Write the Nernst equation and emf of the following cells at 298 K:
(i) `Mg(s) | Mg^(2+) (0.001 M)|| Cu^(2+) (0.0001 M) | Cu (s)`
(ii) Fe(s) | `Fe^(2+) (0.001 M) || H^(+) ( 1 M) | H_(2) (g) (1 bar) | Pt(s)`
(iii) `Sn(s) | Sn^(2+) (0.050 M) || H^(+) (0.020 M) | H_(2) (g) (1 bar) | Pt(s)`
(iv) `Pt(s) | Br_(2) (l) | Br^(-) (0.010 M) || H^(+) (0.030 M) | H_(2) (g) (1 bar) | Pt(s)`
Given, `E_(Mg^(2+)//Mg)^(@) = -2.37 V, E_(Cu^(2+),Cu)^(@) = +0.34 V, E_(Fe^(2+), Fe)^(@) = -0.44, E_(Sn^(2+)//Sn)^(@) = -0.14 V, E_(1//2 Br^(2),Br^(-))^(@) = +1.08 V`

Explanation

(i) Cell reactionP: `Mg + Cu^(2+) to Mg^(2+) + Cu (n=2)`
Using Nernst equation and substituting the values, we get
`E_("cell") = E_("cell")^(@) - (0.0591)/2 log ([Mg^(2+)])/([Cu^(2+)])`
`therefore E_("cell") = 0.34 - (-2.37) - 0.0591/2 log (10^(-3))/(10^(-4)) = 2.71 - 0.02955 = 2.68 V`
(ii) Cell reaction: `Fe + 2H^(+) to Fe^(2+) + H_(2) (n=2)`
Using nernst equation and substituting the values, we get
`E_("cell") = E_("cell")^(@) - (0.0591)/2 log ([Fe^(2+)])/([H^(+)]^(2))`
`therefore E_("cell")^(@) = 0-(-0.44) - 0.0591/2 log (10^(-3))/(1)^(2)`
`=0.44 - (0.0591)/2 xx (-3) = 0.44 + 0.0887 = 0.523` V
(iii) Cell reaction: `Sn + 2H^(+) to Sn^(2+) + H_(2) (n=2)`
Using Nernst equation and substituing the values, we get
`E_(cell) = E_("cell")^(@) - 0.0591/2 log ([Sn^(2+)])/([[H^(+)])^(2)) = 0- (-0.14) - 0.591/2 log 0.05/(0.02)^(2) = 0.078 V`
(iv) Cell reaction : `2Br^(-) + 2H^(+) to Br_(2) + H_(2)`
Using Nernst equation and substituing the values, we get
`E_("cell") = E_("cell")^(@) - (0.0591)/2 log 1/([Br^(-)]^(2) [H^(+)]^(2))`
`therefore =1.08 - 0.0591/2 log (1.111 xx 10^(7)) = -1.08 - 0.0591/2 (7.04567)`
`=-1.08 - 0.208 = -1.288` V
Thus, oxidation will occur at the hydrogen electrode and reduction on the `Br_(2)` electrode.
`E_(cell) = 1.288 V`.