Cosmos Coaching Centre
Resources
Show All Resources
Year 7
Maths Science
Year 8
Maths Science
Year 9
Maths Science
Year 10
Maths Science
Year 11
Maths Standard Maths Advanced Maths Extension Physics Chemistry
Year 12
Maths Standard Maths Advanced Maths Extension Physics Chemistry
Selective
Apps
ATAR calculator Worksheet Forge Uni StarMap
Pricing Contact
3D

Oxidation, reduction and half-equations

Oxidation and reduction as loss and gain of electrons, oxidising and reducing agents, oxidation numbers, writing and balancing half-equations in acidic solution, and combining half-equations into balanced redox equations.

A breathalyser of the old kind turned orange dichromate crystals green when someone who had been drinking blew through them. A swimming pool turns a cloudy green when its chlorine runs out and iron in the water isn't kept in check. Both are electron transfers, and the same four-step method lets you write a balanced equation for either of them.

Oxidation and reduction as electron transfer

When zinc displaces copper from copper(II) sulfate, each zinc atom gives two electrons to a copper(II) ion. We split the reaction into two half-equations:

$$\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^- \qquad \text{oxidation}$$

$$\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \qquad \text{reduction}$$

Zinc atoms lose electrons (oxidation) and copper(II) ions gain them (reduction).
  • Oxidation is loss of electrons; reduction is gain (OIL RIG).
  • The two always happen together, so these are called redox reactions. Electrons lost must equal electrons gained.
  • The species that is reduced takes electrons from the other, so it is the oxidising agent (oxidant). Here Cu²⁺ is the oxidant.
  • The species that is oxidised is the reducing agent (reductant). Here Zn is the reductant.

Every displacement reaction in lesson 3C, and every metal reacting with acid, water or oxygen, is a redox reaction. So is combustion. Precipitation and acid–base reactions are not: no electrons change hands.

Oxidation numbers

For reactions involving molecules or polyatomic ions it isn't obvious who has gained electrons. Oxidation numbers (oxidation states) are a bookkeeping system that tracks this. Assign them with these rules, in this order of priority:

  1. An uncombined element is 0 (Na, O₂, S₈).
  2. A simple ion equals its charge (Fe³⁺ is +3, Cl⁻ is −1).
  3. In compounds, group 1 metals are +1, group 2 metals +2, and fluorine −1.
  4. Hydrogen is +1 (except in metal hydrides such as NaH, where it is −1).
  5. Oxygen is −2 (except in peroxides such as H₂O₂, where it is −1, and with fluorine).
  6. The oxidation numbers in a neutral compound add to 0; in a polyatomic ion they add to the charge of the ion.
Assigning oxidation numbers

Find the oxidation number of (a) Mn in KMnO₄, (b) Cr in Cr₂O₇²⁻ and (c) S in S₂O₃²⁻.

(a) K is +1 and each O is −2: (+1) + Mn + 4(−2) = 0, so Mn = +7.

(b) 2Cr + 7(−2) = −2, so 2Cr = +12 and Cr = +6.

(c) 2S + 3(−2) = −2, so 2S = +4 and S = +2 (an average over the two sulfur atoms).

Now the definitions can be stated in terms of numbers: oxidation is an increase in oxidation number and reduction is a decrease. Written as a Roman numeral, the oxidation number also appears in names, such as iron(III) chloride and manganese(IV) oxide.

Moving up the oxidation number line is oxidation; moving down is reduction.
Identifying the oxidant and reductant

In the reaction 2Fe²⁺(aq) + Cl₂(g) → 2Fe³⁺(aq) + 2Cl⁻(aq), identify what is oxidised and reduced, the oxidising agent and the reducing agent.

Fe: +2 → +3, an increase, so Fe²⁺ is oxidised and is the reducing agent.

Cl: 0 → −1, a decrease, so Cl₂ is reduced and is the oxidising agent. This is why chlorine keeps dissolved iron in a pool as Fe³⁺, which can then be filtered out as Fe(OH)₃.

Balancing half-equations in acidic solution

Many oxidising agents contain oxygen (MnO₄⁻, Cr₂O₇²⁻, NO₃⁻) and react in acid. Their half-equations need water and H⁺ to balance. Use the same four steps every time:

  1. Balance atoms other than O and H.
  2. Balance O by adding H₂O.
  3. Balance H by adding H⁺.
  4. Balance charge by adding electrons to the more positive side.
The four steps applied to the dichromate ion being reduced to chromium(III).
Balancing the permanganate half-equation

Write a balanced half-equation for MnO₄⁻ being reduced to Mn²⁺ in acid solution.

