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Cu^(+) + e rarr Cu, E^(@) = X(1) volt, ...

`Cu^(+) + e rarr Cu, E^(@) = X_(1)` volt,
`Cu^(2+) + 2e rarr Cu, E^(@) = X_(2)`X_(2) volt
For `Cu^(2+) + e rarr Cu^(+), E^(@)` will be :

A

`x_(1) -2x_(2)`

B

`x_(1)+2x_(2)`

C

`x_(1)-x_(2)`

D

`2x_(2)-x_(1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`Cu^(+)+e^(-)rarr Cu Delta G_(1)= -F E^(@)= -FX_(1)`
`Cu^(2+)+2e^(-)rarr Cu Delta G_(2)= -2F E^(@)= -2 FX_(2)`
`Cu^(2+)+e^(-)rarr Cu^(+) Delta G_(3)= -2F E^(@)= -2 FX_(2)`
`Delta G_(3) = Delta G_(2)-Delta G_(1) -F E^(@)= -2 FX_(2)+FX_(1)`
`E^(@)=2X_(2)-X_(1)`
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