What this dot point is asking

VCAA wants the definition of oxidation and reduction in terms of electron transfer, the writing of balanced half-equations and overall ionic equations, the identification of the oxidant and reductant in a reaction, and the use of the electrochemical series (standard electrode potentials at 25°C) to predict whether a reaction is spontaneous.

The answer

Redox in terms of electrons

A redox reaction involves the transfer of electrons between species.

• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.

Memory aid: OIL RIG ("Oxidation Is Loss, Reduction Is Gain") or LEO the lion says GER ("Lose Electrons Oxidation, Gain Electrons Reduction").

The species being oxidised is the reducing agent (reductant): it gives electrons away, causing the other species to be reduced. The species being reduced is the oxidising agent (oxidant): it accepts electrons, causing the other species to be oxidised.

Half-equations

A half-equation shows just one half of the redox process (the oxidation OR the reduction), including the electrons transferred. Half-equations must be balanced for atoms and charge.

For a metal/metal-ion couple:
Cu^2+(aq) + 2e^- -> Cu(s) (reduction)
Zn(s) -> Zn^2+(aq) + 2e^- (oxidation)

For a more complex species in acidic solution (e.g. permanganate), the steps are:

1. Balance the atom being oxidised or reduced.
2. Balance O by adding H2O.
3. Balance H by adding H^+.
4. Balance charge by adding electrons (e^-) to the more positive side.

Example. Permanganate reducing to Mn^2+ in acid:
MnO4^-(aq) + 8H^+(aq) + 5e^- -> Mn^2+(aq) + 4H2O(l)

The Mn oxidation number goes from +7 (in MnO4^-) to +2 (in Mn^2+), a 5-electron change.

Combining half-equations into an overall ionic equation

1. Multiply each half-equation so the number of electrons in oxidation equals the number in reduction.
2. Add the two half-equations.
3. Cancel electrons (they should fully cancel).
4. Cancel any species (like H^+ or H2O) that appear on both sides.

Example. Cu^2+ oxidising Zn:

Zn(s) -> Zn^2+(aq) + 2e^- ×1
Cu^2+(aq) + 2e^- -> Cu(s) ×1

Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)

The electrons are equal (2 each) so no multiplication is needed.

The electrochemical series

The electrochemical series (in your VCAA data book) lists half-equations written as reductions with their standard electrode potentials (E°) measured at 25°C, 1 mol L^-1 (or 1 atm for gases), against the standard hydrogen electrode (SHE = 0.00 V).

Reading rules:

• The half-equation higher up the series (more positive E°) is a stronger oxidant. Strong oxidants on the left of the arrow.
• The half-equation lower down the series (more negative E°) is a stronger reductant on the right of the arrow (the metal, the lower oxidation state).
• A redox reaction is spontaneous if the species you want to act as the oxidant sits above the species you want to act as the reductant in the series.

Visualisation: the species on the left of an upper half-equation can oxidise any species on the right of a lower half-equation.

Predicting spontaneity

To check whether a proposed reaction is spontaneous:

1. Identify the oxidant (the species being reduced) and write its half-equation as a reduction.
2. Identify the reductant (the species being oxidised) and write its half-equation as a reduction (to look it up), then reverse it for the oxidation in your overall equation.
3. Compare E° values:
   • If E°(oxidant) > E°(reductant), the reaction is spontaneous.
   • If E°(oxidant) < E°(reductant), the reaction is non-spontaneous (the reverse reaction is spontaneous).
4. Calculate the cell EMF:
   E°_cell = E°(reduction, cathode) - E°(reduction, anode)
   E°_cell > 0 means spontaneous; E°_cell < 0 means non-spontaneous.

Worked example

Will iron(III) ions oxidise iodide ions in aqueous solution?

From the data book:
Fe^3+(aq) + e^- -> Fe^2+(aq) E° = +0.77 V
I2(s) + 2e^- -> 2I^-(aq) E° = +0.54 V

Fe^3+ is the proposed oxidant; I^- is the proposed reductant. E°(Fe^3+/Fe^2+) > E°(I2/I^-), so Fe^3+ is a strong enough oxidant to take electrons from I^-. The reaction is spontaneous.

Half-equations:
Reduction (×2 to balance electrons): 2Fe^3+(aq) + 2e^- -> 2Fe^2+(aq)
Oxidation: 2I^-(aq) -> I2(s) + 2e^-

Overall:
2Fe^3+(aq) + 2I^-(aq) -> 2Fe^2+(aq) + I2(s)

Cell EMF:
E°_cell = 0.77 - 0.54 = +0.23 V (positive, confirming spontaneity).

Fe^3+ is the oxidant; I^- is the reductant. Yellow brown I2 forms in solution.

Common traps

Forgetting to balance electrons before adding half-equations. If oxidation gives 2e^- and reduction needs 3e^-, multiply oxidation by 3 and reduction by 2 so each contributes 6e^-.

Reversing E° when reversing a half-equation. Do not change the sign of E° when writing oxidation instead of reduction in the overall equation. For the cell EMF formula, always use the reduction potential as listed.

Using the reactivity series in place of the electrochemical series. The reactivity series is a rough ordering for metals only. The electrochemical series is the quantitative, full version. VCAA expects the quantitative argument.

Confusing oxidant/oxidised and reductant/reduced. The oxidant causes oxidation of others while itself being reduced. The reductant causes reduction of others while itself being oxidised. Use the data book conventions to stay consistent.

Predicting spontaneity from concentration or surface area. Standard electrode potentials assume 1 mol L^-1 and 25°C. Real systems shift, but VCAA Unit 3 problems assume standard conditions unless told otherwise.

In one sentence

A redox reaction transfers electrons from a reductant (oxidised, electron loser) to an oxidant (reduced, electron gainer), and the electrochemical series predicts spontaneity by comparing the standard electrode potentials of the two half-equations (higher E° on top means the stronger oxidant).