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Given: (i) Cu^(2+)+2e^(-) rarr Cu, E^(...

Given:
(i) `Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V`
(ii) `Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V`
Electrode potential, `E^(@)` for the reaction, `Cu^(+)+e^(-) rarr Cu`, will be

A

`0.90 V`

B

`0.30 V`

C

`0.38 V`

D

`0.52 V`

Text Solution

Verified by Experts

The correct Answer is:
D
`Cu^(+2)+2e^(-) to Cu " " DeltaG_(1)^(@)`
`(Cu^(+2) +e^(-) to Cu^(+) " "DeltaG_(2)^(@))/(Cu^(+) +e^(-) to Cu " " DeltaG_(3)^(@))`
`DeltaG_(3)^(@) =DeltaG_(1)^(@) -DeltaG_(2)^(@)`
`-1xxF xx E =- 2F xx 0.337 -(-1xxF 0.153)`
`rArr E= 0.521 V`
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