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Zn+Cu^(2+)(aq)hArrCu+Zn^(2+)(aq). Reac...

`Zn+Cu^(2+)(aq)hArrCu+Zn^(2+)(aq).`
Reaction quotient is `Q=([Zn^(2+)])/([Cu^(2+)])` . Variation of `E_(cell)` with log `Q` is of the type with `OA=1.10` `V.E_(cell) ` will be `1.1591V` when

A

`[Cu^(2+)]//[Zn^(2+)] = 0.01`

B

`[Zn^(2+)]//[Cu^(2+)] = 0.01`

C

`[Zn^(2+)]//[Cu^(2+)] = 0.1`

D

`[Zn^(2+)]//[Cu^(2+)] = 1`

Text Solution

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The correct Answer is:
B
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