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

Cell representation is,
`Fe|Fe^(2+)||Ag^(+)|Ag`
Cell reactions are
Oxidation: `Fe(s)toFe^(2+)(aq)+2e^(-)`
Reduction: `Ag^(+)(aq)+e^(-) to Ag(s)3xx2`
Overall:
`Fe(s)+2Ag^(+)(aq) to Fe^(2+)(aq)+2Ag(s)`
Here, n=2