(-4) M)//Ag_((s))`
At anode, `Cu rarr Cu^(2+) + 2e^(-)`
At cathode,
`ul(2Ag^(+) + 2e^(-) rarr 2Ag)`
`ul(Cu+ 2Ag^(+) rarr Cu^(2+) + 2Ag)`
According to Nearest Equation,
`E= E^(@)- (0.0591)/(2) "log" ([Cu^(2+)])/([Ag^(+)]^(2))`
`E^(@)= E_("Cathode")^(@) -E_("Anode")^(@)`
`=0.80-0.34`
=0.46V
E.M.F= `0.46 - (0.0591)/(2) "log" ([Cu^(27)])/([Ag^(+)]^(2))`
`=0.46 = (0.059)/(2) "log" ([0.130])/([10^(4)]^(2))`
`=0.46 - (0.059)/(2) [log 13+6log 10]`
`-0.46- (0.059)/(2) xx 7.11`
=0.25V