Question

If If $E_{Cl{O}_3^{-}/Cl{O}_4^{-}}^0=-0.36V\&E_{Cl{O}_3^{-}/Cl{O}_2^{-}}^0=0.33V$ at $300K$ The equilibrium concentration of perchlorate ion $\left(Cl{O}_4^{-}\right)$which was initially $1.0M$ in $Cl{O}_3^{-}$when the reaction starts to attain the equilibrium, $2Cl{O}_3^{-}\rightleftharpoons Cl{O}_2^{-}+Cl{O}_4^{-}$Given : Anti $\log (0.509)=$ $3.329$

• A. 0.0236 M
• B. 0.0190 M
• C. 0.123 M
• D. 0.191 M

Answer 1

Answer: (D)

The redox reaction can be split as

$E_{\text{cell }}^o=E_{\text{red (cathod) }}^o-E_{\text{red (anode) }}^o$
$=0.33-0.36=-0.03V$
as $E_{\text{red (Anode) }}^o=E_{Cl{O}_4^{-1}/Cl{O}_3^{-1}}^o=+0.36V$
$\left(=-E_{Cl{O}_3^{-1}/Cl{O}_4^{-1}}^o\right)$ at
equilibrium $E_{\text{cell }}=0E_{\text{cell }}^o$
$=\frac{0.059}{n}\log K_{eq}=\frac{0.059}{2}\log K_{eq}$
Writing concentration of speces at equilibrium

$E_{\text{cell }}^o=-0.03=\frac{0.059}{2}\log \frac{(x{)}^2}{(1-2x{)}^2}=0.059\log \frac{x}{1-2x}$
$\frac{-0.03}{0.059}=\log \frac{x}{1-2x}$
$\Rightarrow \log \frac{1-2x}{x}=\frac{0.03}{0.059}$
$\log \frac{1-2x}{x}=0.509$
$\frac{1-2x}{x}=3.229$
$1-2x=3.229x$
$\Rightarrow 5.229x=1$
$x=\frac{1}{5.229}=0.191M$