The reduction of `NO_(3)^(-)` occurs as
` No_(3)^(-) + 4H^(+) + 3e^(-) rightarrow NO+ 2H_(2)O, E^(@) = 0.96V`
If the reaction originally starts with 1M of `NO_(3)^(-)` and 5M of `H^(+)` and the reaction goes to 80% completion such that under those conditions of temperature the pressure of`NO_(g)` was 2 bar. Find the reduction potential of the remaining solution.

Options

A. `(0.84V)`
B. `(0.94V)`
C. `(0.90V)`
D. `(0.88V)`

Answer

A

Explanation

`No_(3)^(-)+ 4H^(+)+ 3e^(-)rightarrowNO+2H_(2)O`, `E^(@) = 0.96V`
`NO_(3)^(-)+ 4H^(+)+ 3e^(-)rightarrowNo + 2H_(2)O` , `E^(@) = 0.96V`
1M 5M
After 80% completion
0.2M 1.8M `2("bar")Q_(C)= (2)/(0.2[1.8]^(4)`
`E_("electrode")= 0.96- (0.06)/(3)log Q_(c)`
`= 0.96 - 0.02log (10)/(10.9)= 0.96-0.02log 0.95 = 0.96`