Question

Calculate the standard free-energy change for the following reaction at 25 °C.

2Au^3+(aq) + 3Ni(s) -->2Au(s) + 3Ni^2+(aq)

Answer 1

The answer should be

(-6) (96.485 kj)(1.758)
= -1017.72

Answer 2

Thanks for showing your work.

I think your problem is that you are using 12 for n but there are only 6 e involved. You multiplied Au half by 2 (to make 6) and multiplied Ni half by 3 (to make 6) so 6 goes in for n.

Answer 3

you are little wrong. because 6 e- are gain and lose. Also, E for Au is 1.498 or not 1.50.

So the correct answer is -1019.04kj/mol

I used yours but got wrong and then, i tried again and got this answer correct.

Answer 4

dGo = -nFEo

where n = 6
F = 96,485
and Eo = cathode - anode = (+1.498) - (-0.26) = 1.758

So,
(-6)(96,485)(1.758) = -1,017,720 J or -1017.72 KJ

(Make sure you divide by 1,000 to convert from J to KJ if the question asks for the answer in KJ)

Answer 5

Calculate Eo for the cell then substitute in dGo = -nFEo and solve for dGo.

Answer 6

for n do you have to use the balanced number of before balancing? i would assume balance it first but i wanted to double check

Answer 7

I've done this problem 4 times and keep getting it wrong. Could you please check my work?

3e- + Au^3+ --> Au and E=1.52
Ni--> Ni^2+ + 2e- and E = .26
total E = 1.78

dGo = -12(9.65x10^4)(1.78) = -2061240J or -2061.24kJ