Write the Nernst equation and `EMF` of the following cells at `298K :`
`a.` `Mg(s)|Mg^(2+)(0.001M)||Cu^(2+)(0.0001M)|Cu(s)`
`b.` `Fe(s)|Fe^(2+)(0.001M)||H^(o+)(1M)|H_(2)(g)(1 ba r)|Pt(s)`
`c.` `Sn(s)|Sn^(2+)(0.050M)||H^(o+)(0.020M)|H_(2)(g)(1 ba r)|Pt(s)`
`d.` `Pt(s)|Br_(2)(1)|Br^(-)(0.010M)||H^(o+)(0.030M)|H_(2)(g)(1 ba r)|Pt(s)`

Explanation

(i). Cell reaction:
`Mg+Cu^(2+)toMg^(2+)+Cu(n=2)`
Nearest equation:
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)log(([Mg^(2+)])/([Cu^(2+)]))`
`thereforeE_(cell)=0.34-(-2.37)-(0.0591)/(2)log((10^(-3))/(10^(-4)))`
`=2.71-0.02955=2.68V`.
(ii). Cell reaction
`Fe+2H^(+)toFe^(2+)+H_(2)(n=2)`
nearest equation:
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)log(([Fe^(2+)])/([H^(+)]^(2)))`
`thereforeE_(cell)=0(0*44)-(0.0591)/(2)log((10^(-3))/((1)^(2)))`
`=0*44-(0.0591)/(2)xx(-3)`
`=0.44+0.0887=0.5287V`.
(iii). Cell reaction:
`Sn+2H^(+)toSn^(2+)+H_(2)(n=2)`
nearest equation
`E_(cell)=E_(cell)^(@)-(0.591)/(2)log(([Sn^(2+)])/([H^(+)]^(2)))`
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)log((0.05)/((0.02)^(2)))`
`=0-(-0.14)-(0.0591)/(2)log((0.05)/((0.02)^(2)))`
`=0.14-(0.591)/(2)log125`
`=0.14-(0.0591)/(2)(2.0969)=0*078V`.
(iv). Cell reaction
`2Br^(-)+2H^(+)toBr_(2)+H_(2)(n=2)`
Nearest equation:
`E_(Cell)=E_(cell)^(@)-(0.0591)/(2)log((1)/([Br^(-)]^(2)[H^(+)]^(2)))`
`=(0-1.08)-((0.591)/(2))log((1)/((1.01)^(2)(0.03)^(2)))`
`=-1.08-(0.0591)/(2)log(1.111xx10^(7))`
`=-1.08-(0.0591)/(2)(7.0457)`
`=-1.08-0.208=-1.288V`
Thus, oxidation will occur at the hydrogen electrode and reduction will occur on `Br_(2)` electrode.