← Unit 3: Equilibrium, acids and redox reactions

QLDChemistrySyllabus dot point

Topic 2: Oxidation and reduction

Determine oxidation numbers and use them to identify oxidation and reduction in chemical reactions, and construct balanced half-equations and overall ionic equations for redox reactions in aqueous solution

A focused answer to the QCE Chemistry Unit 3 dot point on assigning oxidation numbers, identifying oxidising and reducing agents, and constructing balanced half-equations and overall ionic equations for redox reactions in aqueous solution. Includes the half-equation balancing protocol QCAA expects in IA1 short response and EA Paper 2.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answer

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What this dot point is asking

QCAA wants you to assign oxidation numbers from molecular and ionic formulas, use them to identify redox reactions and the oxidising and reducing agents, and construct balanced half-equations (and overall ionic equations) for redox reactions in aqueous solution. The half-equation protocol is the highest-yielding redox skill in Unit 3 and feeds directly into the galvanic cell dot point that follows.

The answer

A redox reaction is a chemical reaction in which one species loses electrons (is oxidised) and another gains them (is reduced). The two changes always occur together. Oxidation numbers track the formal "ownership" of electrons in each species; constructing half-equations separates the two processes so they can be balanced and combined.

Assigning oxidation numbers

Six rules in priority order:

  1. Free elements (in any allotrope) have oxidation number 0. Na in metallic sodium, O in O2, P in P4 are all 0.
  2. Monatomic ions have oxidation number equal to the ion charge. Na+ is +1; Cl- is -1; Al3+ is +3.
  3. Fluorine is always -1 in compounds.
  4. Oxygen is -2 in compounds, except in peroxides (O2^2-, -1) and superoxides (-0.5), and when bonded to F (positive).
  5. Hydrogen is +1 in compounds with non-metals, -1 in metal hydrides (NaH, CaH2).
  6. The sum of oxidation numbers equals the overall charge of the species (0 for a neutral molecule, the ion charge for an ion).

Apply rules 1 to 5 first, then solve for the remaining atom using rule 6.

Worked examples:

  • H2SO4. H is +1 (x2 = +2). O is -2 (x4 = -8). Sum = 0. So S = +6.
  • MnO4-. O is -2 (x4 = -8). Sum = -1. So Mn = +7.
  • Cr2O7^2-. O is -2 (x7 = -14). Sum = -2. 2Cr = +12. So Cr = +6.
  • NH4+. H is +1 (x4 = +4). Sum = +1. So N = -3.
  • NaH. Na is +1. Sum = 0. So H = -1 (metal hydride).
  • H2O2. H is +1 (x2 = +2). Sum = 0. So 2O = -2; O = -1 (peroxide).

Identifying oxidation and reduction

A species is oxidised if its oxidation number increases (it loses electrons). A species is reduced if its oxidation number decreases (it gains electrons).

The oxidising agent is the species that is reduced (it causes oxidation in another species by accepting electrons).

The reducing agent is the species that is oxidised (it causes reduction in another species by donating electrons).

Memory aid: OIL RIG. Oxidation Is Loss; Reduction Is Gain (of electrons).

Worked example: Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s).

  • Zn: 0 -> +2. Increases. Oxidised. Zn is the reducing agent.
  • Cu: +2 -> 0. Decreases. Reduced. Cu2+ is the oxidising agent.

If no oxidation numbers change in a reaction, it is not a redox reaction. Most acid-base and precipitation reactions are non-redox.

Constructing half-equations

A half-equation shows either the oxidation or the reduction half of a redox reaction. The protocol for aqueous solution:

  1. Balance the atoms being oxidised or reduced. (Often there is only one; sometimes coefficients are needed.)
  2. Balance oxygen by adding H2O on the side that needs O.
  3. Balance hydrogen by adding H+ on the side that needs H.
  4. Balance charge by adding electrons (e-) to the side with the more positive total.
  5. Check. Atoms balance and charges balance.

Worked example. Oxidation half-equation for Fe2+ -> Fe3+:

Step 1: Fe2+ -> Fe3+ (only one Fe each side).
Step 2: no O.
Step 3: no H.
Step 4: LHS charge +2; RHS charge +3. Add 1 e- to RHS.

