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For the reaction, Cu^(2+)+2e^(-) to Cu ,...

For the reaction, `Cu^(2+)+2e^(-) to Cu , log [Cu^(2+)]` vs E graph is of type as shown in figure where OA = 0.34 V,the electrode potential of the half-cell of `Cu|Cu^(2+)(0.1 M)` will be

A

`-0.34+"0.0591"/2V`

B

0.34+0.0591 V

C

0.34 V

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`E_(Cu//Cu^(2+))=E_(Cu//Cu^(2+))^@-0.059/2"log"[Cu^(2+)]`
If log `[Cu^(2+)]`=0 i.e. `[Cu^(2+)]`=1
then `E_(Cu//Cu^(2+))=E_(Cu//Cu^(2+))^@`
OA=0.34 V =`E_(Cu^(2+)//Cu)^@=-E_(Cu//Cu^(2+))^@`
`rArr E_(Cu//Cu^(2+))^@`=-0.34 V
Now, `E_(Cu//Cu^(2+))`=-0.34-`0.059/2`log0.1 = -0.34+`0.059/2`V
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