Oxidation, reduction and redox equations — AQA A-Level Chemistry
Test yourself on Oxidation, reduction and redox equations with AQA A-Level practice questions.
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Oxidation, reduction and redox equations explained
Oxidation is loss of electrons. A species that is oxidised gives electrons away, and its oxidation state rises.
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The oxidising agent is the other partner in that transfer: it takes the electrons in, so it is itself reduced and its oxidation state falls. The two ideas are often confused because the oxidising agent is the one that ends up reduced. Concentrated sulfuric acid with solid sodium iodide shows it in action: iodide ions are oxidised to iodine, giving up electrons, while the sulfur in the acid is reduced from +6 down to +4 in sulfur dioxide or all the way to −2 in hydrogen sulfide, so the sulfuric acid is acting as the oxidising agent. Powerful oxidising agents include manganate(VII), dichromate(VI) and the halogens.
Reduction is the process of electron gain and reducing agents are electron donors.
Reduction is gain of electrons, and the reducing agent is the species that supplies them, so it is itself oxidised. The rule of thumb is that the agent is always named for what it does to its partner, not for what happens to it. Reduction lowers the oxidation state of the species reduced. Reactive metals are good reducing agents because they lose electrons readily, and among the halides the iodide ion is the strongest reducing agent, which is why it alone is able to reduce sulfur in concentrated sulfuric acid as far as hydrogen sulfide. Disproportionation is the special case where one element is both oxidised and reduced in the same reaction, as chlorine is when it reacts with water to give chloride and chlorate(I) ions.
The rules for assigning oxidation states.
Oxidation states are assigned by a fixed set of rules applied in order. An uncombined element is zero. The states in a species sum to its overall charge, so zero for a neutral compound and the ionic charge for an ion. Group 1 is +1, Group 2 is +2 and fluorine is always −1. Hydrogen is +1 except in metal hydrides, where it is −1, and oxygen is −2 except in peroxides, where it is −1, and when bonded to fluorine. Work out the unknown element last, by difference. In sulfate, for example, four oxygens contribute −8 and the ion carries 2⁻, so sulfur must be +6. Signs are always written, and a value quoted without a sign is read as positive.
Students should be able to: work out the oxidation state of an element in a compound or ion from the formula write half-equations identifying the oxidation and reduction processes in redox reactions combine half-equations to give an overall redox equation.
Half equations separate a redox reaction into the electron loss and the electron gain. Build each one in a fixed order: balance the element being oxidised or reduced, balance oxygen by adding H₂O, balance hydrogen by adding H⁺, then balance the charge by adding electrons to the more positive side. For manganate(VII) this gives MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, and for ethanedioate C₂O₄²⁻ → 2CO₂ + 2e⁻. To combine them, scale each so the electrons cancel, here by two and by five, then add and cancel anything appearing on both sides: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. Check that the atoms and the total charge balance on both sides.
Your focus
- Explain why the oxidising agent is the species that is itself reduced, tracing the electrons from the species oxidised to the agent in a given equation.
- Identify the species oxidised and the oxidising agent in a given equation, quoting oxidation states as evidence.
- Explain, with reference to electrons, why a named reaction is classified as redox.
Show all 16 objectives
- Deduce the role of concentrated sulfuric acid in its reaction with a solid halide from the products formed.
- Deduce the number of electrons gained per mole of the species reduced from its change in oxidation state, for example two when sulfuric acid is reduced to sulfur dioxide and eight when it is reduced to hydrogen sulfide.
- Identify the reducing agent in a given equation and justify the choice using oxidation states.
- Explain why iodide is a stronger reducing agent than chloride, using the products of their reactions with concentrated sulfuric acid.
- Deduce from oxidation states that a given reaction is a disproportionation.
- Deduce the oxidation state of an element in a compound or polyatomic ion by applying the rules in order.
- Identify the exceptions for hydrogen and oxygen and apply them to a metal hydride and to a peroxide.
- Calculate the oxidation state of the central atom in an oxoanion from the charge on the ion.
