Chemistry - Electrochemistry Question with Solution | TestHub

${\mathrm{ClO}}_3^{-}$ + 2H₂O + 4e⁻ → ClO⁻ + 4OH⁻; $\Delta G_1^{\circ }$

$$\begin{array}{rl}{\mathrm{ClO}}^{-}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{e}}^{-}& \to \frac{1}{2}{\mathrm{Cl}}_2+2{\mathrm{OH}}^{-};\Delta {\mathrm{G}}_2^{\circ }\\ \frac{1}{2}{\mathrm{Cl}}_2+{\mathrm{e}}^{-}& \to {\mathrm{Cl}}^{-};\Delta {\mathrm{G}}_3^{\circ }\\ {\mathrm{ClO}}_3^{-}+3{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}& \to {\mathrm{Cl}}^{-}+6{\mathrm{OH}}^{-};\Delta {\mathrm{G}}^{\circ }\\ \Delta {\mathrm{G}}^{\circ }& =\Delta {\mathrm{G}}_1^{\circ }+\Delta {\mathrm{G}}_2^{\circ }+\Delta {\mathrm{G}}_3^{\circ }\\ -6\times \mathrm{F}\times {\mathrm{E}}^{\circ }& =-4\times \mathrm{F}\times 0.54-1\times \mathrm{F}\times 0.45-1\times \mathrm{F}\times 1.07\\ {\mathrm{E}}^{\circ }& =\frac{3.68}{6}=0.61\mathrm{V}\end{array}$$

Option: (B)