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"):}`

Options

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`

Answer

A

Explanation

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`