Lesson 2 · 30 min

Balancing redox equations: the half-reaction method

The equation for MnO₄⁻ oxidising Fe²⁺ conjures up 8 H⁺ and 4 H₂O. Split the reaction into halves, balance atoms and charge step by step, and electron conservation tells you the rest.

In this lesson

Split a redox reaction into an oxidation half and a reduction half
Balance atoms and charge in acidic solution with H₂O, H⁺ and e⁻, in that order
Convert an acidic result into the equation for basic solution
Use the balanced equation for redox-titration stoichiometry

Titrating iron(II) with potassium permanganate is a classic way to measure iron: purple $\ce{MnO4-}$ drips into pale-green $\ce{Fe^2+}$ and loses its colour at once, until a faint pink lingers at the endpoint. Yet when you write $\ce{MnO4-} + \ce{Fe^2+} \mathrel{\ce{->}} \ce{Mn^2+} + \ce{Fe^3+}$ , manganese and iron balance while the oxygen has nowhere to go and the charge reads +1 on the left and +5 on the right — trial and error stops here.

Predict first

In acidic solution, how many $\ce{Fe^2+}$ can one $\ce{MnO4-}$ oxidise to $\ce{Fe^3+}$ ?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

1 3 5 8

Split the reaction in two

The half-reaction (ion–electron) method writes the oxidised and reduced species separately, balances each, then combines them by their electron counts. In acidic solution:

Write two skeleton halves, such as $\ce{MnO4-} \mathrel{\ce{->}} \ce{Mn^2+}$ and $\ce{Fe^2+} \mathrel{\ce{->}} \ce{Fe^3+}$ ;
Balance atoms other than O and H (e.g. $\ce{Cr2O7^2-} \mathrel{\ce{->}} \ce{2Cr^3+}$ );
Add $\ce{H2O}$ to the side that lacks oxygen;
Add $\ce{H+}$ to the side that lacks hydrogen;
Add $\ce{e-}$ to the more positive side until the charges match;
Multiply the halves so electrons lost equal electrons gained;
Add them, cancel the $\ce{e-}$ , $\ce{H+}$ and $\ce{H2O}$ that appear on both sides, and check atoms and charge once more.

a\,n_\text{red} = b\,n_\text{ox} = \operatorname{lcm}(n_\text{red},\,n_\text{ox})

All of step 6: the electrons gained by the reduction half times a equal those lost by the oxidation half times b. The least common multiple gives the smallest coefficients.

$n_\text{red}$: electrons in the reduction half (5 for MnO₄⁻)
$n_\text{ox}$: electrons in the oxidation half (1 for Fe²⁺)
$a,\ b$: multipliers of the two halves

Ion–electron method · atoms and charge checked at every step

Half-reaction balancing tutor

Reaction acidic solution

Step 1 of 7 · Write the two skeleton half-reactions

Find what goes up in oxidation state (oxidation) and what goes down (reduction); write each as “reactant → product”.

Reduction half

\ce{Cr2O7^2-}

→

\ce{Cr^3+}

	Cr	O	Charge
Left	2	7	-2
Right	1	0	3

Not balanced yet

Oxidation half

\ce{I-}

→

\ce{I2}

	I	Charge
Left	1	-1
Right	2	0

Not balanced yet

Every step has an atom-and-charge ledger; red cells are not yet balanced. Try dichromate with iodide and permanganate with oxalate (step 2 matters), then “hydrogen peroxide disproportionates”, where one substance appears in both halves. At the multiply step, type your own multipliers into the boxes first.

Basic solution: one more step

Basic solution contains almost no $\ce{H+}$ , so it should not appear in the equation. Balance as if acidic, then add as many $\ce{OH-}$ to both sides as there are $\ce{H+}$ : $\ce{H+} + \ce{OH-} \mathrel{\ce{->}} \ce{H2O}$ ; finally cancel the surplus water. Permanganate oxidising sulfite in base, for example, gives

\ce{2MnO4- + 3SO3^2- + H2O -> 2MnO2 + 3SO4^2- + 2OH-}

Manganese only falls to +4 (a brown $\ce{MnO2}$ precipitate) instead of +2 as in acid — the same oxidising agent gives different products in different media.

Worked example

Copper and dilute nitric acid

Copper dissolves in dilute nitric acid, giving colourless NO (which turns red-brown in air). Write the ionic equation and then the full equation.

Concept check

Which ionic equation is fully balanced?

\ce{Fe^3+} + \ce{Cu} \mathrel{\ce{->}} \ce{Fe^2+} + \ce{Cu^2+}

\ce{2Fe^3+} + \ce{Cu} \mathrel{\ce{->}} \ce{2Fe^2+} + \ce{Cu^2+}

\ce{Fe^3+} + \ce{2Cu} \mathrel{\ce{->}} \ce{Fe^2+} + \ce{2Cu^2+}

\ce{2Fe^3+} + \ce{2Cu} \mathrel{\ce{->}} \ce{2Fe^2+} + \ce{Cu^2+}

Try the calculation

In sulfuric-acid solution, how many millilitres of 0.0200 M $\ce{KMnO4}$ oxidise all the $\ce{Fe^2+}$ in 25.00 mL of 0.100 M solution?

Key takeaways

Split the reaction into an oxidation half and a reduction half, balance each, then combine.
In acid: other atoms first, then O with H₂O, H with H⁺, charge with e⁻.
Multiply so electrons match, add, and cancel what appears on both sides; in base, neutralise H⁺ with OH⁻.
The balanced equation gives the titration ratio, e.g. MnO₄⁻ : Fe²⁺ = 1 : 5.

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Data and references

D. C. Harris, Quantitative Chemical Analysis, 9th ed., W. H. Freeman (2016), chs. 16 (redox titrations; permanganate, dichromate and iodometry).
S. S. Zumdahl, S. A. Zumdahl, Chemistry, 10th ed., Cengage (2018), §18.1 (balancing oxidation–reduction equations).
N. N. Greenwood, A. Earnshaw, Chemistry of the Elements, 2nd ed., Butterworth-Heinemann (1997), ch. 11 (copper and nitric acid) and ch. 24 (permanganate).