redox reactions in aqueous solution including the assignment of oxidation numbers, identification of the species oxidised and reduced, and the construction and balancing of half-equations and overall ionic equations in acidic solution

What this dot point is asking

VCAA wants you to assign oxidation numbers, identify the species oxidised and reduced (and the corresponding reductant and oxidant), and to balance half-equations and overall ionic equations for redox reactions in acidic solution.

The answer

Oxidation numbers (rules)

The oxidation number (or oxidation state) is the hypothetical charge an atom would have if every bond were ionic. Use the rules in order; later rules yield to earlier ones if there is a conflict.

1. Elements in their standard state: ox. no. = 0 (Na, O2, S8, Hg(l)).
2. Monatomic ions: ox. no. = charge (Na^+ is +1, S^2- is -2).
3. Group 1 metals in compounds: +1. Group 2 metals in compounds: +2.
4. Fluorine in compounds: -1.
5. Hydrogen: usually +1 (in H2O, HCl), but -1 in metal hydrides (NaH, CaH2).
6. Oxygen: usually -2 (in H2O, CO2), but -1 in peroxides (H2O2), -1/2 in superoxides, and +2 in OF2.
7. Sum of oxidation numbers = overall charge of the species (0 for a neutral compound, the ion charge for an ion).

Quick examples:

• MnO4^-: O is -2 (4 x -2 = -8). Total must be -1, so Mn = +7.
• Cr2O7^2-: O is -2 (7 x -2 = -14). 2Cr + (-14) = -2, so Cr = +6 each.
• SO3^2- (sulfite): S = +4.
• CH4: C = -4 (using H = +1, 4 x +1 = +4, so C = -4 for total of 0).
• CH3COOH: average C = 0 (but the C in COOH is +3 and the C in CH3 is -3; the average matters for the overall balance).

Oxidation and reduction

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

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

The oxidant (or oxidising agent) accepts electrons; it is itself reduced.
The reductant (or reducing agent) donates electrons; it is itself oxidised.

For Zn + Cu^2+ -> Zn^2+ + Cu: Zn goes 0 to +2 (oxidised, the reductant); Cu^2+ goes +2 to 0 (reduced, the oxidant).

Balancing half-equations in acidic solution

A half-equation is a balanced equation showing one half of the redox reaction with electrons explicitly. Use this procedure for acidic solution:

1. Balance the redox atom (the element whose ox. no. is changing).
2. Balance O by adding H2O to the side that needs O.
3. Balance H by adding H^+ to the side that needs H.
4. Balance charge by adding electrons (e^-) to the side that needs more negative charge.
5. Cancel any species that appear on both sides; simplify.

Example: convert MnO4^- to Mn^2+ in acidic solution.

1. Mn balanced: 1 MnO4^- on left, 1 Mn^2+ on right.
2. O balance: 4 O on left, 0 O on right. Add 4 H2O on the right: MnO4^- -> Mn^2+ + 4H2O.
3. H balance: 8 H on right, 0 on left. Add 8 H^+ on the left: MnO4^- + 8H^+ -> Mn^2+ + 4H2O.
4. Charge balance: left = -1 + 8(+1) = +7. Right = +2. Add 5 e^- on the left: MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O.

The full reduction half is now balanced.

Combining half-equations

To get the overall equation:

1. Multiply each half by the appropriate factor so the electrons cancel (find the LCM of the electrons in each half).
2. Add the two halves.
3. Cancel any species (H^+, H2O) that appear on both sides.
4. Check that atoms and charge balance.

The final equation will not contain any electrons.

Quick checklist before submitting

• Every atom balanced on both sides.
• Every charge balanced on both sides.
• Electrons cancelled in the overall equation (only present in half-equations).
• States included where the question expects them.

Worked example

Acidic permanganate oxidises sulfite (SO3^2-) to sulfate (SO4^2-). Write the overall ionic equation.

Half 1 (reduction, MnO4^- to Mn^2+): derived above.
MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O

Half 2 (oxidation, SO3^2- to SO4^2-)

S balanced. O: add H2O on left (1 O needed): SO3^2- + H2O -> SO4^2-. H: add 2 H^+ on right: SO3^2- + H2O -> SO4^2- + 2H^+. Charge: left = -2, right = -2 + 2(+1) = 0. Add 2 e^- on right: SO3^2- + H2O -> SO4^2- + 2H^+ + 2e^-.

Combine (LCM of 5 and 2 is 10): multiply reduction by 2 and oxidation by 5.

2MnO4^- + 16H^+ + 10e^- -> 2Mn^2+ + 8H2O
5SO3^2- + 5H2O -> 5SO4^2- + 10H^+ + 10e^-

Add and cancel 10e^-; cancel 10 H^+ (so net 6 H^+ on left); cancel 5 H2O (so net 3 H2O on right):

2MnO4^-(aq) + 6H^+(aq) + 5SO3^2-(aq) -> 2Mn^2+(aq) + 3H2O(l) + 5SO4^2-(aq)

Check: Mn 2/2, S 5/5, O 8+15 = 23 on left vs 3 + 20 = 23 on right, H 6/6, charge -2 + 6 -10 = -6 left, +4 -10 = -6 right. Balanced.

Common traps

Forgetting to include H2O and H^+. In acidic solution they are part of the half-equation and must appear if O or H balances require them.

Balancing in basic solution by accident. This dot point is acidic solution only. Adding OH^- to your half-equation in this course will lose marks.

Putting electrons on the wrong side. Oxidation produces electrons (e^- on the right). Reduction consumes electrons (e^- on the left).

Forgetting to cancel H^+ and H2O when adding halves. The overall equation should not contain electrons, and the H^+/H2O on the same side should be combined and simplified.

Reading too much into "oxidation". "Oxidised" means lost electrons; it does not require oxygen as a reactant (Zn -> Zn^2+ + 2e^- is oxidation despite no oxygen anywhere).

Calling the substance reduced "the reductant". The oxidant is reduced; the reductant is oxidised. Practise saying both pairs out loud.

In one sentence

Oxidation numbers track which atom lost or gained electrons, the oxidant is reduced and the reductant is oxidised, and balanced half-equations in acidic solution come from balancing the redox atom first, then O with H2O, then H with H^+, then charge with electrons, before scaling and adding to give the overall ionic equation.