Copper reduced `NO_(3)^(-)` into `NO` and `NO_(2)` depending upon conc.of `HNO_(3)` in solution. Assuming `[Cu^(2+)] = 0.1M`, and `P_(NO) = P_(NO_(2)) = 10^(-3)` atm and using data answer the following questions:
`E_(Cu^(2+)//Cu)^(@) =+ 0.34` volt, at `298K (RT)/(F) (2.303) = 0.06` volt
`E_(cell)` for reduction of `NO_(3)^(-) rarr NO` by `Cu(s)` when `[HNO_(3)] = 1M` is [At `t = 298]`

Options

A. `10^(1.23)M`
B. `10^(0.56)M`
C. `10^(0.66)M`
D. `10^(0.12)M`

Answer

C

Explanation

`(e^(-) +2H^(+) +NO_(3)^(-) rarr NO_(2) +H_(2)O) xx 2`
`Cu rarr Cu^(+2) +2e^(-)`
`4H^(+) +2NO_(3)^(-) +Cu rarr Cu^(+2) +2NO_(2) +2H_(2)O`
`E = 0.79-0.34 -(0.0591)/(2) log.([NO_(2)]^(2)[Cu^(+2)])/([NO_(3)^(-)]^(2)[H^(+)]^(4))`..(ii)
Let `[HNO_(3)] = xM` so `[H^(+)] = [NO_(3)^(-)] = x`
equation (i) & (ii)
`0.62 -(0.0591)/(2) log.(10^(-9))/(x^(10)) = 0.45 -(0.0591)/(2)log.(10^(-7))/(x^(6))`
`0.17 =(0.591)/(6) [-9-10 logx] -(0.0591)/(2)[-7-6logx]`
`0.0518 = 0.0788 logx`
`x = 10^(0.657)M`