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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

`0.343+(0.0591)/(2)`

B

`0.34+(0.0591)/(2)`

C

0.34

D

`(0.34)`

Text Solution

Verified by Experts

The correct Answer is:
D
`Cu ^(2+) + 2e^(-) rightarrow Cu`
`E_(Cu^(2+)//Cu) = E_(Cu^(2+) + Cu)^(@) - (0.059)/(2) log 10 (1)/([Cu^(2+)])=E_(Cu ^(2+)//Cu)^(@)-(Rt)/(2F)` in `([Cu^(2+)])`
Intercept = 3.4 V `Rightarrow E_(Cu^(2+)//Cu) ^(@) = 0.34V`
`Rightarrow E_(Cu^(2+)//Cu) = 0.343 + (0.059)/(2) log _(10) 0.1 = 031V Rightarrow E_(Cu//Cu^(2+)) = -E(Cu^(2+) //Cu) = - 0.34 + (0.059)//(2) V`
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