Question

Make the electrons match in these half-reactions:

Cr₂O₇⁻² + 6e⁻ + 14H⁺ = 2Cr⁺³ + 7H₂O
C₂O₄⁻² = 2CO₂ + 2e⁻

Answer 1

To balance the given half-reactions, multiply the oxidation reaction by 3 and the reduction reaction by 2 to equalize the number of electrons. The balanced overall equation is 3Cr₂O₇⁻² + 42e⁻ + 42H⁺ + 6C₂O₄²⁻ = 6Cr⁺³ + 21H₂O + 12CO₂.

Step-by-step explanation:

To balance the given half-reactions, we need to make the number of electrons equal on both sides. The oxidation half-reaction has 6 electrons on the reactant side and the reduction half-reaction has 2 electrons on the product side. The least common multiple of these numbers is 6, so we can multiply the oxidation half-reaction by 3 and the reduction half-reaction by 3 to balance the electrons.

The balanced half-reactions are:

Oxidation: 3(Cr₂O₇⁻² + 6e⁻ + 14H⁺ = 2Cr⁺³ + 7H₂O)

Reduction: 3(C₂O₄²⁻ = 2CO₂ + 2e⁻)

By multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2, we can make the number of electrons equal on both sides and balance the charges.

The balanced overall equation is:

3Cr₂O₇⁻² + 42e⁻ + 42H⁺ + 6C₂O₄²⁻ = 6Cr⁺³ + 21H₂O + 12CO₂