The calculate the e.mf. Of the following cell at 298:
`Fe|Fe^(2+)(0.1M)||Ag^(2+)(0.1M)|Ag`
`E_((Fe^(2+)//Fe))^(@)=-0.44V,E_((Ag^(+)//Ag))^(@)=0.80V`.

Explanation

At anode, `Fe rarr Fe^(2+) + 2e`
At cathode, `2Ag^(+) + 2e^(-) rarr 2Ag`
`Fe + 2Ag^(+) rarr Fe^(2+) + 2Ag`
Our to Nernst, `E= E^(@)- (0.059)/(2) "log" ([Fe^(2+)])/([Ag^(+)]^(2))`
`E^(@)= 0.80 + 0.44=1.24V`
`E=1.24 - (0.059)/(2) "log" ((0.1))/((0.1)^(2))`
`E= 1.24 - (0.059)/(2) log 10`
=1.21V