Construct redox equations from half-equations
Combine oxidation and reduction half-equations by balancing electrons, atoms and charge.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 24. Treat all tabulated electrode potentials as reduction potentials; state conditions, phases, signs and units. Written practice is self-checked
Before you begin
- Use AS redox half-equations, oxidation states, mole ratios, current and charge.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What must be equal before oxidation and reduction half-equations are added?
What you will learn
- Write oxidation and reduction half-equations
- Scale half-equations to cancel electrons
- Check the overall redox equation for atoms and charge
Balance electrons before adding the halves
Write each half-equation in the direction required by the cell. Multiply equations so the number of electrons lost equals the number gained, then add and cancel electrons.
For acid solution, balance O with H₂O and H with H⁺, then balance charge with e⁻. Keep state symbols and any specified medium; do not leave electrons in the overall equation.
Check conservation after cancellation
After combining, count every atom and total charge on both sides. The electron count must cancel exactly, and coefficients should be reduced to the smallest whole-number ratio.
For a cell reaction, oxidation occurs at the anode and reduction at the cathode. The electron direction used in the half-equations must agree with the cell polarity and E°cell sign.
Pause and explain
What should happen to electrons in the overall redox equation?
Worked example
Combine Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu into an overall equation.
Show the worked solution
- The half-equations already contain two electrons each.
- Add them and cancel electrons on both sides.
- Check atoms and total charge.
Answer Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Atoms and total charge are balanced.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Combine Fe²⁺ → Fe³⁺ + e⁻ and MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O in acidic solution.
Give me a hint
Multiply the iron half-equation by five, then add and cancel electrons.
Compare my reasoning
- 5Fe²⁺ → 5Fe³⁺ + 5e⁻.
- Add the permanganate reduction half-equation.
- Cancel 5e⁻ and check atoms and charge.
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺.
Look for these in your work
- Electron multiplier correct
- Electrons cancel
- Atoms and charge balanced
Combine Al → Al³⁺ + 3e⁻ with Ag⁺ + e⁻ → Ag.
Give me a hint
Multiply the silver reduction by three before adding.
Compare my reasoning
- 3Ag⁺ + 3e⁻ → 3Ag.
- Add Al oxidation.
- Cancel electrons and verify total charge.
Al(s) + 3Ag⁺(aq) → Al³⁺(aq) + 3Ag(s).
Look for these in your work
- Electron count matched
- Overall ratio correct
- Charge and atoms balanced
A proposed overall equation for Zn and Cu²⁺ leaves 2e⁻ on both sides. Explain the correction and give the net equation.
Give me a hint
Electrons are intermediates that cancel when the half-equations are added.
Compare my reasoning
- Add Zn oxidation and Cu²⁺ reduction.
- Cancel the two electrons on each side.
- Retain only net species and states.
Cancel the electrons to obtain Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
Look for these in your work
- Electrons treated as intermediates
- Net equation supplied
- States retained
Common mistakes
- Adding half-equations without matching electron numbers
- Leaving electrons in the overall equation
- Balancing atoms but not charge
- Changing a half-equation direction without tracking its role
What can you explain now?
What must be equal before oxidation and reduction half-equations are added?
Compare with the explanation
Electrons lost and gained
Electron transfer is conserved.
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Make it stick
How do you balance electrons first and then verify atoms and charge in the net redox equation?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.