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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
`DeltaG^(@)=nFE^(@)`
For, `Cu^(2+)+2e^(-)toCu,DeltaG_(1)^(@)=-2F(0.337)` . . . . (i)
`Cu^(2+)+e^(-)toCu^(+),DeltaG_(2)^(@)=-F(0.153)` . . . (ii)
Subtracting (ii) from (i),
`Cu^(+)+e^(-)toCu`
`DeltaG^(@)=-0.674F+0.153F` . . . (iii)
`nFE^(@)=-0.521F`
`because`For (iii), n=1, `thereforeE^(@)=0.521V`.
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