Question

The $E^{\circ }$ in the given diagram is

• A. 0.52 V
• B. 0.61 V
• C. 0.74 V
• D. 0.83 V

Answer 1

The correct option is B

0.61 V

$\begin{array}{cc}Cl{O}_3^{-}+2H_{2}O+4e^{-}\to Cl{O}^{-}+4O{H}^{-}\Delta G_1^O& Cl{O}_3^{-}+3H_{2}O+3e^{-}\to C{l}^{-}+6O{H}^{-}\Delta G^{\circ }\\ Cl{O}^{-}+H_{2}O+e^{-}\to \frac{1}{2}C{l}_2+2O{H}^{-}\Delta G_2^{\circ }& \therefore \Delta G^{\circ }=\Delta G_1^{\circ }+\Delta G_2^{\circ }+\Delta G_3^{\circ }\\ \frac{1}{2}C{l}_2+l{e}^{-}\to C{l}^{-}\Delta G_3^{\circ }& -66F{E}^{\circ }=-4F\times 0.54-1F\times 0.45-1F\times 1.07\\ --------------------& E^{\circ }=+\frac{3.68}{6}=+0.61V\end{array}$