Write The Nernst equation and calculate the e.m.f. of the following cell at 298K.
`Fe_((s))|Fe_((0.001M))^(2+)||H_((1M))^(+)|H_(2(1atm).Pt)`
Given `E_(Fe^(2+)//Fe)^(@)=-0.44V`

Explanation

`Fe(s)|Fe^(2+)(0.001M)||H^(+)(1M)|H_(2)(1atm)Pt`.
Cell reactions are
`Fe(s)toFe^(2+)(aq)+2e^(-)`
`2H^(+)(aq)+2e^(-) to H_(2)(g)`
`2H^(+)(aq)+Fe(s) to Fe^(2+)(aq)+H_(2)(g)`
Here, `n=2 and Q_(C)=([Fe^(2+)])/([H^(+)]^(2))`
Nernst equation at 298 K is
`EMF=EMF^(@)-(2.303RT)/(nF)logQ_(C)`
`EMF=(E_(H^(+)//1//2H_(2))^(@)-E_(Fe^(2+)//Fe)^(@))`
`(0.0591V)/(2)"log"([Fe^(2+)])/([H^(+)]^(2))`
`=[0-(-0.44V)]`
`(0.0591V)/(2)"log"(10^(-3))/((1)^(2))`
`=0.44V(0.0591xx(-3)V)/(2)`
`=[0.44+0.09]V=0.53V`