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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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  • Cu^(2+)+2e^(-)rarrCu, log[Cu^(2+)] vs. E_(red) graph is of the type as shown in figure where OA=0.34V then electrode potential of the half cell of Cu|Cu^(2+)(0.1M) will be

    A
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    B
    `0.34+0.0591V`
    C
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    D
    `-0.34V`

  • A reaction, Cu^(2+)+2e^(-) rarr Cu is given. For this reaction, graph between E_(red) versus ln[Cu^(2+)] is a straight line of intercept 0.34 V, then the electrode oxidation potential of the half-cell Cu//Cu^(2+) (0.1 M) will be

    A
    0.34
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    A
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    B
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