Write Nernst equation and calculate the e.m.f of the following cell at 298 K:
`Zn|Zn^(2+)(0.01M)||Fe^(2+)(0.005M)|Fe`
Given that:
`(Zn^(2+)//Zn)=-0.763V and E_(Fe^(2+)//Fe)^(@)=-0.44V" "log2=0.3010`.

Explanation

Cell representation is
`Az|Zn^(2+)(10^(-2)M)||Fe^(2+)(5xx10^(-3)M)De`
Cell reaction are,
Oxidation : `Zn(s) to Zn^(2+)(aq)+2e^(-)`
Reduction : `Fe^(2+)(aq)+Ze^(-) to Fe(s)`
Overall : `Zn(s)+Fe^(2+)(aq)toFe(s)+Zn^(2+)(aq)`
Here `n=2, and Q_(C)=([Zn^(2+)])/([Fe^(2+)])`