Question

$E_{Cl{O}_4^{-}/Cl{O}_3^{-}}^o$ and $E_{Cl{O}_3^{-}/Cl{O}_2^{-}}^o$ are$+0.36V$ and $+0.33V$respectively. The equilibrium concentration of per chlorate ion for the given reaction would be:

$2Cl{O}_3^{-}\rightleftharpoons Cl{O}_2^{-}+Cl{O}_4^{-}$
$t=0$ 0.1 M 0 0

• A. $0.19M$
• B. $0.024$
• C. $0.24$
• D. $0.019$

Answer 1

The correct option is D $0.019$

$2Cl{O}_3^{-}\rightleftharpoons Cl{O}_2^{-}+Cl{O}_4^{-}$
$t=0$ 0.1 M 0 0
at eqm.

$\underset{(0.1-x)}{2Cl{O}_3^{-}}\rightleftharpoons \underset{\frac{x}{2}}{Cl{O}_2^{-}}+\underset{\frac{x}{2}}{Cl{O}_4^{-}}$

$C{l}^{5+}\to C{l}^{7+}+2e$

$2e+C{l}^{5+}\to C{l}^{3+}$

According to Nernst equation:

$E=E^o-\frac{0.0591}{n}logQ$

$E_{cell}=E_{Cl{O}_3^{-}/Cl{O}_4^{-}}^o-\frac{0.059}{2}log\frac{\left[Cl{O}_4^{-}\right]}{\left[Cl{O}_3^{-}\right]}+E_{Cl{O}_3^{-}/Cl{O}_2^{-}}^o+\frac{0.059}{2}log\frac{\left[Cl{O}_3^{-}\right]}{\left[Cl{O}_2^{-}\right]}$

$=-0.36+0.33+\frac{0.059}{2}log\frac{[Cl{O}_3^{-}{]}^2}{\left[Cl{O}_4^{-}\right]\left[Cl{O}_2^{-}\right]}$

As emf is zero at eqm so,

At Eq. $0=-0.03+\frac{0.059}{2}log\frac{(0.1-x{)}^2}{x^2}\times 2^2$.

$log{2}^2\times \frac{(0.1-x{)}^2}{x^2}=\frac{0.03\times 2}{0.059}=1$

$\frac{2(0.1-x)}{x}=\sqrt{10}$

$\therefore x=0.038$

$\therefore \left[Cl{O}_4^{-}\right]=\frac{0.038}{2}=0.019$