Question

The free energy change for the following reaction at 25 °C, when [Cu2+] = 1.19 M and [Zn2+] = 5.28×10-3 M, is -226 kJ:

Cu2+(1.19 M) + Zn(s) = Cu(s) + Zn2+(5.28×10-3 M) ΔG = -226 kJ

What is the cell potential for the reaction as written under these conditions?

Answer:______ V

Would this reaction be spontaneous in the forward or the reverse direction?

Answer 1