Grade by Grade
F. Electrochemistry · Electrochemistry

Balancing Redox Equations in Basic Solution, and the Oxidation-Number Method

F2.3 F3.1About 75 minutes
Learning goals
  • I can balance redox half-reactions and equations in basic solution.
  • I can balance a disproportionation reaction.
  • I can balance a redox equation with the oxidation-number method.
  • I can choose the more efficient method for a given equation.

Words to know

basic solution
A solution in which hydroxide ions, OH⁻(aq), are plentiful and H⁺(aq) is scarce, so the final balanced equation should contain OH⁻ and H₂O, not H⁺.
disproportionation
A redox reaction in which the same element is both oxidized and reduced, e.g. Cl₂ → Cl⁻ + ClO⁻.
oxidation-number method
A way to balance redox equations by making the total increase in oxidation number equal the total decrease.
hypochlorite ion
ClO⁻, the active ingredient in household bleach (as sodium hypochlorite, NaClO).
comproportionation
The opposite of disproportionation: two forms of an element, one higher and one lower, react to give one in-between form.

When the solution is basic

Many important redox reactions happen in basic solution: making bleach, the reactions inside alkaline batteries, and treating cyanide in mine wastewater. In a basic solution, OH⁻ ions are everywhere and H⁺ ions are extremely scarce. So a final equation with H⁺ in it would describe something that does not happen.

The good news: you do not need a new method. Balance exactly as you learned in the last lesson, Balancing Redox Equations: The Half-Reaction Method in Acidic Solution, then convert.

Converting to basic solution

After balancing a half-reaction or a full equation as if it were acidic:

  1. For every H⁺ in the equation, add the same number of OH⁻ to both sides. (Adding the same thing to both sides keeps everything balanced.)
  2. On the side that has both H⁺ and OH⁻, combine them into water: H⁺ + OH⁻ → H₂O.
  3. Cancel water molecules that appear on both sides.
  4. Check atoms and charge again.
1. Balance as if acidicMnO₄⁻ + 4H⁺ + 3e⁻→ MnO₂ + 2H₂O2. Add OH⁻ to BOTH sidesMnO₄⁻ + 4H⁺ + 4OH⁻ + 3e⁻→ MnO₂ + 2H₂O + 4OH⁻3. Join H⁺ + OH⁻ = H₂OMnO₄⁻ + 4H₂O + 3e⁻→ MnO₂ + 2H₂O + 4OH⁻4. Cancel extra waterMnO₄⁻ + 2H₂O + 3e⁻→ MnO₂ + 4OH⁻
Converting the permanganate-to-manganese(IV)-oxide half-reaction from acidic to basic. The 4H⁺ become 4H₂O after adding 4OH⁻ to both sides; then 2 of the waters cancel. Both the final atoms and the final charge (−4 on each side) balance.
Worked example

Example 1: Permanganate and iodide in base

Balance in basic solution: MnO₄⁻(aq) + I⁻(aq) → MnO₂(s) + I₂(aq).

  1. Reduction (balanced as if acidic): MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O.
  2. Oxidation: 2I⁻ → I₂ + 2e⁻.
  3. Equalize electrons (LCM of 3 and 2 is 6): 2MnO₄⁻ + 8H⁺ + 6e⁻ → 2MnO₂ + 4H₂O and 6I⁻ → 3I₂ + 6e⁻.
  4. Add: 2MnO₄⁻ + 8H⁺ + 6I⁻ → 2MnO₂ + 3I₂ + 4H₂O.
  5. Add 8OH⁻ to both sides. The left now has 8H⁺ + 8OH⁻ = 8H₂O: 2MnO₄⁻ + 8H₂O + 6I⁻ → 2MnO₂ + 3I₂ + 4H₂O + 8OH⁻.
  6. Cancel 4H₂O from each side: 2MnO₄⁻ + 4H₂O + 6I⁻ → 2MnO₂ + 3I₂ + 8OH⁻.
  7. Check: O 8 + 4 = 12 and 4 + 8 = 12; H 8 = 8; charge −2 − 6 = −8 and −8. ✓

2MnO₄⁻(aq) + 6I⁻(aq) + 4H₂O(l) → 2MnO₂(s) + 3I₂(aq) + 8OH⁻(aq)

Worked example

Example 2: Making bleach (disproportionation)

Household bleach is made by bubbling chlorine gas through sodium hydroxide solution. Balance in basic solution: Cl₂(g) → Cl⁻(aq) + ClO⁻(aq).

