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The reduction potential for the two half...

The reduction potential for the two half cell reactions are :
`Cu^(2+) + e^(-) to Cu^(+), E^(@) = 0.15 V `
`Cu^(+) + e^(-) to Cu, E^(@) = 0.50 V`
Calculate reduction potential for the following reaction :
`Cu^(2+) + 2e^(-) to Cu`

Text Solution

Verified by Experts

This can be solved in terms of their free energy changes .
(i) `Cu^(2+) + e^(-) to Cu^(+) " " E_(1)^(@) = 0.15` V
`Delta G_(1)^(@) = - 1 xx F xx 0.15 = -0.15 F`
(ii) `Cu^(+) + e^(-) to Cu " " E_(2)^(@) = 0.50` V
`Delta G_(2)^(@) =-1 xx F xx 0.50 = -0.50 F`
(iii) `Cu^(2+) + 2e^(-) to Cu " " E_(3)^(@) = ?`
`Delta G_(3)^(@) = - 2 xx F xx E_(3)^(@) = - 2 E_(3)^(@) F`
Now `Delta G_(3)^(@) = Delta G_(1)^(@) + Delta G_(2)^(@) = - 0.15 F - 0.50 F = -0.65 F`
`therefore E_(3)^(@) = (0.65 F)/(2 F) = 0.325 V`
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