Question

Balance each of the following redox reactions occurring in basic solution.

Express your answer as a chemical equation. Identify all of the phases in your answer.

H2O2(aq)+ClO2(aq)?ClO?2(aq)+O2(g)

Al(s)+MnO?4(aq)?MnO2(s)+Al(OH)?4(aq)

Cl2(g)?Cl?(aq)+ClO?(aq)

Answer 1

H₂O₂ +ClO₂ = ClO₂^- + O₂

Step 1: Split into two half-reactions

H₂O₂ = O₂

ClO₂ = ClO₂^-

Step 2: Balance elements (except H and O). In this case, Cl is already balanced.

H₂O₂ = O₂

ClO₂ = ClO₂^-

Step 3: Balance O by adding H₂O. In this case, O is already balanced in both half-reactions.

H₂O₂ = O₂

ClO₂ = ClO₂^-

Step 4: Balance H by adding H⁺. In this case, add 2H⁺ on the right side of the first equation to balance H. The second equation doesn't have H so it doesn't require adding anything.

H₂O₂ = O₂ + 2H⁺

ClO₂ = ClO₂^-

Step 5: Balance charges by adding e^-. In this case, in the first half-reaction, the charge on the left is 0 and the charge on the right is 2+ so add 2e^- on the right side, making the charge on both sides 0. In the second half-reaction, the charge on the left is 0 and the charge on the right is 1-, so add 1e^- on the left side, making the charge 1- for both sides.

H₂O₂ = O₂ + 2H⁺ + 2e^-

e^- + ClO₂ = ClO₂^-

Step 6: Multiply the equations (if needed) so that they have the same number of electrons, in this case, multiply the second half reaction times 2 so that both half-reactions have 2e^-.

H₂O₂ = O₂ + 2H⁺ + 2e^-

2e^- + 2ClO₂ = 2ClO₂^-

Step 7: Add the half-reactions.

H₂O₂ + 2ClO₂ = O₂ + 2H⁺ + 2ClO₂^-

Step 8: Add OH^- per every H⁺ present, in this case, add 2OH^- on each side of the equation.

H₂O₂ + 2ClO₂ + 2OH^- = O₂ + 2H⁺ + 2ClO₂^- + 2OH^-

Step 9 (Final): Combine H⁺ and OH^- into water.

H₂O₂ + 2ClO₂ + 2OH^- = O₂ + 2H₂O + 2ClO₂^-