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

At anode, `Cu rarr Cu^(2+) + 2e^(-)`
At cathode,
`ul(2Ag^(+) + 2e^(-) rarr Ag)`
`ul(Cu+2 Ag^(+) rarr Cu^(2+) + 2Ag)`
According to Nernst equation, `E= E^(@)- (0.059)/(2) "log" ([Cu^(2+)])/([Ag^(+)]^(2))`
`E= 0.46 - (0.059)/(2) "log" ([Cu^(2+)])/([Ag^(+)]^(2))`
`=0.46 - (0.059)/(2) (log 13 + 6 log 10)`
`=0.46- (0.059)/(2) xx 7.11`
= 0.25V