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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

0.52 V

B

0.90 V

C

0.30 V

D

0.38 V

Text Solution

Verified by Experts

The correct Answer is:
A
(a) `DeltaG^(@)=-nFE^(@)`
For reaction, `Cu^(2+)+2e^(-) to Cu, ` …(i)
`DeltaG^(@)=-2xxFxx0.337`
For reaction, `Cu^(+) to Cu^(2+)+e^(-)`, ….(ii)
`DeltaG^(@)=-1xxFxx(-0.153)`
`=+0.153 F`
Adding Eqs. (i) and (ii) , we get
`Cu^(+)+e^(-) to Cu, DeltaG^(@)=-0.521 F`
`DeltaG^(@)=-nFE^(@)`
`:. -0.521 F=-nFE^(@)`
`:. E^(@)=0.52 V`
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