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

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.

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.

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):

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:

Add to the dichromate half-equation:

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.

• 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.

Step 3: add half-equations.

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

Examples in context

Example 1. Permanganate titration of cane juice iron at Mackay sugar laboratory. Mill labs determine $\text{Fe}^{2+}$ contamination in cane juice by titration against acidified $\text{KMnO}_4$. Half-equations: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (oxidation, Fe $+2 \rightarrow +3$) and $\text{MnO}_4^- + 8 \text{H}^+ + 5 e^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2 \text{O}$ (reduction, Mn $+7 \rightarrow +2$). Overall: $\text{MnO}_4^- + 5 \text{Fe}^{2+} + 8 \text{H}^+ \rightarrow \text{Mn}^{2+} + 5 \text{Fe}^{3+} + 4 \text{H}_2 \text{O}$. The faint pink endpoint at the first unreacted drop of $\text{KMnO}_4$ marks the titration completion.

Example 2. Chromate-to-chromium reduction at Townsville plating shop. Plating waste contains $\text{Cr(VI)}$ as $\text{Cr}_2 \text{O}_7^{2-}$ which is highly toxic. Iron(II) sulfate reduces it to non-toxic $\text{Cr}^{3+}$ before lime precipitation. Half-equation: $\text{Cr}_2 \text{O}_7^{2-} + 14 \text{H}^+ + 6 e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2 \text{O}$ (Cr $+6 \rightarrow +3$). Combine with six $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ to balance: $\text{Cr}_2 \text{O}_7^{2-} + 6 \text{Fe}^{2+} + 14 \text{H}^+ \rightarrow 2 \text{Cr}^{3+} + 6 \text{Fe}^{3+} + 7 \text{H}_2 \text{O}$. Real industry confirms theory: $1 \, \text{g}$ Cr(VI) needs $6.5 \, \text{g}$ $\text{FeSO}_4$.

Try this

Q1. Assign oxidation numbers to all elements in (a) $\text{KMnO}_4$, (b) $\text{Cr}_2 \text{O}_7^{2-}$, (c) $\text{H}_2 \text{O}_2$. [3 marks]

• Cue. (a) K $+1$, Mn $+7$, O $-2$. (b) Cr $+6$, O $-2$. (c) H $+1$, O $-1$ (peroxide).

Q2. Balance the half-equation $\text{NO}_3^- \rightarrow \text{NO}$ in acidic solution. [3 marks]

• Cue. $\text{NO}_3^- + 4 \text{H}^+ + 3 e^- \rightarrow \text{NO} + 2 \text{H}_2 \text{O}$. Check atoms and charge.

Q3. Combine half-equations for $\text{MnO}_4^-$/Mn$^{2+}$ and $\text{C}_2 \text{O}_4^{2-}$/$\text{CO}_2$ in acid. (a) Write both. (b) Combine into a balanced overall equation. (c) Identify oxidant and reductant. [3+3+2 marks]

• Cue. (a) Mn$^{7+} \rightarrow$Mn$^{2+}$ (5e); oxalate C$^{3+} \rightarrow$ C$^{4+}$ (2e). (b) $2 \text{MnO}_4^- + 5 \text{C}_2 \text{O}_4^{2-} + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2 \text{O}$. (c) MnO$_4^-$ oxidant, oxalate reductant.