F. Electrochemistry · Electrochemistry

Balancing Redox Equations: The Half-Reaction Method in Acidic Solution

F2.3 F2.1 F3.1About 75 minutes

Learning goals

I can balance half-reactions in acidic solution using H₂O, H⁺ and e⁻.
I can combine half-reactions so that electrons lost equal electrons gained.
I can check a redox equation for both atom balance and charge balance.
I can use a balanced redox equation in a titration calculation.

Words to know

half-reaction method
A way to balance redox equations by balancing the oxidation and reduction half-reactions separately, then adding them so the electrons cancel. Also called the ion-electron method.

acidic solution
A solution with plenty of H⁺(aq) (really H₃O⁺(aq)) available, so H⁺ and H₂O can be used to balance hydrogen and oxygen.

charge balance
The total charge on the left side of an equation equals the total charge on the right side.

redox titration
A titration in which the reaction between the solutions is a redox reaction, for example permanganate with iron(II).

endpoint
The point in a titration where an indicator (or, for permanganate, the permanganate itself) changes colour.

Why the old method breaks down

Try to balance this by inspection, the way you balanced equations in Grade 10 and Grade 11:

MnO₄⁻(aq) + Fe²⁺(aq) → Mn²⁺(aq) + Fe³⁺(aq)

Manganese is balanced. Iron is balanced. But there are 4 oxygen atoms on the left and none on the right, and the charges are −1 + 2 = +1 on the left and +5 on the right. In acidic solution, the missing pieces are water molecules, hydrogen ions and electrons. The half-reaction method tells you exactly how many of each to add. It works for any redox reaction, no matter how messy.

The half-reaction method in eight steps. Steps 1 to 5 balance each half-reaction separately; steps 6 to 8 combine them. Learn the order: atoms, then O, then H, then charge.

The method, step by step (acidic solution)

Split the reaction into two half-reactions. Put each element that changes oxidation number in its own half-reaction.
Balance all atoms except O and H.
Balance O by adding H₂O molecules to the side that needs oxygen.
Balance H by adding H⁺ ions to the side that needs hydrogen.
Balance charge by adding electrons (e⁻) to the more positive side. In an oxidation the electrons end up on the right; in a reduction they end up on the left.
Multiply one or both half-reactions so the number of electrons lost equals the number gained.
Add the half-reactions. Cancel the electrons and any H₂O or H⁺ that appear on both sides.
Check that atoms balance and the total charge is the same on both sides.

Why can we add H₂O and H⁺? In an acidic aqueous solution, water is everywhere and hydrogen ions are plentiful. They are real participants in the reaction; the unbalanced equation just left them out.

Worked example

Example 1: Permanganate and iron(II)

Balance in acidic solution: MnO₄⁻(aq) + Fe²⁺(aq) → Mn²⁺(aq) + Fe³⁺(aq).

Split. Reduction: MnO₄⁻ → Mn²⁺. Oxidation: Fe²⁺ → Fe³⁺. (Mn goes from +7 to +2; Fe from +2 to +3.)
Atoms other than O, H: Mn and Fe are already balanced.
O: MnO₄⁻ has 4 O, so add 4H₂O to the right: MnO₄⁻ → Mn²⁺ + 4H₂O.
H: now 8 H on the right, so add 8H⁺ to the left: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O.
Charge: left (−1) + (+8) = +7; right +2. Add 5e⁻ to the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. The iron half: Fe²⁺ → Fe³⁺ + e⁻.
Multiply: 5 electrons gained, 1 lost per Fe²⁺. Multiply the iron half-reaction by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻.
Add and cancel 5e⁻: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
Check: Mn 1 = 1, Fe 5 = 5, O 4 = 4, H 8 = 8. Charge: (−1) + 8 + 10 = +17; (+2) + 15 = +17. ✓

MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

Step 5 for the permanganate half-reaction. After O and H are balanced, the left side has a charge of +7 and the right side +2. Adding 5 electrons to the left balances the charge. The 5 also matches the change in oxidation number of Mn (+7 to +2).

Two built-in checks

The number of electrons you add in step 5 should equal the change in oxidation number (times the number of atoms that change). For Mn: +7 → +2 is a change of 5, and you added 5e⁻. If these two numbers disagree, look for an error before going on.

Worked example

Example 2: Copper in nitric acid

When copper is placed in dilute nitric acid, colourless nitrogen monoxide gas forms. Balance: Cu(s) + NO₃⁻(aq) → Cu²⁺(aq) + NO(g).

