InterviewSolution
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InterviewSolution
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Consider the reaction fo extraction of gold from its ore `Au +2CN^(-) (aq)+(1)/(4)O_(2) (g)+(1)/(2)H_(2)O rarr Au(CN)_(2)^(-) +OH^(-)` Use the following data to calculate `DeltaG^(@)` for the reaction `K_(f) [Au(CN)_(2)^(-)] = X` `{:(O_(2)+2H_(2)O +4e^(-)rarr4OH^(-),:,E^(@) =+0.41 "volt"),(Au^(3+)+3e^(-)rarr Au,:,E^(@) = +1.5 "volt"),(Au^(3+)+2e^(-)rarr Au^(+),:,E^(@) =+1.4 "volt"):}`A. `-RT In X +1.29 F`B. `-RT In X - 2.11 F`C. `-RT In(1)/(X) +2.11F`D. `-RT In X -1.29 F` |
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Answer» Correct Answer - A Given `Au^(+3) +3e^(-) rarr E_(1)^(@) = 1.5` `Au^(+3) +2e^(-) rarr Au^(+) rarr E_(2)^(@) = 1.4` Ao for reaction `Au rarr Au^(+) +e^(-) E^(@) = 2E_(2)^(@) - 3E_(1)^(@) rArr E^(@) - 1.7` `Au +2CN^(-) rarr Au(CN)_(2)^(-) +e^(-) E^(@) = x` `Au^(+) +e^(-) rarr E^(@) = 1.7` `O = x +1.7 -(RT)/(F) In X - 1.7` `x = (RT)/(F) In - 1.7` so for reaction `Au +2CN^(-) +(1)/(4)O_(2) +(1)/(2)H_(2)O rarr Au(CN)_(2)^(-) +OH^(-)` `E^(@) = 0.41 +(RT)/(F) In x - 1.7 =- 1.29 +(RT)/(F) In X` `DeltaG^(@) =- nFE^(@) (n =1)` `DeltaG^(@) = 1.29 F - RT In x` |
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