The problem is: Calculate the standard free-energy change for the following reaction at 25 degrees Celsius.
2Au^+3(aq) + 3Cr(s)-> 2Au(s) + 3Cr^+2(aq)
I worked it out, getting half reaction:
Au^3+ + 3e- -> Au; E = 1.498
Cr^3+ + 3e- -> Cr; E = -0.74
Ecell = 1.498 - (-0.74) = 2.238
n = 6
F = 96485
I plugged it into equation changeG = -nFEcell and got -1295.6 kJ (answer is supposed to be in kJ, so I converted it from J). However, the problem says I'm wrong, and I'm wondering at which point I messed up?
4 answers
3e not 6.
Thanks for your help! I got it now. I messed up with second half reaction, it was supposed to be Cr^2+ + 2e- -> Cr; E = -0.91.
It actually is 6 electrons when you plug it in Bob. You need 3 for each Au that is fully oxidized.
If anyone needs this, the correct answer is -1394 kJ/mol. n=6 and Ecell=2.408
2 answers