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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) 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^(-)toCu,E^(+)+x_(1)"volt"`
`Cu^(2+)+2e^(-)toCu,x_(2)"volt"`
`CutoCu^(+)+e^(-)-x_(1)"volt"`
`Cu^(2+)+e^(-)toCu^(+)`
`-2xx x_(2) xx f+1xx x_(1)xx f=-1xxE^(@)xxf`
`E^(@)=2x_(2)-x_(1)`
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