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How do we identify electron transfer and balance redox reactions?

Assign oxidation numbers, identify oxidants and reductants, and balance redox half-equations and overall equations.

Oxidation numbers, the meaning of oxidation and reduction as electron transfer, identifying oxidants and reductants, disproportionation, and balancing redox half-equations and full ionic equations, with worked TASC-style examples.

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What this dot point is asking

TASC expects you to assign oxidation numbers, identify what is oxidised and reduced, name the oxidant and reductant, recognise disproportionation, and balance redox half-equations and overall ionic equations.

Oxidation and reduction as electron transfer

Redox reactions involve transfer of electrons:

  • Oxidation is loss of electrons; the oxidation number increases.
  • Reduction is gain of electrons; the oxidation number decreases.

The oxidant (oxidising agent) is the species that is reduced, because it takes electrons from something else. The reductant (reducing agent) is the species that is oxidised, because it supplies the electrons. The two always occur together: there is no oxidation without a matching reduction.

Assigning oxidation numbers

Apply these rules in order until every atom is assigned:

  • Free elements have an oxidation number of 00 (for example Cu\text{Cu}, O2\text{O}_2).
  • A monatomic ion equals its charge (for example Na+\text{Na}^+ is +1+1, Cl\text{Cl}^- is 1-1).
  • Oxygen is usually 2-2 (but 1-1 in peroxides such as H2O2\text{H}_2\text{O}_2).
  • Hydrogen is usually +1+1 (but 1-1 in metal hydrides such as NaH\text{NaH}).
  • The sum of oxidation numbers equals the overall charge of the species.

Writing and combining half-equations

Split the reaction into an oxidation half and a reduction half. For each half, balance the atoms other than O and H first, then in acidic aqueous conditions balance oxygen with H2O\text{H}_2\text{O}, balance hydrogen with H+\text{H}^+, and finally balance charge by adding electrons.

Oxidation number versus charge

Oxidation number is a bookkeeping device, not necessarily the real charge on an atom. In a covalent molecule the shared electrons are assigned, by convention, to the more electronegative atom, giving each atom a notional oxidation number even though no full transfer of electrons has occurred. For a monatomic ion the oxidation number does equal the real charge, but for atoms within molecules and polyatomic ions it is an assigned value used to track which atoms gain or lose electron density during a reaction. Keeping this distinction clear avoids confusion when an atom such as carbon shows a wide range of oxidation numbers across its compounds.

Redox in everyday chemistry

Redox reactions are everywhere: combustion of fuels, respiration in cells, photosynthesis, the rusting of iron, bleaching, and the operation of every battery. Recognising redox quickly through oxidation-number changes lets you analyse all of these with one framework. In Unit 4 the same electron-transfer ideas reappear in the oxidation of alcohols to aldehydes, ketones and carboxylic acids, so the skill of assigning oxidation numbers and identifying the oxidant and reductant carries directly into organic chemistry.

State the oxidant and reductant by name or formula in your answer, since identifying them is often a separate mark from balancing the equation.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20236 marks(a) Determine the oxidation number of vanadium in V3+\text{V}^{3+}, VO2+\text{VO}^{2+}, V2O4\text{V}_2\text{O}_4 and NH4VO3\text{NH}_4\text{VO}_3. (b) With reference to oxidation states, explain whether vanadium is oxidised or reduced in 2V2O4(s)+O2(g)2V2O5(s)2\text{V}_2\text{O}_4(s) + \text{O}_2(g) \rightarrow 2\text{V}_2\text{O}_5(s).
Show worked answer →

(a) Oxygen is 2-2 and the charges must balance. V3+\text{V}^{3+}: V=+3\text{V} = +3. VO2+\text{VO}^{2+}: V+(2)=+2\text{V} + (-2) = +2, so V=+4\text{V} = +4. V2O4\text{V}_2\text{O}_4: 2V+4(2)=02\text{V} + 4(-2) = 0, so V=+4\text{V} = +4. NH4VO3\text{NH}_4\text{VO}_3: the NH4+\text{NH}_4^+ part is +1+1, so V+3(2)=1\text{V} + 3(-2) = -1, giving V=+5\text{V} = +5. (4 marks)

(b) In V2O4\text{V}_2\text{O}_4 vanadium is +4+4; in V2O5\text{V}_2\text{O}_5, 2V+5(2)=02\text{V} + 5(-2) = 0 gives V=+5\text{V} = +5. The oxidation number rises from +4+4 to +5+5, so vanadium is oxidised (loses electrons). The oxygen gas is reduced from 00 to 2-2. (2 marks)

TCE 20223 marksWhen melted, potassium chlorate KClO3\text{KClO}_3 disproportionates, the chlorate ion ClO3\text{ClO}_3^- forming both perchlorate ClO4\text{ClO}_4^- and chloride. (a) Give the oxidation state of chlorine in ClO3\text{ClO}_3^- and ClO4\text{ClO}_4^-. (b) Write the half-equation for chlorate reacting to chloride in acidic solution and state whether it is oxidation or reduction.
Show worked answer →

(a) In ClO3\text{ClO}_3^-, Cl+3(2)=1\text{Cl} + 3(-2) = -1, so Cl=+5\text{Cl} = +5. In ClO4\text{ClO}_4^-, Cl+4(2)=1\text{Cl} + 4(-2) = -1, so Cl=+7\text{Cl} = +7. (2 marks)

(b) Going from ClO3\text{ClO}_3^- (+5+5) to Cl\text{Cl}^- (1-1), chlorine gains 66 electrons, so this is reduction: ClO3(aq)+6H+(aq)+6eCl(aq)+3H2O(l)\text{ClO}_3^-(aq) + 6\text{H}^+(aq) + 6e^- \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l). Charge check: left =1+66=1= -1 + 6 - 6 = -1, matching the right; O and H balance via three water molecules. (1 mark)

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