Step 1: Mn is already balanced: MnO₄⁻ → Mn²⁺

Step 2: 4 O on the left, so add 4H₂O on the right: MnO₄⁻ → Mn²⁺ + 4H₂O

Step 3: 8 H on the right, so add 8H⁺ on the left: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Step 4: charge on the left = −1 + 8 = +7; on the right = +2. Add 5e⁻ to the left: $$\text{MnO}_4^-\text{(aq)} + 8\text{H}^+\text{(aq)} + 5\text{e}^- \rightarrow \text{Mn}^{2+}\text{(aq)} + 4\text{H}_2\text{O(l)}$$ Electrons on the left: a reduction. Mn goes from +7 to +2, a change of 5, which matches the 5e⁻.

Combining half-equations

To get the overall equation, multiply each half-equation so the electrons lost equal the electrons gained, add them, and cancel anything that appears on both sides (electrons always cancel completely; H⁺ and H₂O sometimes partly cancel).

Dichromate and iron(II)

Acidified dichromate oxidises Fe²⁺ to Fe³⁺. Write the overall balanced equation.

Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Oxidation: Fe²⁺ → Fe³⁺ + e⁻. Multiply by 6 so it releases 6e⁻: 6Fe²⁺ → 6Fe³⁺ + 6e⁻

Add and cancel the 6e⁻: $$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$$ Check charge: −2 + 14 + 12 = +24 on the left; 6 + 18 = +24 on the right.

When H⁺ cancels partly

Acidified permanganate oxidises sulfite ions, SO₃²⁻, to sulfate ions. Write the overall equation.

Oxidation: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

LCM of 2 and 5 is 10: multiply the oxidation by 5 and the reduction by 2.

Sum: 2MnO₄⁻ + 16H⁺ + 5SO₃²⁻ + 5H₂O → 2Mn²⁺ + 8H₂O + 5SO₄²⁻ + 10H⁺

Cancel 10H⁺ and 5H₂O from both sides: $$2\text{MnO}_4^- + 6\text{H}^+ + 5\text{SO}_3^{2-} \rightarrow 2\text{Mn}^{2+} + 5\text{SO}_4^{2-} + 3\text{H}_2\text{O}$$

InteractiveHalf-equation balancer
Try this
  • Choose Cr₂O₇²⁻ → Cr³⁺ and press Next step four times. Watch the check table: which step makes the chromium column turn green, and which makes the charge column match?
  • Choose the reductant SO₃²⁻ → SO₄²⁻. After step 4 the electrons are on the right. Why does that make it an oxidation?
  • Set oxidant MnO₄⁻ → Mn²⁺ and reductant Fe²⁺ → Fe³⁺, then pick Overall. You should get MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O with 5 electrons transferred.
  • Change the reductant to C₂O₄²⁻ → CO₂. How many electrons are transferred now, and why are the multipliers 2 and 5?

Test yourself

  1. What is the oxidation number of nitrogen in the nitrate ion, NO₃⁻?

  2. In the reaction Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq), the reducing agent is

  3. Which is the correctly balanced half-equation for nitrate being reduced to nitrogen monoxide in acid?

  4. Which reaction is NOT a redox reaction?

  5. When the half-equations for MnO₄⁻ → Mn²⁺ and H₂O₂ → O₂ (in acid) are combined, how many electrons are transferred in the smallest whole-number overall equation?

Name each step in balancing the half-equation.

SkeletonCr₂O₇²⁻ → Cr³⁺Balance CrCr₂O₇²⁻ → 2Cr³⁺Add H₂O for OCr₂O₇²⁻ → 2Cr³⁺ + 7H₂OAdd H⁺ for HCr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂OAdd e⁻ for chargeCr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂Ocharge: +12 − 6 = +6 = 2 × (+3)

Is each change an oxidation or a reduction?

Oxidation is the of electrons. In an oxidation the oxidation number . The species that is reduced is called the . In acidic solution, oxygen atoms are balanced by adding and hydrogen atoms by adding .

5 marks

Acidified potassium dichromate oxidises ethanol to ethanoic acid. The half-equation for the ethanol is C₂H₅OH + H₂O → CH₃COOH + 4H⁺ + 4e⁻. (a) Show that chromium is reduced, using oxidation numbers. (b) Write the half-equation for Cr₂O₇²⁻ forming Cr³⁺ and use it to write the balanced overall equation.