1.

`Cu^(2+)+2e^(-) rarr Cu`. For this, graph between `E_(red)` versus `ln[Cu^(2+)]` is a straight line of intercept `0.34V`, then the electrode oxidation potential of the half cell `Cu|Cu^(2+)(0.1M)` will beA. `0.34+(0.0591)/(2)`B. `-0.34-(0.0591)/(2)`C. `0.34`D. `-0.34+(0.0591)/(2)`

Answer» Correct Answer - d
`Cu^(2+)+2e^(-) rarr Cu`
`E_(Cu^(2+)|Cu)=E^(c-)._(Cu^(2+)|Cu)-(0.059)/(2) log .(1)/([Cu^(2+)])`
` =E^(c-)._(Cu^(2+)|Cu)-(RT)/(2F)ln[Cu^(2+)]`
Intercept`=0.34impliesE^(c-)._(Cu^(2+)|Cu)=0.34V`
`implies E_(cu^(2+)|Cu)=0.34+(0.059)/(2) log 0.1=0.31V`
`impliesE_(Cu|Cu^(2+))=-E_(Cu^(2+)|Cu)=-0.34+(0.059)/(2)V`


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