Question

The standard potential for the reaction is
$Hg _{2} Cl _{2}+ Cl _{2} \rightarrow 2 Hg ^{2+}+4 Cl ^{-}$
Given,
(i) $Hg _{2} Cl _{2}+2 e \rightarrow 2 Hg +2 Cl ^{-}; \,\, E^{\circ}=0.270\, V$
(ii) $Hg _{2}^{2+} \rightarrow 2 Hg ^{2+}+2 e; \,\,\,E^{\circ}=-0.92\, V$
(iii) $2 Hg \rightarrow Hg _{2}^{2+}+2 e; \,\,E ^{\circ}=-0.79\, V$
(iv) $Cl _{2}+2 e \rightarrow 2 Cl ^{-}; \,\,\,E^{\circ}=-1.36\, V$

Answer 1

Just as two half reaction can be combined to yield third, three half reactions can be combined to yield fourth.

$Hg _{2} Cl _{2}+2 e \rightarrow 2 Hg +2 Cl ;\,\, E ^{\circ}=0.270\, V$

$Hg _{2}^{2+} \rightarrow 2 Hg ^{2+}+2 e ;\,\, E ^{\circ}=-0.92\, V$

$2 Hg \rightarrow 2 Hg _{2}^{2+}+2 e ;\,\, E ^{\circ}=-0.79\, V$

On adding these three

$Hg _{2} Cl _{2} \rightarrow 2 Hg ^{2+}+2 Cl ^{-}+2 e$

$\Delta G _{4} =\Delta G _{1}+\Delta G _{2}+\Delta G _{3}$

$- nE _{4} F =[-2 \times 0.270+2 \times 0.92-2 \times 0.79] F$

$=2.88 F$

$E _{4}=-1.44\, V$

Also, $Cl _{2}+2 e \rightarrow 2 Cl ^{-} \,\, E ^{\circ}=1.36\, V$

Adding (i) and (ii)

$Hg _{2} Cl _{2}+ Cl _{2} \rightarrow 2 Hg ^{2+}+4 Cl ^{-}$

and $E^{\circ}=E_{( i )}^{\circ}+E_{( ii )}^{\circ}$

$=1.44+1.360=-0.08\, V$