Representation of reactions at electrodes as half equations (HT only): 4OH⁻ – 4e⁻ → O₂ + 2H₂O

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Q: Is this Representation of reactions at electrodes as half equations (HT only): 4OH⁻ – 4e⁻ → O₂ + 2H₂O guide aligned to the AQA specification?

Yes. This guide is linked to the Representation of reactions at electrodes as half equations (HT only): 4OH⁻ – 4e⁻ → O₂ + 2H₂O topic in the AQA GCSE specification.

This half-equation describes oxidation at the anode during the electrolysis of aqueous solutions.

2 key terms

Study notes

From the checked teaching notes for this specification point.

The idea

What earns the marks

Recognise that OH⁻ ions are oxidised because they lose electrons, and that loss of electrons is oxidation.
Balance the oxygen atoms: four OH⁻ contain four oxygen atoms, which appear as one O₂ (two oxygen atoms) plus two H₂O (two oxygen atoms).
Balance the hydrogen atoms: four OH⁻ contain four hydrogen atoms, which appear as four hydrogen atoms in two H₂O molecules.
Balance charge by subtracting four electrons on the reactant side (– 4e⁻) to give a net zero charge on both sides, or adding them to the product side (+ 4e⁻) to give a net 4⁻ charge on both sides.
Link the half-equation to the anode process in aqueous electrolysis, where hydroxide ions may be discharged instead of other anions.

Common mistakes and how to put them right

Writing 4OH⁻ + 4e⁻ → O₂ + 2H₂O.
Put it right: Is to subtract electrons on the left (– 4e⁻) or add them on the right (+ 4e⁻) to show oxidation.
Using 2OH⁻ instead of 4OH⁻.
Put it right: Is to use four hydroxide ions so that one O₂ and two H₂O balance.
Omitting the minus sign on hydroxide.
Put it right: Is to write OH⁻ with a Unicode superscript minus sign.

Exam tips

Check the charge on each side after balancing atoms: 4OH⁻ → O₂ + 2H₂O + 4e⁻ has a net charge of 4⁻ on both sides, ensuring charge is conserved.
If the question asks for the anode reaction in aqueous solution, consider hydroxide discharge as well as other possible anions, and choose the half-equation that matches the information given.
Write the equation exactly as required, using Unicode superscripts for charges and subscripts for formulae, and do not add state symbols unless they are requested.

Try the lesson

The start of the Representation of reactions at electrodes as half equations (HT only): 4OH⁻ – 4e⁻ → O₂ + 2H₂O lesson, made from checked teaching notes for this specification point.

Have a go first

In the half equation 4OH⁻ – 4e⁻ → O₂ + 2H₂O, which substance is shown as losing electrons?

Tip: press 1 to 4 to choose

Listen

This half-equation describes oxidation at the anode during the electrolysis of aqueous solutions. Hydroxide ions, OH⁻, lose electrons. Four OH⁻ ions each lose one electron, represented by subtracting 4e⁻ on the left side (– 4e⁻). The remaining atoms rearrange to form one O₂ molecule and two H₂O molecules, balancing the oxygen and hydrogen atoms. Loss of electrons is oxidation, confirming this is an anodic process. To check the charge balance: if written as 4OH⁻ – 4e⁻ → O₂ + 2H₂O, the left side has a 4⁻ charge minus a 4⁻ charge, giving zero, matching the neutral products. If written as 4OH⁻ → O₂ + 2H₂O + 4e⁻, both sides have a net charge of 4⁻. This reaction occurs when OH⁻ is discharged instead of sulfate or nitrate ions. A strong answer earns marks for these points. Recognise that OH⁻ ions are oxidised because they lose electrons, and that loss of electrons is oxidation. Balance the oxygen atoms: four OH⁻ contain four oxygen atoms, which appear as one O₂ (two oxygen atoms) plus two H₂O (two oxygen atoms). Balance the hydrogen atoms: four OH⁻ contain four hydrogen atoms, which appear as four hydrogen atoms in two H₂O molecules. Balance charge by subtracting four electrons on the reactant side (– 4e⁻) to give a net zero charge on both sides, or adding them to the product side (+ 4e⁻) to give a net 4⁻ charge on both sides. A common slip: Writing 4OH⁻ + 4e⁻ → O₂ + 2H₂O: the error is adding electrons to the reactant side, which implies reduction. Instead, is to subtract electrons on the left (– 4e⁻) or add them on the right (+ 4e⁻) to show oxidation. Exam tip: Check the charge on each side after balancing atoms: 4OH⁻ → O₂ + 2H₂O + 4e⁻ has a net charge of 4⁻ on both sides, ensuring charge is conserved.

Key words

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Oxidation

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Anode process

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See every definition

Oxidation: Loss of electrons, as when hydroxide ions lose electrons at the anode.
Anode process: The reaction at the anode in aqueous electrolysis, where hydroxide ions may be discharged instead of other anions.

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