Question

The chlorate ion can disproportionate in basic solution according to reaction,
$2Cl{O}_3^{-}\rightleftharpoons Cl{O}_2^{-}+Cl{O}_4^{-}$
what is the equilibrium concentration of perchlorate ions from a solution initially at $0.1M$ in chlorate ions at $298K$?
Given that :$E_{Cl{O}_4^{-}|Cl{O}_3^{-}}^{\circ }=0.36Vand{E}_{Cl{O}_3^{-}|Cl{O}_2^{-}}^{\circ }=0.33Vat298K$.

• A. $0.019M$
• B. $0.024M$
• C. $0.1M$
• D. $0.19M$

Answer 1

Answer: (A)

$0.019M$

$H_{2}O\left(l\right)+Cl{O}_3^{-}\left(aq\right)\to Cl{O}_4^{-}\left(aq\right)+2H^{+}\left(aq\right)+2e^{-}$
$2H^{+}\left(aq\right)+Cl{O}_3^{-}\left(aq\right)+2e^{-}\to 2Cl{O}_2^{-}\left(aq\right)+H_{2}O\left(l\right)$
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$2Cl{P}_3^{-}\left(aq\right)\to Cl{O}_2^{-}\left(aq\right)+Cl{O}_4^{-}\left(aq\right)$
$E_{cell}^{\circ }=0.33-0.36=-0.03$; $E_{cell}^{\circ }=\frac{RT}{2F}lnK$
$-0.03=\frac{0.06}{2}logK$ or $K=0.1$
$2Cl{O}_3^{-}\rightleftharpoons Cl{O}_4^{-}+Cl{O}_2^{-}$
$0.1-2x$ $x$ $x$
$\frac{x^2}{(0.1-2x{)}^2}=\frac{1}{10};3.16x=0.1-2x$

$5.16x=0.1$

$\Rightarrow$ $x=\frac{0.1}{5.16}=0.0193\cong 1.9\times {10}^{-2}$

Hence option (A) is correct.