Question

What is the overall voltage for a redox reaction with the half-reactions Mg(s) Mg2+ + 2e- and Cu2+ + 2e- Cu(s)?

A. -2.37 - 0.34
B. 0.34 - 2.37
C. 0.34 - (-2.37)
D. 2.37 - 0.34

Answer 1

Well, first you need to identify both the reduction and oxidation half-reactions. Reduction is the gain of electrons, so therefore Cu 2+ + 2e- Cu(s) is the reduction half.

Oxidation is the loss of electrons, thus Mg(s) -> Mg2+ + 2e- is the oxidation half.

The voltage of the cell is \[E ^{o}_{cathode} - E ^{o}_{anode}\]

If you use your standard reduction potentials table, you will see that the reduction 1/2 (which occurs at the cathode) gives a voltage of 0.34. The oxidation half (which occurs at the anode) gives a voltage of -2.37.

Thus, (0.34) - (-2.37) = 2.71 V or C)

Hope this helps!

Answer 2

thanks! really helpful!