Copper reduced `NO_(3)^(-)` into NO and `NO_(2)` depending upon cone. Of `HNO_(3)` in solution. Assuming `[Cu^(2+)]=0.1M` and `P_(NO)=P_(NO_(2)=10^(-3)` atm and using given data answer the following question:
`E_(Cu^(2+)|Cu)^(@)=+0.34` volt
`E_(NO_(3)^(-)|NO)^(@)=+0.96` volt
`E_(NO_(#)^(-)|NO_(3))^(@)=+0.79` volt
at 298K `(RT)/(F)(2.303)=0.06` volt
`E_("cell")` for reduction of `NO_(3)^(-)rarrNO` 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