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^(c-)(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)`
nernst eqn.: `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Mg^(2+)])/([Cu^(2+)])`
`E_(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)`
Nernst eqn.: `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=0.523V`.
(iii) Cell reaction: `Sn+2H^(+)toSn^(2+)+H_(2)(n=2)`
Nernst eqn. : `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Sn^(2+)])/([H^(+)]^(2))=0-(-0.14)-(0.591)/(2)"log"(0.05)/((0.02)^(2))`
`=0.14-(0.0591)/(2)log125=0.14-(0.0591)/(2)(2.0969)=0.078V`
(iv) Cell reaction: `2Br^(-)+2H^(+)toBr_(2)+H_(2)` (Note carefully) ltrgt Nernst eqn. : `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"(1)/([Br^(-)]^(2)[H^(+)]^(2))`
`therefoerE_(cell)=(0-1.08)-(0.0591)/(2)"log"(1)/((0.01)^(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 occurs at the hydrogen electrode and reduction on the `Br_(2)` electrode. `E_(cell)=1.288V`.