In the following process of disproportionation:
`underset(underset("ion")("Chlorate"))(2 ClO_(3)^(-))hArr underset(underset("ion")("Per chlorate"))(ClO_(4)^(-))` `{:(E_(ClO_(4)^(-)//ClO_(3)^(-))^(@),= + 0.36 V),(E_(ClO_(3)^(-)//ClO_(2)^(-))^(@),= + 0.33 V):}`
Initial concentration of chlorate ion was 0.1 M. The equilibrium concentration of per chlorate ion will be :

Options

A. 0.19 V
B. 0.1 M
C. 0.024 M
D. 0.019 M

Answer

D

Explanation

`{:(ClO_(3)^(-) (aq.) +H_(2)O (l) rarr ClO_(4)^(-) (aq.)+2H^(+) (aq.) +2e^(-)),(2H^(+)+ClO_(3)^(-) (aq.) +2e^(-) rarr ClO_(2)^(-) (aq.)+H_(2)O(l)),(ulbar(2ClO_(3)^(-) (aq.) hArr ClO_(2)^(-) (aq.) +ClO_(4)^(-) (aq.))):}`
`E_(cell)^(@)=0.33-0.36=-0.03 V`
`E=E^(@)-0.059/n log Q`
At equilibrium, `E=0, n=2, Q=K`
`0=-0.03-0.059/2 log K`
`log K=-1`
`K=1/10` ...(1)
`{:(,2ClO_(3)^(-) (aq.),hArr,ClO_(2)^(-) (aq.),+,ClO_(4)^(-) (aq.)),(t_(0),0.1,,0,,0),(t_(eq.),0.1-2x,,x,,x):}`
`K=(x xx x)/((0.1-2x)^(2))=1/10`
`x=0.019`