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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C

`E_(cell)=_(cell)^(@)-(0.0591)/(2)log.([Zn^(2+)])/([Cu^(2+)])`
`1.1591=1.10-(0.0591)/(2)log.(Zn^(+2))/(Cu^(+2))`
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