Write the Nernst equation and emf of the following cells at 298 K:
(i) `Mg_((S))|Mg^(2+)(0.001M)||Cu^(2+)(0.0001M)|Cu_((S))`
(ii) `Fe_((S))|Fe^(2+)(0.001M)||H^(+)(1M)|H_(2(g))(1" bar")|Pt_((S))`
(iii) `Sn_((S))|Sn^(2+)(0.050M)||H^(+)(0.020M)|H_(2(g))(1" bar")|Pt_((S))`
(iv) `Pt_((S))|Br^(-)(0.010M)|Br_(2(l))||H^(+)(0.030M)|H_(2(g))(1" bar")|Pt_((S))`.

Explanation

For the given reaction, the Nernst equation can be given as:
`E_(Mg^(2+)|Mg)^(Theta)=-2.37V`
`E_(Cu^(2+)|Cu)^(Theta)=+0.34V,E_(Sn^(2+)|Sn)^(Theta)=-0.14V`
`E_(Br^(2+)|Br^(-))^(Theta)=+1.08V,E_(Fe^(2+)|Fe)^(Theta)=-0.44V`.
(i) Standard electrode potential calculation:
Cell reactioin : `Mg_((S))+underset((0.0001M))(Cu^(2+))toMg^(2+)+underset((0.001M))(Cu_((S)))`
* In above cell reaction the concentration are not standard and its Nernst equation will be
`therefore E_(cell)^(Theta)=E_(Cu^(2+)|Cu)^(Theta)-E_(Mg^(2+)|Mg)^(Theta)`
`=0.34-(-2.37V)`
`=2.71V`
Where `[Mg^(2+)]0.001M and [Cu^(2+)]=0.0001M`
`therfore E_(cell)=2.71V+((0.0591)/(2))xxlog((0.0001M)/(0.001M))`
`=2.71+((0.0591)/(2))(log" "0.1)`
`=2.71+0.02955xx(-1.0)`
`=2.71-0.02955`
`=2.68045V~~2.68V`
`E_(cell)=E_(cell)^(Theta)-(0.0591)/(n)log(([Mg^(2+)])/([Cu^(2+)]))` where, n=2
`E_(cell)=E_(cell)^(Theta)+(0.0591)/(2)log(([Cu^(2+)])/([Mg^(2+)]))`