- Explain how a change in oxidation state between two species shows that oxidation or reduction has occurred.
- Construct a balanced half equation for a species in acidic solution using H₂O, H⁺ and electrons in the correct order.
- Identify which half equation represents oxidation and which represents reduction from the side the electrons appear on.
- Combine two half equations into an overall redox equation by scaling so that the electrons cancel.
- Verify a redox equation by checking that atoms and total charge balance on both sides.
Oxidation, reduction and redox equations exam tips
Marking Points
- one mark for defining oxidation as loss of electrons
- one mark for defining an oxidising agent as an electron acceptor
- one mark for identifying the species oxidised and the species reduced in a given equation
- one mark for naming the oxidising agent in a reaction and supporting it with a change in oxidation state
- one mark for defining reduction as gain of electrons
- one mark for defining a reducing agent as an electron donor
- one mark for identifying a reducing agent from a fall in the oxidation state of the species it reacts with
- one mark for explaining disproportionation as one element being both oxidised and reduced in the same reaction
- one mark for using the rule that oxidation states sum to the overall charge on the species
- one mark for assigning the fixed values first, such as oxygen at −2 and hydrogen at +1
- one mark for the correct value of the unknown element, written with its sign
- one mark for identifying an exception correctly, such as oxygen at −1 in a peroxide or hydrogen at −1 in a metal hydride
- one mark for each correctly balanced half equation, with H₂O and H⁺ used to balance oxygen and hydrogen
- one mark for electrons added to the correct side so that the charges balance
- one mark for scaling the half equations so the electrons cancel exactly
- one mark for the combined equation with electrons cancelled and both atoms and charge balanced
Examiner Tips
- 💡Remember OIL RIG, then add the twist: the oxidising agent is the one that gets reduced.
- 💡Support any claim about an agent with the oxidation state before and after; bare assertions rarely score.
- 💡Use the phrase with reference to electrons if the question asks you to explain why a reaction is redox.
- 💡Write the two half equations out; the one showing electrons on the left is the reduction.
- 💡For disproportionation, quote the oxidation state of the element in all three species to prove both changes.
- 💡Say electron donor rather than gives away electrons when a definition is asked for.
- 💡Always write the sign, even for a positive value; an unsigned number is taken to be positive.
- 💡Set the calculation out as an equation, for example x + 4(−2) = −2, so the method is visible.
- 💡Check the halogens: a halogen bonded to oxygen or to a more electronegative halogen is positive.
- 💡Work through the order every time: element, oxygen with water, hydrogen with H⁺, charge with electrons.
- 💡Write the number of electrons under each half equation and find the lowest common multiple before adding.
- 💡Finish by totalling the charge on both sides; it is the fastest check that the equation is right.
Common Mistakes
- calling the species that is oxidised the oxidising agent
- defining oxidation as gain of oxygen only, which fails for reactions with no oxygen present
- writing that the oxidising agent donates electrons
- quoting an oxidation state change without saying which species was oxidised and which reduced
- assuming a species with a high oxidation state must always be oxidised further
- swapping the agents, so the species reduced is called the reducing agent
- defining reduction as loss of oxygen only, which fails for electron transfer between ions
- writing that a reducing agent gains electrons
- missing disproportionation because the element appears twice on the product side
- assuming the strongest reducing agent is the most reactive metal in every context, without reference to the reaction
- setting the total to zero for a polyatomic ion instead of to the ionic charge
- giving oxygen −2 in hydrogen peroxide, where it is −1
- multiplying the charge of an ion by the number of atoms, so the sulfate ion is treated as 8⁻
- omitting a minus sign, which is then read as a positive value
- confusing the oxidation state of an element with the charge on a whole polyatomic ion
- adding OH⁻ to balance oxygen in an acidified reaction, where H₂O and H⁺ are required
- putting the electrons on the wrong side, so an oxidation is written as a reduction
- adding two half equations without scaling them first, leaving electrons in the final equation
- balancing the atoms but not checking the total charge on each side
- leaving H⁺ and H₂O uncancelled when they appear on both sides of the combined equation