1.

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

Answer»

0.52 V
0.90 V
0.30 V
0.38 V

Solution :`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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