Q. Consider the following disproportionation $2{ClO}_{3}^{-} \rightleftharpoons {ClO}_{2}^{-} + {ClO}_{4}^{-}$ If the initial concentration of perchlorate ion is 0.1 M what it would be at equilibrium at 298 K? $\left( E_{{ClO}_{4}^{-}\ /\ {ClO}_{3}^{-}}^{{^\circ}} = 0.36\ V\ and\ E_{{ClO}_{3}^{-}\ /\ {ClO}_{2}^{-}}^{{^\circ}} = 0.33\ V \right)$
A) 0.1 M
B) 0.05 M
C) 0.07 M
D) 0.19 M
Explanation
(d)
Oxidation half-cell
${ClO}_{3}^{-} \rightarrow {ClO}_{2}^{-} + 2e^{-};\ \ \ E{^\circ}_{cell} = - 0.36\ V$
Reduction half-cell
${ClO}_{3}^{-} + 2e^{-} \rightarrow {ClO}_{2}^{-};\ E{^\circ}_{cell} = 0.33\ V$
$E{^\circ}_{cell} = 0.33 - 0.36 = - 0.03 = \frac{RT}{2F}{\ \ln}{\ K}$
or$\ - 0.03 = \frac{0.059}{2}\log K\$or$\ K = 0.1$
$2ClO_{3}^{-} \rightleftharpoons ClO_{4}^{-} + ClO_{2}^{-}$
$0.1 - 2x$ $x\ \ \ \ \ \ \ \ \ \ \ \ x$
$\frac{x^{2}}{(0.1 - 2x)^{2}} = 0.1$
or$\ \ x = 1.9 \times 10^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.