  1. Chlorine starts at 0 and ends at −1 (in Cl⁻) and +1 (in ClO⁻). The same element is reduced and oxidized: a disproportionation. Write Cl₂ in both half-reactions.
  2. Reduction: Cl₂ + 2e⁻ → 2Cl⁻.
  3. Oxidation (as if acidic): Cl₂ + 2H₂O → 2ClO⁻ + 4H⁺ + 2e⁻.
  4. Electrons are already equal (2 each). Add: 2Cl₂ + 2H₂O → 2Cl⁻ + 2ClO⁻ + 4H⁺.
  5. Add 4OH⁻ to both sides; 4H⁺ + 4OH⁻ = 4H₂O on the right: 2Cl₂ + 2H₂O + 4OH⁻ → 2Cl⁻ + 2ClO⁻ + 4H₂O.
  6. Cancel 2H₂O: 2Cl₂ + 4OH⁻ → 2Cl⁻ + 2ClO⁻ + 2H₂O. Divide every coefficient by 2.

Cl₂(g) + 2OH⁻(aq) → Cl⁻(aq) + ClO⁻(aq) + H₂O(l). With sodium ions as spectators, this is Cl₂ + 2NaOH → NaCl + NaClO + H₂O.

Bleach and acid: the reverse reaction

In acid, the bleach reaction runs the other way (a comproportionation): ClO⁻ + Cl⁻ + 2H⁺ → Cl₂ + H₂O. That is why bleach labels warn never to mix bleach with acidic cleaners (some toilet-bowl and rust removers): the mixture releases toxic chlorine gas. Mixing bleach with ammonia is also dangerous; it forms other toxic gases.

The oxidation-number method

The half-reaction method is the most reliable for ionic equations in solution. For some equations, especially ones written with whole compounds, the oxidation-number method is faster.

  1. Assign oxidation numbers and find the atoms that change.
  2. Find the increase per atom oxidized and the decrease per atom reduced. Multiply by the number of those atoms in each formula.
  3. Choose coefficients so the total increase equals the total decrease. (This is the same as electrons lost = electrons gained.)
  4. Balance the other atoms by inspection. For an ionic equation in solution, balance charge and O and H using H₂O and H⁺ (acidic) or H₂O and OH⁻ (basic).
Worked example

Example 3: Oxidation-number method with compounds

Balance: H₂S + HNO₃ → S + NO + H₂O.

  1. Oxidation numbers that change: S goes from −2 (in H₂S) to 0: an increase of 2. N goes from +5 (in HNO₃) to +2 (in NO): a decrease of 3.
  2. Make the totals equal: 3 S atoms × (+2) = +6 and 2 N atoms × (−3) = −6. So: 3H₂S + 2HNO₃ → 3S + 2NO + H₂O.
  3. Balance H and O by inspection. Left H: 6 + 2 = 8, so 4H₂O on the right. O: left 6; right 2 + 4 = 6. ✓

3H₂S + 2HNO₃ → 3S + 2NO + 4H₂O

Worked example

Example 4: Oxidation-number method in basic solution

Balance in basic solution: MnO₄⁻(aq) + SO₃²⁻(aq) → MnO₂(s) + SO₄²⁻(aq).

  1. Mn: +7 → +4, a decrease of 3. S: +4 → +6, an increase of 2.
  2. Total increase = total decrease = 6: 2MnO₄⁻ + 3SO₃²⁻ → 2MnO₂ + 3SO₄²⁻.
  3. Charge: left 2(−1) + 3(−2) = −8; right 3(−2) = −6. In basic solution, add 2OH⁻ to the right: charge −8 on both sides.
  4. Oxygen: left 8 + 9 = 17; right 4 + 12 + 2 = 18. Add 1 H₂O to the left. Hydrogen: 2 = 2. ✓

2MnO₄⁻(aq) + 3SO₃²⁻(aq) + H₂O(l) → 2MnO₂(s) + 3SO₄²⁻(aq) + 2OH⁻(aq)

012345678Mn: −3 each, × 2 Mn = −6
Oxidation-number method, part 1. Each manganese drops from +7 (in MnO₄⁻) to +4 (in MnO₂): it gains 3 electrons. Two MnO₄⁻ ions gain 6.
012345678S: +2 each, × 3 S = +6
Part 2. Each sulfur rises from +4 (in SO₃²⁻) to +6 (in SO₄²⁻): it loses 2 electrons. Three SO₃²⁻ ions lose 6, which matches the 6 gained by manganese.
SituationUsually easiest method
Ions in acidic solution (MnO₄⁻, Cr₂O₇²⁻)half-reaction method, acidic
Ions in basic solutionhalf-reaction method, then add OH⁻
Disproportionationhalf-reaction method (write the element in both halves)
Whole compounds, e.g. combustion, smelting, KMnO₄ + HCloxidation-number method
You need the half-reactions for a cell (next lessons)half-reaction method
Both methods give the same answer. Choose the one that suits the equation, and always finish with an atom and charge check.
Just for fun

Bromine is a dramatic element: in hot basic solution it splits itself into bromide (−1) and bromate (+5). Chemists call that disproportionation. Everyone else calls it "not being able to make up your mind".

