Question

The standard potential for the reaction is
$H{g}_{2}C{l}_2+C{l}_2\to 2H{g}^{2+}+4C{l}^{-}$
Given,
(i) $H{g}_{2}C{l}_2+2e\to 2Hg+2C{l}^{-};E^{\circ }=0.270V$
(ii) $H{g}_2^{2+}\to 2H{g}^{2+}+2e;E^{\circ }=-0.92V$
(iii) $2Hg\to H{g}_2^{2+}+2e;E^{\circ }=-0.79V$
(iv) $C{l}_2+2e\to 2C{l}^{-};E^{\circ }=-1.36V$

Answer 1

Answer: -0.08

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

$H{g}_{2}C{l}_2+2e\to 2Hg+2Cl;E^{\circ }=0.270V$

$H{g}_2^{2+}\to 2H{g}^{2+}+2e;E^{\circ }=-0.92V$

$2Hg\to 2H{g}_2^{2+}+2e;E^{\circ }=-0.79V$

On adding these three

$H{g}_{2}C{l}_2\to 2H{g}^{2+}+2C{l}^{-}+2e$

$\Delta G_4=\Delta G_1+\Delta G_2+\Delta G_3$

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

$=2.88F$

$E_4=-1.44V$

Also, $C{l}_2+2e\to 2C{l}^{-}{E}^{\circ }=1.36V$

Adding (i) and (ii)

$H{g}_{2}C{l}_2+C{l}_2\to 2H{g}^{2+}+4C{l}^{-}$

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

$=1.44+1.360=-0.08V$