FInd Ecell for reduction of NO3−→NO by Cu(s), when [HNO3]=1M,PNO=10−3atm,[Cu2+]=0.1M (Given: at 298 K RTF(2.303)=0.06volt;E∘NO−3/NO=+0.96volt;E∘Cu2+/Cu=+0.34volt)
A
≈0.61
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B
≈0.71
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C
≈0.51
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D
≈0.81
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Solution
The correct option is A≈0.61 2NO3−+3Cu→NO+Cu2+n=6Ecell=Eocell−0.0591nlog(Cu2+)(NO−3)=(0.96−0.34)−0.05916log(0.11)Ecell=0.62+0.00985Ecell≈0.62