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For the cell given below Zn|Zn^(2+)||C...

For the cell given below
`Zn|Zn^(2+)||Cu^(2+) | Cu, (E_("cell")-E_("cell")^(0)) ` is -0.12 V . It will be when :

A

`[Zn^(2+)]//[Cu^(2+)]=10^(2)`

B

`[Zn^(2+)]//[Cu^(2+)]=10^(-2)`

C

`[Zn^(2+)]//[Cu^(2+)]=10^(4)`

D

`[Zn^(2+)]//[Cu^(2+)]=10^(-4)`

Text Solution

Verified by Experts

The correct Answer is:
C, D
`E = E^(0)-(0.0591)/(2) log""([Zn^(+2)])/([Cu^(+2)]) , (E_("cell") - E_("cell")^(0)) =(-0.86)/(2) log""([Zn^(+2)])/([Cu^(+2)])-0.12 = (-0.06)/(2) log""([Zn^(+2)])/([Cu^(+2)])`
`log""([Zn^(+2)])/([Cu^(+2)]) = (0.12 xx 2)/(0.06) = (0.24)/(0.06) = 4, ([Zn^(+2)])/([Cu^(+2)]) = 10^(4)`
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