Oxidation: Cu → Cu²⁺ + 2e⁻ (already balanced in atoms; 2e⁻ balance the charge).
Reduction: NO₃⁻ → NO. O: add 2H₂O to the right: NO₃⁻ → NO + 2H₂O. H: add 4H⁺ to the left: NO₃⁻ + 4H⁺ → NO + 2H₂O.
Charge: left −1 + 4 = +3, right 0. Add 3e⁻ to the left: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O. (N: +5 → +2, a change of 3. ✓)
Electrons: 2 and 3, so the lowest common multiple is 6. Multiply the oxidation by 3 and the reduction by 2.
3Cu → 3Cu²⁺ + 6e⁻ and 2NO₃⁻ + 8H⁺ + 6e⁻ → 2NO + 4H₂O.
Add: 3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O. Charge: −2 + 8 = +6 and +6. ✓

3Cu(s) + 2NO₃⁻(aq) + 8H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 4H₂O(l)

Worked example

Example 3: The dichromate breath-alcohol test

Older breath-alcohol testers used orange dichromate, which turns green when it oxidizes ethanol to acetic acid. Balance in acidic solution: Cr₂O₇²⁻ + C₂H₅OH → Cr³⁺ + CH₃COOH.

Reduction: Cr₂O₇²⁻ → 2Cr³⁺ (balance Cr first). O: add 7H₂O on the right. H: add 14H⁺ on the left. Charge: left −2 + 14 = +12, right +6, so add 6e⁻ on the left: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
Oxidation: C₂H₅OH → CH₃COOH. Carbon is balanced. O: 1 on the left, 2 on the right, so add H₂O on the left: C₂H₅OH + H₂O → CH₃COOH. H: 8 on the left, 4 on the right, so add 4H⁺ on the right. Charge: left 0, right +4, so add 4e⁻ on the right: C₂H₅OH + H₂O → CH₃COOH + 4H⁺ + 4e⁻.
Electrons: 6 and 4, so the lowest common multiple is 12. Multiply the reduction by 2 and the oxidation by 3.
2Cr₂O₇²⁻ + 28H⁺ + 12e⁻ → 4Cr³⁺ + 14H₂O and 3C₂H₅OH + 3H₂O → 3CH₃COOH + 12H⁺ + 12e⁻.
Add and cancel: 28H⁺ − 12H⁺ = 16H⁺ left on the left; 14H₂O − 3H₂O = 11H₂O left on the right.
Check: Cr 4 = 4; C 6 = 6; H 18 + 16 = 34 and 12 + 22 = 34; O 14 + 3 = 17 and 6 + 11 = 17. Charge: −4 + 16 = +12 and +12. ✓

2Cr₂O₇²⁻(aq) + 3C₂H₅OH(aq) + 16H⁺(aq) → 4Cr³⁺(aq) + 3CH₃COOH(aq) + 11H₂O(l). (Modern roadside devices in Canada typically use an electrochemical fuel-cell sensor instead; you will meet fuel cells in Fuel Cells, Electric Vehicles and Electrochemical Technologies, near the end of this unit.)

Using the balanced equation: redox titrations

A balanced redox equation gives you mole ratios, just like the stoichiometry you did in SCH3U. In a redox titration, a solution of an oxidizing agent of known concentration is added to a reducing agent (or the reverse) until the reaction is complete.

Permanganate is a favourite titrant because it is its own indicator: MnO₄⁻ is deep purple and Mn²⁺ is nearly colourless. While Fe²⁺ remains, each drop of purple loses its colour. The first drop that stays pale pink marks the endpoint.

Worked example

Example 4: Iron in a sample

A 25.00 mL sample of an Fe²⁺ solution is acidified and titrated with 0.0200 mol/L KMnO₄(aq). The endpoint is reached after 18.40 mL. Find the concentration of Fe²⁺.

Moles of permanganate: n = cV = 0.0200 mol/L × 0.01840 L = 3.68 × 10⁻⁴ mol MnO₄⁻.
From the balanced equation in Example 1, 1 MnO₄⁻ reacts with 5 Fe²⁺: n(Fe²⁺) = 5 × 3.68 × 10⁻⁴ = 1.84 × 10⁻³ mol.
Concentration: c = n/V = 1.84 × 10⁻³ mol ÷ 0.02500 L = 0.0736 mol/L.

[Fe²⁺] = 0.0736 mol/L (three significant figures, limited by 0.0200 mol/L).

Just for fun

Potassium permanganate is so intensely coloured that a few small crystals can turn a whole litre of water purple. In a titration that makes it the rare chemical that announces "I'm done!" all by itself.

Hands-on activity

Paper-tile balancing

You need: scissors (use them carefully), two colours of paper or sticky notes, a pen, and a table.

