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
UGHH

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.

none of these are right fu

got it! thanks!

To calculate the change in Gibbs free energy (ΔG) for a reaction, you need to use the equation: ΔG = -nFE, where n is the number of moles of electrons transferred in the balanced equation, F is the Faraday constant (96,485 J/mol e-volt), and E is the cell potential.

Let's break down the calculation step by step:

1. First, determine the number of moles of electrons transferred in the reaction. Looking at the balanced equation, we can see that 3 moles of electrons are transferred.

2. Next, substitute the values into the equation: ΔG = -(3 mol e-) * (96,485 J/mol e-volt) * (1.27 V)

3. Now, calculate the product of the moles of electrons, the Faraday constant, and the cell potential: -(3) * (96,485) * (1.27) = -387,982.95 J = -387.98 kJ (rounded to two decimal places)

Based on the calculations, the ΔG of the reaction should be -387.98 kJ, not -1.22 x 10^2 kJ. It seems there might have been an error in substituting the values. Double-check your calculations and ensure that you have used the correct values for moles of electrons, Faraday constant, and cell potential to get the correct answer.