Final answer:
For the overall reaction 2 Cr(s) + 3 Cu²+ (aq) → 2 Cr³+ (aq) + 3 Cu(s), the oxidation half-reaction is Cr(s) → Cr³+ (aq) + 3e¯ and occurs at the anode, while the reduction half-reaction is Cu²+ (aq) + 2e¯ → Cu(s) and occurs at the cathode.
Step-by-step explanation:
When balancing half-reactions for oxidation and reduction, we must consider both the transfer of electrons and the conservation of mass and charge. For the given overall reaction 2 Cr(s) + 3 Cu²+ (aq) → 2 Cr³+ (aq) + 3 Cu(s), the oxidation and reduction half-reactions are:
- Oxidation (occurs at the anode): Cr(s) → Cr³+ (aq) + 3e¯
- Reduction (occurs at the cathode): Cu²+ (aq) + 2e¯ → Cu(s)
To balance them, we multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2 to get the same number of electrons in both half-reactions. Then, we add them together and simplify to get the overall balanced equation:
- 2 Cr(s) → 2 Cr³+ (aq) + 6e¯ (oxidation)
- 3 × (Cu²+ (aq) + 2e¯ → Cu(s)) → 3 Cu²+ (aq) + 6e¯ → 3 Cu(s) (reduction)
Adding the half-reaction equations and simplifying yields the same overall reaction as given.
The reaction at the anode is the oxidation while the reaction at the cathode is the reduction.