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. `~0.61`
B. `~0.71`
C. `~0.51`
D. `~0.81`

Answer

B

Explanation

`(3e^(-) +4H^(+) +NO_(3)^(-) rarr NO +2_(2)O) xx2`
`(Cu rarr Cu^(+2) +2e^(-)) xx 3`
`E = 0.96 -0.34 -(0.0591)/(6) log.([NO]^(2)[Cu^(+2)]^(3))/([NO_(3)^(-)]^(2)[H^(+)]^(8))` ..(i)
since `[HNO_(3)] = 1M` so `[H^(+)] = [NO_(3)^(-)] = 1`
`E = 0.62 -(0.0591)/(6) log (10^(-3))^(2) (0.1)^(3) = 0.70865`
`E = 0.62 -(0.0591)/(6)log (10^(-3))^(2) (0.1)^(3) = 0.70865`