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Cu^(2+)+2e^(-) rarr Cu. For this, graph ...

`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 be

A

`0.34+(0.0591)/(2)`

B

`-0.34-(0.0591)/(2)`

C

`0.34`

D

`-0.34+(0.0591)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
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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