In the following process of disproportionation,
$\underset{\begin{array}{l} C h l o r a t e \\ i o n \end{array}}{2 C l O_{3}^{-}} ⇌ C l O_{2}^{-} + \underset{\begin{array}{l} P e r c h l o r a t e \\ i o n \end{array}}{C l O_{4}^{-}}; \begin{matrix} E_{C l O_{4}^{-} / C l O_{3}^{-}}^{o} = + 0.36 V \\ E_{C l O_{3}^{-} / C l O_{2}^{-}}^{o} = + 0.33 V \end{matrix}$
Initial concentration of chlorate ion was 0.1 M. The equilibrium concentration of per chlorate ion will be $1.9 \times 10^{- x}$ . Hence x is

Moderate

Solution

$C l O_{3}^{-} (a q) + H_{2} O (l) \to C l O_{4}^{-} (a q) + 2 H^{+} (a q) + 2 e^{-}$
$2 H^{+} + C l O_{3}^{-} (a q) + 2 e^{-} \to C l O_{2}^{-} (a q) + H_{2} O (l)$
$2 C l O_{3}^{-} (a q) ⇌ C l O_{2}^{-} (a q) + C l O_{4}^{-} (a q)$
$E_{c e l l}^{o} = 0.33 - 0.36 = - 0.03 V$
$E = E^{o} - \frac{0.059}{n} \log Q$
At equilibrium, E=0, n=2, Q=K
$0 = - 0.03 - \frac{0.059}{2} \log K$
$\log K = - 1$
$K = \frac{1}{10}$
$K = \frac{x \times x}{{(0.1 - 2 x)}^{2}} = \frac{1}{10}$
$x = 0.019$

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