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

Question

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

Infinity Learn · Accepted Answer

$$ClO3$$−$$aq+H2Ol$$→$$ClO4$$−$$aq+2H+aq+2e$$−$$2H++ClO3$$−$$aq+2e$$−→$$ClO2$$−$$aq+H2Ol2ClO3$$−aq⇌$$ClO2$$−$$aq+ClO4$$−$$aqEcello=0.33$$−$$0.36=$$−$$0.03VE=Eo$$−$$0.059nlogQAt$$ equilibrium, $$E=0$$, $$n=2$$, $$Q=K0=$$−$$0.03$$−$$0.0592logKlogK=$$−$$1K=110K=x$$×$$x0.1$$−$$2x2=110x=0.019$$

Electrochemical cells

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Electrochemical cells

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In the following process of disproportionation,2ClO3−Chlorate ion⇌ClO2−+ClO4−Perchlorate ion;EClO4−/ClO3−o=+0.36VEClO3−/ClO2−o=+0.33VInitial concentration of chlorate ion was 0.1 M. The equilibrium concentration of per chlorate ion will be 1.9×10−x. Hence x is

ClO3−aq+H2Ol→ClO4−aq+2H+aq+2e−2H++ClO3−aq+2e−→ClO2−aq+H2Ol2ClO3−aq⇌ClO2−aq+ClO4−aqEcello=0.33−0.36=−0.03VE=Eo−0.059nlogQAt equilibrium, E=0, n=2, Q=K0=−0.03−0.0592logKlogK=−1K=110K=x×x0.1−2x2=110x=0.019

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