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A reaction, Cu^(2+)+2e^(-) rarr Cu is gi...

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

B

`0.34+0.0591/2`

C

`-0.34-0.0591/2`

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_(Cu^(2+)//Cu)^(@)-(0.059)/2"log "1/([Cu^(2+)])`
`=E_(Cu^(2+)//Cu)^(@)+(RT)/(2F)ln [Cu^(2+)]`
`E_(Cu^(2+)//Cu)=0.34+0.059/2 log 0.1=0.31 V`
`E_(Cu//Cu^(2+))=-E_(Cu^(2+)//Cu)=-0.34+0.059/2 V`
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Knowledge Check

  • 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)`

  • Cu^(2+) + 2e rarr Cu(s) . For the graph between log[Cu^(2+)] versus E_("red") a straight line of intercept 0.34 V is obtained. Then electrode potential of the half-cell Cu// Cu^(2+)(0.1 M) will be :

    A
    `[0.34+(0.0591)/2]V`
    B
    `[-0.34-(0.0591)/2]V`
    C
    `0.34 V`
    D
    `[-0.34+(0.0591)/2]V`

  • 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
    `-0.34+(0.0591)/(2)V`
    B
    `0.34+0.0591V`
    C
    `0.34V`
    D
    `-0.34V`