posted by Katherine on .
Calculate delta G of the reaction at 25 degrees C: 2Au^3+ (aq) + 3Ni(s) <=> 2Au(s) + 3Ni^2+(aq)
Au3+ + 3e- ==> Au Eo = +1.50V
Ni2+ + 2e- ==> Ni Eo = -0.23 V
2Au3+ + 3Ni ==> 3Ni2+ + 2Au
The answer is -1.00 x 10^3 kJ
but i'm not getting that answer!!!!!!! please tell me what i'm substituting incorrectly.
Delta G = -nFE
= - (1 moles e-) (96,485 J/mol e-volt) (1.27 volts)
dG = -122,535.95 J = -122.53 kJ = -1.22 x 10^2 kJ
n = 6 (Balance the two half equations--one oxdn and the other redn-- to see you must multiply each half equation by 2 and 3 respectively which gives six total electrons transferred.)
F = 96,485 which you have.
E = 1.50+0.23 = 1.73 (You add the oxidn half to the redn half OR Ecell = E1-E2 = 1.50 - (-0.23) = ?
Answer is in joules. Change to kJ.
got it! thanks!
none of these are right fu