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

Copper reduced `NO_(3)^(-)` into NO and `NO_(2)` depending upon cone.

1.	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]A. `~0.61`B. `~0.71`C. `~0.51`D. `~0.81`
Answer» Correct Answer - B

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