O-Level Chemistry · Secondary 3 and 4
Redox Reactions
Redox reactions are reactions in which oxidation and reduction happen together. They include burning, rusting, the extraction of metals, displacement reactions and electrolysis. In O-Level Chemistry you learn four ways to describe oxidation and reduction, how to work out oxidation states, how to pick out the oxidising and reducing agents, and the standard laboratory tests for them. These ideas carry straight into electrolysis and into the organic reactions where alcohols are oxidised.
What the syllabus expects
- Define oxidation and reduction in terms of oxygen, hydrogen, electrons and oxidation state
- Work out the oxidation state of an element in an element, compound or ion
- Identify the substances oxidised and reduced, and the oxidising and reducing agents, in a reaction
- Write half-equations showing electron loss and gain
- Describe the use of acidified potassium manganate(VII) to test for reducing agents and aqueous potassium iodide to test for oxidising agents
Four ways to define oxidation and reduction
Oxidation is the gain of oxygen, the loss of hydrogen, the loss of electrons or an increase in oxidation state. Reduction is the opposite: loss of oxygen, gain of hydrogen, gain of electrons or a decrease in oxidation state. The phrase OIL RIG helps with electrons: Oxidation Is Loss, Reduction Is Gain. In CuO + H₂ gives Cu + H₂O, copper(II) oxide loses oxygen and is reduced, while hydrogen gains oxygen and is oxidised. The electron definition works even when no oxygen is involved. In Zn + Cu²⁺ gives Zn²⁺ + Cu, zinc loses electrons, Zn gives Zn²⁺ + 2e⁻, so it is oxidised. Copper(II) ions gain electrons, Cu²⁺ + 2e⁻ gives Cu, so they are reduced. Because electrons lost by one species are gained by another, oxidation and reduction always happen together.
Working out oxidation states
Oxidation states let you track electrons in any reaction. Use these rules. An uncombined element has an oxidation state of 0, for example Mg, O₂ and Cl₂. A simple ion has an oxidation state equal to its charge, so Fe³⁺ is +3 and Cl⁻ is −1. In compounds, oxygen is usually −2 and hydrogen is usually +1. The oxidation states in a compound add up to zero, and in a polyatomic ion they add up to the charge on the ion. For sulfur in SO₄²⁻: x + 4(−2) = −2, so x = +6. For manganese in MnO₄⁻: x + 4(−2) = −1, so x = +7, which is why potassium manganate(VII) has VII in its name. For nitrogen in NH₃: x + 3(+1) = 0, so x = −3. An increase in oxidation state is oxidation and a decrease is reduction.
Oxidising agents and reducing agents
An oxidising agent causes another substance to be oxidised, and is itself reduced in the process. A reducing agent causes another substance to be reduced, and is itself oxidised. In Zn + Cu²⁺ gives Zn²⁺ + Cu, zinc is the reducing agent and Cu²⁺ is the oxidising agent. Common oxidising agents include oxygen, chlorine and acidified potassium manganate(VII). Common reducing agents include hydrogen, carbon, carbon monoxide and reactive metals. Displacement reactions are redox reactions: a more reactive metal reduces the ions of a less reactive metal, and a more reactive halogen oxidises the ions of a less reactive one, for example Cl₂ + 2Br⁻ gives 2Cl⁻ + Br₂. Spectator ions, such as sulfate in zinc reacting with copper(II) sulfate, keep the same oxidation state and are neither oxidised nor reduced.
Tests for oxidising and reducing agents
Acidified potassium manganate(VII) is a strong oxidising agent and is used to test for reducing agents. If a reducing agent is present, the purple solution turns colourless, because MnO₄⁻ (manganese +7) is reduced to Mn²⁺ (manganese +2). Iron(II) ions and sulfur dioxide both give this result. Aqueous potassium iodide is used to test for oxidising agents. If an oxidising agent is present, the colourless solution turns brown, because iodide ions are oxidised to iodine: 2I⁻ gives I₂ + 2e⁻. Chlorine and hydrogen peroxide both give this result. Remember that iodine in water is brown, not purple.
Worked example
Chlorine gas is bubbled into aqueous potassium iodide. Cl₂ + 2KI gives 2KCl + I₂. State what you would observe, use oxidation states to show which species is oxidised and which is reduced, name the oxidising agent and write the two half-equations.
- Observation: iodine is produced, so the colourless solution turns brown.
- Chlorine changes from 0 in Cl₂ to −1 in KCl, a decrease, so chlorine is reduced.
- Iodine changes from −1 in KI to 0 in I₂, an increase, so iodide is oxidised.
- Chlorine causes the iodide to be oxidised and is itself reduced, so chlorine is the oxidising agent and iodide is the reducing agent.
- Reduction half-equation: Cl₂ + 2e⁻ gives 2Cl⁻.
- Oxidation half-equation: 2I⁻ gives I₂ + 2e⁻. Potassium ions keep an oxidation state of +1 and are spectators.
Answer: The solution turns from colourless to brown. Chlorine (0 to −1) is reduced and is the oxidising agent, iodide (−1 to 0) is oxidised. Half-equations: Cl₂ + 2e⁻ gives 2Cl⁻ and 2I⁻ gives I₂ + 2e⁻.
Common mistakes
- Saying the oxidising agent is oxidised. The oxidising agent is reduced, because it takes electrons from the other substance.
- Forgetting that the oxidation states in a polyatomic ion add up to the charge on the ion, not to zero.
- Treating spectator ions such as SO₄²⁻ or K⁺ as being oxidised or reduced. Check that their oxidation state actually changes.
- Writing electrons on the wrong side of a half-equation. In an oxidation half-equation the electrons are on the product side, in a reduction they are on the reactant side.
- Describing iodine solution as purple. In aqueous potassium iodide, the iodine formed makes the solution brown.
- Writing the oxidation state of iron in Fe³⁺ as 3+ in an oxidation state question. Oxidation states are written with the sign first, +3.
Exam tips
- When asked whether a reaction is redox, write the oxidation state above each element on both sides. If any element changes, it is redox, and you have your evidence written down.
- When naming the oxidising or reducing agent, name the whole species (for example 'chlorine' or 'Cu²⁺ ions'), not just 'it'.
- For the manganate(VII) and iodide tests, always give both the starting colour and the final colour, such as 'purple to colourless'.
Practice questions
1. Easy: Work out the oxidation state of nitrogen in the nitrate ion, NO₃⁻.
Show answer
+5. Let nitrogen be x. Then x + 3(−2) = −1, so x − 6 = −1 and x = +5.
2. Medium: Acidified potassium manganate(VII) is added to a solution of iron(II) sulfate. State what you would observe and explain what happens to the iron(II) ions.
Show answer
The purple colour of the manganate(VII) disappears. The iron(II) ions are oxidised to iron(III) ions, Fe²⁺ gives Fe³⁺ + e⁻, so the oxidation state of iron increases from +2 to +3. Iron(II) is acting as a reducing agent.
3. Harder: Burning magnesium continues to burn in carbon dioxide. 2Mg + CO₂ gives 2MgO + C. Use oxidation states to identify the species oxidised and reduced, and name the oxidising agent.
Show answer
Magnesium goes from 0 in Mg to +2 in MgO, an increase, so magnesium is oxidised. Carbon goes from +4 in CO₂ (x + 2(−2) = 0) to 0 in C, a decrease, so carbon dioxide is reduced. Carbon dioxide is the oxidising agent and magnesium is the reducing agent.
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