Defining oxidation and reduction

Oxidation and reduction can be defined in several ways:

1. Oxygen transfer:

• Oxidation involves a substance gaining oxygen.
• Reduction involves a substance losing oxygen.

For example, in the reaction:

2CuO_(s) + C_(s) ➔ 2Cu_(s) + CO_2(g),

C is oxidised to CO₂ by gaining oxygen while CuO is reduced to Cu by losing it.

2. Hydrogen transfer:

• Oxidation involves a substance losing hydrogen.
• Reduction involves a substance gaining hydrogen.

For example, in the reaction:

N₂H_4(l) + O_2(g) ➔ N_2(g) + 2H₂O_(g),

N₂H₄ is oxidised to N₂ by losing hydrogen while O₂ is reduced to H₂O by gaining it.

3. Electron transfer:

• Oxidation involves a substance losing electrons.
• Reduction involves a substance gaining electrons.

A useful mnemonic to remember this is OIL RIG:

Oxidation

Is

Loss (of electrons),

Reduction

Is

Gain (of electrons)

For example, in the reaction:

Na_(s) + 1⁄2Cl_2(l) ➔ NaCl_(s)

• Na is oxidised to Na⁺ by losing electrons: Na ➔ Na⁺ + e^-
• Cl₂ is reduced to Cl^- by gaining electrons: 1⁄2Cl₂ + e^- ➔ Cl^-

4. Oxidation state change:

• Oxidation involves an increase in oxidation state.
• Reduction involves a decrease in oxidation state.

For example, in the reaction:

Mg_(s) + 2HCl_(aq) ➔ MgCl_2(aq) + H_2(g)

• Mg is oxidised because its oxidation number increases from 0 (in Mg) to +2 (in MgCl₂).
• H is reduced because its oxidation number decreases from +1 (in HCl) to 0 (in H₂).

In all redox reactions, one species is oxidised while another is simultaneously reduced.

Identifying oxidising and reducing agents

In a redox reaction:

• Oxidising agents are electron acceptors - they oxidise other substances and are themselves reduced.
• Reducing agents are electron donors - they reduce other substances and are themselves oxidised.

For example, in the reaction:

Na_(s) + 1⁄2Cl_2(g) ➔ NaCl_(s)

• Na is the reducing agent because it is oxidised: Na ➔ Na⁺ + e^-
• Cl is the oxidising agent because it is reduced: 1⁄2Cl₂ + e^- ➔ Cl^-

Disproportionation is simultaneous oxidation and reduction

Disproportionation is a unique type of redox reaction where a single substance is both oxidised and reduced.

For example, in the reaction:

Cl₂ + 2OH⁻ ➔ ClO⁻ + Cl⁻ + H₂O

The oxidation state of Cl:

• Increases from 0 (in Cl₂) to +1 (in ClO^-) so Cl is oxidised.
• Decreases from 0 (in Cl₂) to -1 (in Cl^-) so Cl is reduced.

Balancing redox equations in aqueous solution

The steps for writing balanced redox equations in aqueous solutions are:

1. Assign oxidation numbers and identify what is oxidised and reduced.
2. Split the reaction into 2 separate oxidation and reduction half-equations.
3. Balance any atoms being oxidised or reduced.
4. Add electrons to the more positive side of each half-equation. For oxidation, electrons appear as products whilst for reduction, electrons appear as reactants. The number of electrons equals the change in oxidation state.
5. Multiply the half-equations to equalise the electrons.
6. Combine the half-equations.
7. Check the final equation is fully balanced in terms of atoms and charge.

For reactions in acidic solution, balance oxygen atoms by adding H₂O and hydrogen atoms by adding H⁺.

Worked example 1 - Balancing a redox reaction

Write the balanced redox equation that occurs when Cu_(s) and NO₃^–_(aq) react in acid to form Cu²⁺_(aq) and NO_2(g).

