Question

If If $E_{ ClO _{3}^{-} / ClO _{4}^{-}}^{0}=-0.36 V \& E _{ ClO _{3}^{-} / ClO _{2}^{-}}^{0}=0.33 V$ at $300 K$ The equilibrium concentration of perchlorate ion $\left( ClO _{4}^{-}\right)$which was initially $1.0 M$ in $ClO _{3}^{-}$when the reaction starts to attain the equilibrium, $2 ClO _{3}^{-} \rightleftharpoons ClO _{2}^{-}+ ClO _{4}^{-}$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

$E _{\text {cell }}^{ o }= E _{\text {red (cathod) }}^{ o }- E _{\text {red (anode) }}^{ o }$
$=0.33-0.36=-0.03 V$
as $E _{\text {red (Anode) }}^{ o }= E _{ ClO _{4}^{-1} / ClO _{3}^{-1}}^{ o }=+0.36 V$
$ \left(=- E _{ ClO _{3}^{-1} / ClO _{4}^{-1}}^{ o }\right)$ at
equilibrium $E _{\text {cell }}=0 E _{\text {cell }}^{ o }$
$=\frac{0.059}{ n } \log K _{ eq }=\frac{0.059}{2} \log K _{ eq }$
Writing concentration of speces at equilibrium

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