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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)V`

B

`0.34+0.0591" V"`

C

`0.34" V"`

D

none of these.

Text Solution

Verified by Experts

The correct Answer is:
A
(a) `E_(Cu//Cu^(2+))=E_(Cu//Cu^(2+))^(@)-(0.0591)/(2)log[Cu^(2+)]`
It `log[Cu^(2+)]=0`, then `Cu^(2+)=1`
`E_(Cu//Cu^(2+))=E_(Cu//Cu^(2+))^(@)`
`:. OA=E_(Cu//Cu^(2+))^(@)=-E_(Cu//Cu^(2+))^(@)=-0.34" V"`
Hence `E_(Cu//Cu^(2+))=-0.34-(0.0591)/(2)log 0.1=-0.34+(0.0591)/(2)`
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