1.

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

Answer»

SOLUTION :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)^(@) =-1XX 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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