Question

Which equation represents the reduction half-reaction for this redox equation?

Mg + CuSO4 MgSO4 + Cu

A. Cu Cu
B. Cu + 2e- Cu
C. Cu Cu + 2e-
D. Cu + 3e- Cu,

Answer 1

The oxidation number of Cu in CuSO₄ is +2 and 0 in Cu, the free elemental form. As the oxidation of Cu decreases from +2 to 0, we are looking at the reduction half-reaction for the given redox reaction:

Cu²⁺ + 2e⁻ → Cu

The correct answer is option B.

Answer 2

The oxidation state of Copper here will be 2+ because of the sulfate moiety (2-). The correct answer therefore should be C, Cu to Cu +2e-. The 2+ oxidation state results in 2 free electrons on the right side of the equation, a reduction.