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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

Verified by Experts

The correct Answer is:
b
`E_(cell)=E^(c-)._(cell)-(0.0591)/(2)log.([Zn^(2+)])/([Cu^(2+)])`
From line `OA=E^(c-)._(cell)=1.10V`
If `([Zn^(2+)])/([Cu^(2+)])=10^(-2)M,` then `E_(cell)=1.1591V`
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