Question

Study this chemical reaction: Cu + Cl2 → CuCl2 Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction oxidation: reduction: ク

Answer 1

The oxidation half-reaction is Cu → Cu²+ + 2e-, and the reduction half-reaction is Cl2 + 2e- → 2Cl-. Cu is oxidized at the anode, and Cl2 is reduced at the cathode.

Step-by-step explanation:

The oxidation and reduction half-reactions for the chemical reaction Cu + Cl2 → CuCl2 are as follows:

Oxidation half-reaction: Cu → Cu²+ + 2e-

Reduction half-reaction: Cl2 + 2e- → 2Cl-

The anode is where oxidation occurs, so in this case, Cu is oxidized at the anode. The cathode is where reduction occurs, so in this case, Cl2 is reduced at the cathode.

Answer 2

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:

1. 2 Cr(s) → 2 Cr³+ (aq) + 6e¯ (oxidation)
2. 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.