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`Cu^(+) + e rarr Cu, E^(@) = X_(1)` volt, `Cu^(2+) + 2e rarr Cu, E^(@) = X_(2)`X_(2) volt For `Cu^(2+) + e rarr Cu^(+), E^(@)` will be :A. `X_(1) - 2X_(2)`B. `X_(1) + 2X_(2)`C. `X_(1) - X_(2)`D. `2X_(2) - X_(1)` |
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Answer» Correct Answer - D `{:(Cu^(+)+erarrCu,-Delta G_(1)^(@)=1xxX_(1)F),(underset(-)(Cu)^(2+)+2erarr underset(-)(Cu),-Delta G_(2)^(2)=underset(-)(2)X_(2)F),(bar(Cu^(+)rarr Cu^(2+)+e, :. -Delta G_(3)^(@)=(-2X_(2)+X_(1))F)),(orCu^(+)+e rarr Cu^(1+), :. Delta G^(@)=(-X_(1)+2X_(2))F):}` or `1xxE_(3)^(@)xxF=(2X_(2)-X_(1))F` or `E_(3)^(@)=(2X_(2)-X_(1))V`. |
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