Hands-on activity

Read the warning label, then balance it

You need: a bottle of household bleach with the cap on (do not open it), paper and a pencil. A grown-up should be present.

  1. Find the active ingredient on the label (sodium hypochlorite, NaClO) and the warnings about mixing.
  2. Write the half-reactions for bleach meeting an acid: ClO⁻ is reduced to Cl₂ and Cl⁻ is oxidized to Cl₂. Balance each in acidic solution.
  3. Add them to get the overall equation. Identify the oxidizing agent and the reducing agent.
  4. Write two sentences, as if for a younger sibling, explaining why the bottle says not to mix it with other cleaners.
  5. Find the hazard symbols on the label. Which hazards do they show?

Model answer: 2ClO⁻ + 4H⁺ + 2e⁻ → Cl₂ + 2H₂O and 2Cl⁻ → Cl₂ + 2e⁻. Added and divided by 2: ClO⁻ + Cl⁻ + 2H⁺ → Cl₂ + H₂O. ClO⁻ is the oxidizing agent and Cl⁻ is the reducing agent. For step 5, write down each symbol you find and what it means (for example, a corrosive symbol warns that the product can damage skin and eyes). Never test this reaction. Chlorine gas damages the lungs.

Your turn

1

Balance in basic solution: MnO₄⁻ → MnO₂.

2

Balance in basic solution: Cr(OH)₃ → CrO₄²⁻.

3

Balance in basic solution: ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + I₂(aq).

4

Balance in basic solution: ClO₂(aq) + H₂O₂(aq) → ClO₂⁻(aq) + O₂(g).

5

Use the oxidation-number method: Fe₂O₃ + CO → Fe + CO₂.

6

Use the oxidation-number method: SnCl₂ + HgCl₂ → SnCl₄ + Hg₂Cl₂.

7

Why should a balanced equation for a reaction in basic solution not contain H⁺?

More practice: warm-up

1

In basic solution, what ion do you add to both sides to remove H⁺?

2

What do H⁺ and OH⁻ combine to form?

3

What is a disproportionation reaction?

4

In the oxidation-number method, what must be equal?

5

Is MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ charge-balanced?

More practice: on your own

1

Balance in basic solution: CrO₄²⁻(aq) + SO₃²⁻(aq) → Cr(OH)₃(s) + SO₄²⁻(aq).

2

Balance the disproportionation in basic solution: Br₂ → Br⁻ + BrO₃⁻.

3

Use the oxidation-number method: N₂H₄ + Cu(OH)₂ → N₂ + Cu + H₂O.

4

Use the oxidation-number method: KMnO₄ + HCl → KCl + MnCl₂ + Cl₂ + H₂O.

5

A student balances a basic-solution equation and gets 2MnO₄⁻ + 8H⁺ + 6I⁻ → 2MnO₂ + 3I₂ + 4H₂O. What should they do next?

Challenge

1

Balance in basic solution: Al(s) + NO₃⁻(aq) → Al(OH)₄⁻(aq) + NH₃(aq).

2

Gold mines in Canada destroy leftover cyanide in wastewater by oxidizing it. Balance the permanganate version in basic solution: CN⁻(aq) + MnO₄⁻(aq) → CNO⁻(aq) + MnO₂(s).

3

Use the oxidation-number method: P₄ + HNO₃ + H₂O → H₃PO₄ + NO.

Check your understanding

Quiz

1.

When converting a balanced acidic equation to basic solution, you add OH⁻:

2.

Which is the balanced equation for making bleach in basic solution?

3.

Which reaction is a disproportionation?

4.

In the oxidation-number method for Fe₂O₃ + CO → Fe + CO₂, what is the total decrease in oxidation number for one Fe₂O₃?

5.

Why is mixing bleach with an acidic cleaner dangerous?

6.

Which balanced half-reaction is correct for basic solution?

For parents and teachers

Big idea (F2.3): in basic solution, balance as if acidic, then add OH⁻ to both sides to turn every H⁺ into water. The oxidation-number method balances by making the total increase in oxidation number equal the total decrease. Both methods must end with an atom and charge check.

How to help: Ask your teen to show you the two extra moves for basic solution (add OH⁻ to both sides; combine into water and cancel). Then let them pick the method for a new question and explain why.

Common difficulties:

  • Adding OH⁻ to one side only.
  • Forgetting to cancel water at the end, or cancelling from one side only.
  • In disproportionation, writing the element only once. It belongs in both half-reactions.
  • In the oxidation-number method, forgetting to multiply by the number of atoms in a formula (2 Fe in Fe₂O₃, 2 N in N₂H₄).

Safety: the activity only reads a bleach label. Keep the cap on, and never mix bleach with other cleaners.