Redox Reactions and Half-Equations

Learning Objectives (OCR Spec 2.1.5)

By the end of this lesson you should be able to:

• Describe redox in terms of electron transfer and oxidation number changes
• Identify oxidising agents and reducing agents in a chemical equation
• Construct balanced half-equations for oxidation and reduction
• Combine half-equations to obtain an overall balanced ionic equation
• Apply these ideas to typical reactions of metals with acids, halogen displacement and common redox titrations

Redox: A Reminder

A redox reaction is one in which electrons are transferred from one species to another. Using the OIL RIG mnemonic:

• Oxidation = loss of electrons; oxidation number increases
• Reduction = gain of electrons; oxidation number decreases

Both processes must occur together - electrons released by one species are accepted by another.

Oxidising and Reducing Agents

• An oxidising agent (oxidant) accepts electrons from another species, thereby oxidising it. The oxidising agent is itself reduced.
• A reducing agent (reductant) donates electrons to another species, thereby reducing it. The reducing agent is itself oxidised.

Species	Typical Role
O2, Cl2, Br2, H2O2, KMnO4, K2Cr2O7, Fe^3+	Oxidising agents
Metals (Na, Mg, Fe), H2, C, CO, I-, S2O3^2-, Fe^2+	Reducing agents

Note: Some species (e.g. H2O2) can act as both, depending on conditions.

Writing Half-Equations

A half-equation shows either the oxidation or the reduction part of a redox reaction, explicitly including the electrons transferred.

Systematic Procedure for Half-Equations

1. Write the species being oxidised/reduced on each side
2. Balance all atoms other than H and O
3. Balance O by adding H2O
4. Balance H by adding H+ (for acidic conditions) or OH- (for alkaline)
5. Balance charge by adding electrons (e-) on the appropriate side
6. Check total charge and atoms balance

Example 1: Mg -> Mg^2+

Mg -> Mg^2+ + 2e- (oxidation)

Mg starts at oxidation state 0 and becomes +2; it has lost 2 electrons.

Example 2: Cl2 -> Cl-

Cl2 + 2e- -> 2Cl- (reduction)

Chlorine is reduced from 0 to -1 per atom; two electrons gained per molecule.

Example 3: MnO4- -> Mn^2+ (acidic conditions)

Mn goes from +7 in MnO4- to +2 in Mn^2+ - a 5-electron reduction.

1. MnO4- -> Mn^2+
2. Balance O by adding 4 H2O on the right: MnO4- -> Mn^2+ + 4H2O
3. Balance H by adding 8 H+ on the left: MnO4- + 8H+ -> Mn^2+ + 4H2O
4. Balance charge: left = (-1 + 8) = +7; right = +2. Add 5e- on left:

MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O

Example 4: Cr2O7^2- -> Cr^3+ (acidic conditions)

Cr goes from +6 in dichromate to +3. Two Cr atoms, so 6 electrons total.

Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O

Combining Half-Equations

To form an overall equation:

1. Multiply each half-equation so the number of electrons is equal
2. Add the equations and cancel electrons (and any species that appear on both sides)

Example: Mg + HCl

Oxidation: Mg -> Mg^2+ + 2e- Reduction: 2H+ + 2e- -> H2

Electrons already balance (2 each). Adding:

Mg + 2H+ -> Mg^2+ + H2

Adding spectator Cl- gives the full equation:

Mg + 2HCl -> MgCl2 + H2

Example: MnO4- + Fe^2+

Oxidation: Fe^2+ -> Fe^3+ + e- Reduction: MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O

Multiply oxidation x5: 5Fe^2+ -> 5Fe^3+ + 5e-

Add:

MnO4- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+

This is the central reaction in the manganate(VII)-iron(II) titration used to find the percentage of iron in an iron ore or iron tablet.

Halogen Displacement

More reactive halogens displace less reactive ones from their halide solutions. This is a redox reaction:

Cl2 + 2Br- -> 2Cl- + Br2

Half-equations:

• Cl2 + 2e- -> 2Cl- (reduction; Cl is the oxidant)
• 2Br- -> Br2 + 2e- (oxidation; Br- is the reductant)

Reactivity order: F2 > Cl2 > Br2 > I2

Worked Example

Q: A 1.00 g sample of iron ore is dissolved to give Fe^2+. The solution is titrated against 0.0200 mol dm-3 KMnO4, requiring a mean titre of 24.50 cm3. Calculate the percentage of iron in the ore.

• Moles MnO4- = (0.0200 x 24.50)/1000 = 4.90 x 10^-4 mol
• From stoichiometry (1:5), moles Fe^2+ = 5 x 4.90 x 10^-4 = 2.45 x 10^-3 mol
• Mass Fe = 2.45 x 10^-3 x 55.8 = 0.1367 g
• %Fe = (0.1367 / 1.00) x 100 = 13.7%

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