Cut 42 small squares. On one colour, write atoms: 8 × O, 16 × H and 2 × Mn. On the other colour, write charges: 10 × "+" and 6 × "−". (An electron tile is a "−" tile on its own.)
Build MnO₄⁻ on the left side of the table: 1 Mn tile, 4 O tiles and one "−" tile. Build Mn²⁺ on the right: 1 Mn tile and two "+" tiles.
Follow steps 3 to 5 of the method with real tiles. Add H₂O as groups of H, H, O. Add H⁺ as an H tile with a "+" tile. Add electrons as single "−" tiles.
When finished, count: are the atom tiles the same on both sides? Pair each "+" with a "−" on the same side and remove the pairs. The leftover charge on each side should match.
Now do Cr₂O₇²⁻ → 2Cr³⁺ on paper, using tally marks instead of tiles (it needs more pieces). You should need 7 water molecules, 14 H⁺ and 6 electrons.

Why this helps: seeing the 5 extra "−" tiles you must add makes it clear why electrons go on the more positive side.

Your turn

Balance the half-reaction in acidic solution: Cr₂O₇²⁻ → Cr³⁺.

Balance the half-reaction in acidic solution: SO₂ → SO₄²⁻.

Balance the half-reaction in acidic solution: H₂O₂ → H₂O.

Balance in acidic solution: H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l).

Balance in acidic solution: Zn(s) + NO₃⁻(aq) → Zn²⁺(aq) + NH₄⁺(aq).

In step 5, why do the electrons in a reduction half-reaction always end up on the left?

Balance in acidic solution: MnO₄⁻(aq) + Cl⁻(aq) → Mn²⁺(aq) + Cl₂(g).

More practice: warm-up

In acidic solution, what do you add to balance oxygen?

In acidic solution, what do you add to balance hydrogen?

What do you add to balance charge?

Balance: Fe³⁺ → Fe²⁺.

Balance: I⁻ → I₂.

More practice: on your own

Balance in acidic solution: Cr₂O₇²⁻(aq) + I⁻(aq) → Cr³⁺(aq) + I₂(aq).

Balance in acidic solution: MnO₄⁻(aq) + H₂C₂O₄(aq) → Mn²⁺(aq) + CO₂(g). (H₂C₂O₄ is oxalic acid.)

Balance in acidic solution: Ag(s) + NO₃⁻(aq) → Ag⁺(aq) + NO₂(g).

A 20.00 mL sample of hydrogen peroxide solution is titrated with 0.0500 mol/L KMnO₄ in acid. The endpoint is at 24.60 mL. The balanced equation is 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O. Find [H₂O₂].

A student balances a half-reaction as MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O + 5e⁻. What is wrong, and how could a charge check have caught it?

Challenge

Balance in acidic solution: As₂O₃(s) + NO₃⁻(aq) → H₃AsO₄(aq) + NO(g).

In a lab, 0.200 g of an iron ore is dissolved in acid and all its iron is converted to Fe²⁺. Titration needs 21.50 mL of 0.0180 mol/L KMnO₄. What is the percentage of iron by mass in the ore? (M of Fe = 55.85 g/mol)

Explain why you cannot balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ correctly by "just making the atoms equal", and show what the equation 'MnO₄⁻ + Fe²⁺ + 8H⁺ → Mn²⁺ + Fe³⁺ + 4H₂O' gets wrong.

Check your understanding

Quiz

When balancing a half-reaction in acidic solution, what do you add to balance oxygen atoms?

How many electrons appear in the balanced half-reaction for Cr₂O₇²⁻ → Cr³⁺ in acid?

Which half-reaction is correctly balanced in acidic solution?

In the balanced equation 3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O, how many electrons are transferred?

A balanced redox equation must have equal:

In the titration MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, 2.00 × 10⁻⁴ mol of MnO₄⁻ is used. How many moles of Fe²⁺ reacted?

For parents and teachers

Big idea (F2.3): a balanced redox equation must balance atoms and charge, which means electrons lost equal electrons gained. The half-reaction method does this in a fixed order: atoms other than O and H, then O with water, then H with H⁺, then charge with electrons.

How to help: Have your teen talk through each step aloud while balancing ("Now I balance oxygen with water..."). Most mistakes happen when a step is skipped or done out of order. Ask: "Did you check the charge?" at the end of every question.

Common difficulties:

Balancing O with O₂ or H with H₂. In acidic solution only H₂O and H⁺ are used.
Adding electrons to the wrong side. Electrons always go on the more positive side.
Forgetting to multiply every species in a half-reaction when equalizing electrons.
Not cancelling H⁺ and H₂O that appear on both sides at the end.
In titrations, forgetting the mole ratio (5 Fe²⁺ per MnO₄⁻) or using mL instead of L.

Safety: the paper-tile activity uses scissors.