Fe2+→Fe3++e−Fe^{2+} \rightarrow Fe^{3+} + e^-

Worked example. Reduction half-equation for Cr2O7^2- -> Cr3+ in acidic solution:

Step 1: Cr2O7^2- -> 2 Cr3+.
Step 2: 7 O on LHS. Add 7 H2O on RHS.
Step 3: 14 H on RHS. Add 14 H+ on LHS.
Step 4: LHS charge -2 + 14 = +12. RHS charge +6. Add 6 e- to LHS.

Cr2O72−+14H++6e−→2Cr3++7H2OCr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

Half-equations in basic solution

When QCAA states the reaction occurs in basic conditions, do the acidic-conditions protocol first, then add OH- to both sides to neutralise the H+, and simplify the resulting H2O on each side.

Example. Reduction of MnO4- to MnO2 in basic solution:

Acidic protocol: MnO4- + 4 H+ + 3 e- -> MnO2 + 2 H2O.

Add 4 OH- to both sides: MnO4- + 4 H2O + 3 e- -> MnO2 + 2 H2O + 4 OH-.

Simplify (subtract 2 H2O from both sides):

MnO4−+2H2O+3e−→MnO2+4OH−MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

QCE Chemistry stays mostly in acidic solution, but this conversion appears occasionally in EA Paper 2.

Combining half-equations into an overall ionic equation

The number of electrons must balance. Multiply each half-equation so that the electrons cancel when added.

Worked example. Cr2O7^2- (gains 6 e-) reacting with Fe2+ (loses 1 e-):

Multiply Fe2+ half-equation by 6:

6Fe2+→6Fe3++6e−6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-

Add to the dichromate half-equation:

Cr2O72−+14H++6Fe2+→2Cr3++7H2O+6Fe3+Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 7H_2O + 6Fe^{3+}

Final check: atoms balance, charge balances ((-2) + 14 + 12 = 24 on left; 6 + 18 = 24 on right). The electrons have cancelled completely; they do not appear in the overall equation.

Disproportionation

A special case where the same species is both oxidised and reduced. Classic example: copper(I) in acid.

2Cu(aq)+→Cu(s)+Cu(aq)2+2Cu^+_{(aq)} \rightarrow Cu_{(s)} + Cu^{2+}_{(aq)}

  • Cu (+1 -> 0): reduced.
  • Cu (+1 -> +2): oxidised.

QCAA may ask you to identify disproportionation in a stimulus reaction; the test is one element with two different oxidation-number changes.

Common oxidising and reducing agents to recognise

Oxidising agents (commonly examined).

  • MnO4- (acidic): reduced to Mn2+, deep purple to colourless.
  • Cr2O7^2- (acidic): reduced to Cr3+, orange to green.
  • H2O2: reduced to H2O (or oxidised to O2; it is amphoteric in redox terms).
  • O2 (oxygen gas).
  • Cl2, Br2, I2 (halogens, decreasing strength down the group).

Reducing agents (commonly examined).

  • Active metals (Zn, Fe, Mg, Al, Na, K).
  • H2 (hydrogen gas).
  • I- (and Br-, Cl- to a much lesser extent).
  • Fe2+: oxidised to Fe3+.
  • SO3^2-: oxidised to SO4^2-.

The standard reduction potential table (covered in the galvanic cell dot point) is the quantitative version of this list.

Worked example: putting the pieces together

A copper coin is placed in concentrated nitric acid. Nitric acid is reduced to NO2 gas. Construct the overall ionic equation.

Step 1: identify the half-equations.

  • Cu oxidised: 0 -> +2. Cu -> Cu2+ + 2 e-.
  • HNO3 reduced (N: +5 in NO3-, +4 in NO2). NO3- + 2 H+ + e- -> NO2 + H2O.

Step 2: balance electrons. Multiply reduction by 2.

2NO3−+4H++2e−→2NO2+2H2O2NO_3^- + 4H^+ + 2e^- \rightarrow 2NO_2 + 2H_2O

Step 3: add half-equations.