Step 1: Assign oxidation numbers and identify species oxidised and reduced

Cu is oxidised: 0 in Cu ➔ +2 in Cu²⁺

N is reduced: +5 in NO₃^– ➔ +4 in NO₂

Step 2: Write separate oxidation and reduction equations

Oxidation: Cu ➔ Cu²⁺

Reduction: NO₃^– ➔ NO₂

Step 3: Balance O and H atoms

Only the reduction half-equation requires O and H balancing:

NO₃^– + 2H⁺ ➔ NO₂ + H₂O

Step 4: Add electrons

Cu ➔ Cu²⁺ + 2e^–

NO₃^– + 2H⁺ + e^– ➔ NO₂ + H₂O

Step 5: Equalise electrons

Multiply the reduction half-equation (NO₃^– + 2H⁺ + e^– ➔ NO₂ + H₂O) by 2 so both half-equations contain 2e^-:

2NO₃^– + 4H⁺ + 2e^– ➔ 2NO₂ + 2H₂O

Step 6: Combine half-equations

2NO₃^– + 4H⁺ + 2e^– + Cu ➔ 2NO₂ + 2H₂O + Cu²⁺ + 2e^–

Cancelling the electrons gives the final balanced equation:

2NO₃^– + 4H⁺ + Cu ➔ 2NO₂ + 2H₂O + Cu²⁺

Worked example 2 - Balancing a redox reaction

Write the balanced redox equation that occurs when MnO₄^–_(aq) and C₂O₄^2–_(aq) react in acid to form Mn²⁺_(aq) and CO_2(g).

Step 1: Assign oxidation numbers and identify what is oxidised and reduced

C is oxidised: +3 in C₂O₄^2– ➔ +4 in CO₂

Mn is reduced: +7 in MnO₄^– ➔ +2 in Mn²⁺

Step 2: Write separate oxidation and reduction equations

Oxidation: C₂O₄^2– ➔ CO₂

Reduction: MnO₄^– ➔ Mn²⁺

Step 3: Balance O and H atoms

C₂O₄^2– ➔ 2CO₂

MnO₄^– + 8H⁺ ➔ Mn²⁺ + 4H₂O

Step 4: Add electrons

C₂O₄^2– ➔ 2CO₂ + 2e^–

MnO₄^– + 8H⁺ + 5e^– ➔ Mn²⁺ + 4H₂O

Step 5: Equalise electrons

Multiply the oxidation half-equation (C₂O₄^2– ➔ 2CO₂ + 2e^–) by 5 and the reduction half-equation (MnO₄^– + 8H⁺ + 5e^– ➔ Mn²⁺ + 4H₂O) by 2 so both half-equations contain 10e^-:

5C₂O₄^2– ➔ 10CO₂ + 10e^–

2MnO₄^– + 16H⁺ + 10e^– ➔ 2Mn²⁺ + 8H₂O

Step 6: Combine half-equations

5C₂O₄^2– 2MnO₄^– + 16H⁺ + 10e^– ➔ 10CO₂ + 10e^– + 2Mn²⁺ + 8H₂O

Cancelling the electrons gives the final balanced equation;

5C₂O₄^2– 2MnO₄^– + 16H⁺ ➔ 10CO₂ + 2Mn²⁺ + 8H₂O

Worked example 3 - Balancing a redox reaction

Write the balanced redox equation that occurs when I^–_(aq) and IO₃^–_(aq) react in acid to form I_2(s).

Step 1: Assign oxidation numbers and identify species oxidised and reduced

I is oxidised: -1 (in I^–) ➔ 0 in I₂

I is reduced: +5 in IO₃^– ➔ 0 in I₂

Step 2: Write separate oxidation and reduction equations

Oxidation: I^– ➔ I₂

Reduction: IO₃^– ➔ I₂

Step 3: Balance I atoms

I^– ➔ 1⁄2I₂

IO₃^– ➔ 1⁄2I₂

Step 4: Balance O and H atoms

Only the reduction half-equation requires O and H balancing:

IO₃^– + 6H⁺ ➔ 1⁄2I₂ + 3H₂O

Step 5: Add electrons

I^– ➔ 1⁄2I₂ + e^-

IO₃^– + 6H⁺ + 5e^- ➔ 1⁄2I₂ + 3H₂O

Step 6: Equalise electrons

Multiply the oxidation half-equation (I^– ➔ 1⁄2I₂ + e^–) by 5 so both half-equations contain 5e^-:

5I^– ➔ 5⁄2I₂ + 5e^–

Step 7: Combine half-equations

5I^– + IO₃^– + 6H⁺ + 5e^- ➔ 5⁄2I₂ + 5e^– + 1⁄2I₂ + 3H₂O

Cancelling the electrons and combining I₂ gives the final balanced equation:

5I^– + IO₃^– + 6H⁺ ➔ 3I₂ + 3H₂O