The reduction of `NO_(3)^(-)` occurs as :
`NO_(3)^(-)+ 4H^(+) + 3e^(-) to NO+ 2H_(2)O, E^(@)= 0.96 V`
The electrons are provided by Cd till that the solution originally having `0.1 M NO_(3)^(-)` and `0.4 M H^(+)` shows that `80%` of `NO_(3)^(-)` ions are converted to NO showing 1 bar pressure. The reduction potential of remaining solution. [log 8.192 = 0.96]

Options

A. 0.84 V
B. 1.36 V
C. 1.08 V
D. 1.56 V

Answer

A

Explanation

`underset(0.1-x)(underset(0.1)((NO_(3)^(-)))) + underset(0.4-0.4x)(underset(0.4)(4H^(+))) + 3e^(-) to underset(-)(underset(0)(NO))+ underset(-)(underset(0)(2H_(2)O))`
`x=(80)/(200) xx 0.1= 0.08`
After reduction,
`[NO_(3)^(-)]= 0.1- 0.08= 0.02`
`[H^(+)]= 0.4- 0.32= 0.08`
`:. E_(RP)= E_(RP)^(@)+ (0.059)/(3)log(([NO_(3)^(-)][H^(+)]^(4))/([P_(NO)]))`
`= 0.96 + (0.059)/(3)log(( 0.02 xx (0.08)^(4))/(1))`
0.96- 0.12= 0.84 V