Question

Balance these equations for reactions occurring in a basic solution. Don't include physical states for your answer.

CrO₂^- + S₂O₈^2- → CrO₄^2- + SO₄^2-

Au + CN^- + O₂ → Au(CN)₄^- + OH^-

Answer 1

1. First, balance them as if they were in acid. Then add as many OH⁻ ions to both sides as you have H⁺ ions to form water and remove the H⁺ from the equation.

CrO₂⁻ + S₂O₈²⁻ → CrO₄²⁻ + SO₄²⁻

Split the equation into half reactions.

CrO₂⁻ → CrO₄²⁻
S₂O₈²⁻ → SO₄²⁻

Balance the elements, leaving H and O until last. Balance O by adding H₂O, and H by adding H⁺. Finally, balance the electrons.

CrO₂⁻ + 2H₂O → CrO₄²⁻ + 4H⁺ + 3e⁻
S₂O₈²⁻ + 2e⁻ → SO₄²⁻

Now add the half reactions together, making sure the electrons cancel (so here, multiply the top reaction by 2 and the bottom one by 3).

2CrO₂⁻ + 4H₂O → 2CrO₄²⁻ + 8H⁺ + 6e⁻
3S₂O₈²⁻ + 6e⁻ → 3SO₄²⁻
—————————————————
2CrO₂⁻ + 4H₂O + 3S₂O₈²⁻ → 2CrO₄²⁻ + 8H⁺ + 3SO₄²⁻

Now remove the H⁺ by adding an equal number of OH⁻ to both sides.

2CrO₂⁻ + 4H₂O + 3S₂O₈²⁻ + 8OH⁻ → 2CrO₄²⁻ + 8H⁺ + 8OH⁻ + 3SO₄²⁻

Now form water from the H⁺ and OH⁻ and remove excess water molecules to get the final reaction.

2CrO₂⁻ + 3S₂O₈²⁻ + 8OH⁻ → 2CrO₄²⁻ + 3SO₄²⁻ + 4H₂O

2. 4 Au + 16 CN^- + 3 O₂ + 6 H₂O = 4 Au(CN)₄^- + 12 OH^-