Cu+2NO3−+4H+→Cu2++2NO2+2H2OCu + 2NO_3^- + 4H^+ \rightarrow Cu^{2+} + 2NO_2 + 2H_2O

Cu is the reducing agent; NO3- is the oxidising agent.

Common traps

Forgetting that oxygen in peroxides is -1, not -2. H2O2 surprises students who apply rule 4 mechanically.

Counting fluorine as -2 when it bonds to oxygen. F always wins; in OF2 oxygen is +2.

Balancing atoms after balancing charge. Always do atom balance first, then charge balance with electrons.

Calling the oxidising agent "the one that is oxidised". It is the one that is reduced (it causes oxidation in something else).

Confusing oxidation states of polyatomic ions. The oxidation number is per atom, not per ion. In Cr2O7^2-, Cr is +6 each, not +12.

Leaving electrons in the overall equation. Once balanced, electrons must cancel completely.

In one sentence

Assign oxidation numbers using the standard rules to track formal electron ownership; the species whose oxidation number increases is oxidised (the reducing agent) and the one whose oxidation number decreases is reduced (the oxidising agent); construct balanced half-equations using the atoms-then-O-then-H-then-charge protocol; and combine them by multiplying through so that electrons cancel exactly.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2023 QCAA-style5 marksAcidified potassium dichromate reacts with iron(II) sulfate solution according to the unbalanced ionic equation: Cr2O7^2-(aq) + Fe2+(aq) + H+(aq) -> Cr3+(aq) + Fe3+(aq) + H2O(l). (a) Determine the oxidation number of chromium in Cr2O7^2- and in Cr3+. (b) Construct balanced half-equations for the oxidation and reduction processes. (c) Combine to give the balanced overall ionic equation. (d) Identify the oxidising agent and the reducing agent.
Show worked answer →

A 5-mark answer needs both oxidation numbers, both half-equations, the combined equation, and the agent identifications.

(a) Oxidation numbers. In Cr2O7^2-, oxygen is -2 (so 7 x -2 = -14) and overall charge is -2, so 2Cr = -2 + 14 = +12, giving Cr = +6. In Cr3+, Cr = +3. Chromium goes from +6 to +3, gaining electrons; it is reduced.

(b) Half-equations.

Reduction (Cr +6 -> +3, 3 electrons per Cr, 6 electrons per Cr2O7^2-):

Cr2O72−+14H++6e−→2Cr3++7H2OCr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

Oxidation (Fe +2 -> +3, 1 electron per Fe):

Fe2+→Fe3++e−Fe^{2+} \rightarrow Fe^{3+} + e^-

(c) Combined. Multiply the oxidation half-equation by 6 to balance electrons, then sum:

Cr2O72−+14H++6Fe2+→2Cr3++7H2O+6Fe3+Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 7H_2O + 6Fe^{3+}

(d) Oxidising and reducing agents. Cr2O7^2- is the oxidising agent (it is reduced; it accepts electrons). Fe2+ is the reducing agent (it is oxidised; it donates electrons).

Markers reward (1) both oxidation numbers shown with the calculation, (2) both half-equations balanced for atoms, oxygen with water, hydrogen with H+, charge with electrons, (3) the combined equation with cancellation of electrons, (4) correct agent identification with the "is reduced" / "is oxidised" reasoning.

2022 QCAA-style3 marksConstruct a balanced half-equation for the reduction of MnO4-(aq) to Mn2+(aq) in acidic solution.
Show worked answer →

A 3-mark answer needs the protocol applied step by step.

Step 1: balance the species being reduced.

MnO4−→Mn2+MnO_4^- \rightarrow Mn^{2+}

Step 2: balance oxygen with water (4 O on the left, so 4 H2O on the right).

MnO4−→Mn2++4H2OMnO_4^- \rightarrow Mn^{2+} + 4H_2O

Step 3: balance hydrogen with H+ (8 H on the right, so 8 H+ on the left).

MnO4−+8H+→Mn2++4H2OMnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O

Step 4: balance charge with electrons. LHS charge is -1 + 8 = +7. RHS charge is +2. Difference = 5, so add 5 e- to the LHS.

MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

Markers reward the systematic protocol shown step by step and explicit final charge balance check.

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