Question

Write the balanced half-reactions for the given redox reaction: 3I₂(s) + 2Cr³⁺(aq) + 14OH⁻(aq) → 6I⁻(aq) + Cr₂O⁷²⁻(aq) + 7H₂O(l).

a) Oxidation: 3I₂(s) → 6I⁻(aq) + 12e⁻; Reduction: 2Cr³⁺(aq) + 14OH⁻(aq) + 6e⁻ → Cr₂O⁷²⁻(aq) + 7H₂O(l)
b) Oxidation: 3I⁻(aq) → 3I₂(s) + 6e⁻; Reduction: 2Cr²O⁷²⁻(aq) + 7H₂O(l) + 6e⁻ → 2Cr³⁺(aq) + 14OH⁻(aq)
c) Oxidation: 6I⁻(aq) + 12e⁻ → 3I₂(s); Reduction: 2Cr³⁺(aq) + 14OH⁻(aq) + 6e⁻ → Cr₂O⁷²⁻(aq) + 7H₂O(l)
d) Oxidation: 2Cr³⁺(aq) + 14OH⁻(aq) + 6e⁻ → Cr₂O⁷²⁻(aq) + 7H₂O(l); Reduction: 3I²(s) → 6I⁻(aq) + 12e⁻

Answer 1

The balanced half-reactions for the given redox reaction are the oxidation of iodine to iodide ions with the formation of 12 electrons, and the reduction of chromium ions to dichromate ions in the presence of hydroxide ions, resulting in water as a byproduct.

Step-by-step explanation:

To answer the question of writing the balanced half-reactions for the given redox reaction, let's first identify the oxidation and reduction processes separately:

• Oxidation: This process involves the loss of electrons. In this redox reaction, iodine (I2) is oxidized to iodide ions (I−).
• Reduction: In this process, a substance gains electrons. Chromium ions (Cr3+) are reduced to the dichromate ion (Cr2O72−) in the presence of hydroxide ions (OH−).

Applying the steps for balancing redox reactions:

1. Write separate half-reactions for the oxidation and reduction.
2. Balance the atoms other than O and H in each half-reaction.
3. Balance oxygen atoms by adding water (H2O).
4. Balance hydrogen atoms by adding hydrogen ions (H+).
5. Balance charge by adding electrons (e−).
6. Add the balanced half-reactions, and cancel species that appear on both sides.

The correct balanced half-reactions for this redox reaction are:

Oxidation: 3I2(s) → 6I−(aq) + 12e−

Reduction: 2Cr3+(aq) + 14OH−(aq) + 6e− → Cr2O72−(aq) + 7H2O(l)

The answer to the given question is option (a).