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Combining half equations into an overall equation

Covers AS91167

The rule that governs everything

Every electron lost must be gained. No electrons are left over at the end of a real reaction.
So before you can add two half equations together, the number of electrons in each must be the same.
If the two half equations already have the same number of electrons, add them straight away. If they do not, multiply.

The method

Write the oxidation half equation and the reduction half equation.
Find the lowest common multiple of the two electron counts.
Multiply each half equation through by whatever factor gets it to that number.
Add the two equations together.
Cancel the electrons — they must vanish completely.
Cancel anything else that appears identically on both sides (often H⁺ or H₂O).
Check the atoms and the total charge.
Multiplying a half equation means multiplying every species in it, including the H⁺ and the H₂O. Forgetting one is the usual error.

Worked through: permanganate and iron(II)

The two half equations:

$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \qquad (5e^-)$

$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \qquad (1e^-)$

Lowest common multiple of 5 and 1 is 5. The permanganate half stays as it is; the iron half is multiplied by 5:

$5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-$

Add them:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+} + 5e^-$

Cancel the 5e⁻ from both sides:

$\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5\text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) + 5\text{Fe}^{3+}(aq)$

Check the charge: left is $(-1) + (+8) + 5(+2) = +17$ ; right is $(+2) + 0 + 5(+3) = +17$ . Correct.

What the overall equation tells you

The mole ratio. Here, 1 mol of MnO₄⁻ reacts with 5 mol of Fe²⁺ — which is exactly the ratio you would need for a titration calculation.
Whether acid is consumed. Eight H⁺ appear on the left, so the reaction needs a large excess of acid to go to completion.
Which species are spectators — they never appear at all.

Worked ExampleCombining half equations with different electron counts

Acidified potassium dichromate solution reacts with iodide ions. Write the overall balanced ionic equation, and state the mole ratio in which the two react.

Step 1 — Write both half equations

Reduction (dichromate, from the standard set):

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \qquad (6e^-)$

Oxidation (iodide to iodine):

$2\text{I}^- \rightarrow \text{I}_2 + 2e^- \qquad (2e^-)$

Step 2 — Match the electrons

The lowest common multiple of 6 and 2 is 6.

The dichromate half already has 6 electrons, so it stays. The iodide half must be multiplied by 3:

$6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-$

Note that every term was multiplied — the 2I⁻ became 6I⁻ and the I₂ became 3I₂.

Step 3 — Add and cancel

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{I}_2 + 6e^-$

The 6e⁻ cancel from both sides. Nothing else appears on both sides.

$\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6\text{I}^-(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) + 3\text{I}_2(aq)$

Step 4 — Check

Cr: 2 = 2. O: 7 = 7. H: 14 = 14. I: 6 = 6.

Charge: left is $(-2) + (+14) + 6(-1) = +6$ ; right is $2(+3) = +6$ . Correct.

Step 5 — Read off the ratio

Answer: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6I⁻(aq) → 2Cr³⁺(aq) + 7H₂O(l) + 3I₂(aq). One mole of dichromate reacts with six moles of iodide ions.