Reduction Potential Question. Calculating delta G
JL asked:
What is the value of DGo in kJ at 25 oC for the reaction between the pair:
Cu(s) and Cr3 (aq) to give Cr(s) and Cu2 (aq)
Use the reduction potentials for Cr3 (aq) is -0.74 V and for Cu2 (aq) is 0.34 V.
I know that Cu is being reduced since its reduction potential is higher but then that makes Cr 0.74 for the oxidation potential. So E is positive and doesn’t that make G negative since G = -nFE?
I keep getting -6E2. How would it be positive?
Incoming search terms:
- calculate delta g of cu2 2e-
- reduction poteintial G
Cu —–> Cu2+ + 2e- Eo = -0.34 V
Cr3+ + 3e- —-> Cr Eo = -.073 V
3Cu + 2Cr3+ —-> 3Cu2+ + 2Cr Eocell = -1.07 V
This reaction will not occur spontaneously because the Eocell value is negative.
dGo = -nFEocell = -6 mol x 96,500 C/mol x (-1.07 J/C) = 620,000 J = 620 kJ
dGo is positive, meaning that no useful work can be obtained from this cell. The reason is that it is a nonspontaneous reaction.