Redox Reactions • Topic 1 of 3

Oxidation Number & Redox Concepts

A redox reaction is one in which oxidation and reduction occur together. Two views describe the same change.

Classical (oxygen / hydrogen) concept

In the old view, oxidation is addition of oxygen or removal of hydrogen, and reduction is the reverse. For example, in 2Mg + O2 → 2MgO magnesium is oxidised; in CuO + H2 → Cu + H2O the CuO is reduced. This idea fails when no oxygen or hydrogen is involved.

Electronic (electron-transfer) concept

The modern view is general: oxidation is loss of electrons and reduction is gain of electrons. The mnemonic OIL RIG — Oxidation Is Loss, Reduction Is Gain — captures this. In Zn + Cu2+ → Zn2+ + Cu, zinc loses two electrons (oxidised) and Cu2+ gains them (reduced). Loss and gain occur together and in equal number.

Oxidation number (oxidation state)

The oxidation number is the imaginary charge an atom would carry if every bond were fully ionic. The rules are:

  • Free elements (O2, Na, P4) have oxidation number 0.
  • A monatomic ion's oxidation number equals its charge (Na+ = +1, Cl- = −1).
  • Oxygen is usually −2 (but −1 in peroxides, +2 in OF2).
  • Hydrogen is usually +1 (but −1 in metal hydrides such as NaH).
  • Group 1 metals = +1, Group 2 = +2; fluorine is always −1.
  • The sum of oxidation numbers in a neutral molecule is 0, and in an ion equals the ion's charge.

Oxidising and reducing agents

An oxidising agent is itself reduced (gains electrons); a reducing agent is itself oxidised (loses electrons). The species whose oxidation number increases is the reducing agent; the one whose oxidation number decreases is the oxidising agent.

Types of redox reactions

  • Combination: two species combine, e.g. C + O2 → CO2.
  • Decomposition: a compound splits, e.g. 2H2O → 2H2 + O2.
  • Displacement: a more reactive element displaces another, e.g. Zn + CuSO4 → ZnSO4 + Cu.
  • Disproportionation: the same element is simultaneously oxidised and reduced from one intermediate oxidation state, e.g. in 2H2O2 → 2H2O + O2 oxygen goes from −1 to −2 and 0.
  • Comproportionation: the reverse — two species of the same element in different states form one intermediate state, e.g. 5Cl- + ClO3- + 6H+ → 3Cl2 + 3H2O.
OIL RIG: zinc is oxidised (loses electrons), copper ion is reduced (gains electrons)Zn → Zn²⁺OXIDATION (loss)Cu²⁺ → CuREDUCTION (gain)2 e⁻OIL RIGOxidation Is Loss · Reduction Is Gain
Solved Examples
1
Worked Example
Find the oxidation number of sulphur in H2SO4.
Solution
  1. Assign H = +1 and O = −2 by the standard rules.
  2. Let the oxidation number of S be x. The molecule is neutral, so the sum is 0.
  3. 2(+1) + x + 4(−2) = 0 ⇒ 2 + x − 8 = 0.
  4. Solve: x = +6.

Answer: Sulphur has oxidation number +6.

2
Worked Example
Determine the oxidation number of Mn in the permanganate ion MnO4-.
Solution
  1. Oxygen is −2 each; there are four O atoms.
  2. Let Mn be x. The ion's total charge is −1.
  3. x + 4(−2) = −1 ⇒ x − 8 = −1.
  4. Solve: x = +7.

Answer: Mn is in the +7 state.

3
Worked Example
In the reaction Zn + CuSO4 → ZnSO4 + Cu, identify the oxidising and reducing agents.
Solution
  1. Zn goes from 0 to +2: its oxidation number increases, so it is oxidised — it is the reducing agent.
  2. Cu in CuSO4 is +2 and becomes 0 as Cu metal: its oxidation number decreases, so it is reduced.
  3. The species that is reduced is the oxidising agent — here Cu2+ (CuSO4).

Answer: Zn is the reducing agent; CuSO4 (Cu2+) is the oxidising agent.

4
Worked Example
Find the average oxidation number of carbon in glucose C6H12O6.
Solution
  1. Take H = +1 and O = −2.
  2. Let average C = x. Neutral molecule: 6x + 12(+1) + 6(−2) = 0.
  3. 6x + 12 − 12 = 0 ⇒ 6x = 0.
  4. So x = 0.

Answer: The average oxidation number of carbon is 0.

5
Worked Example
Classify the reaction 2H2O2 → 2H2O + O2 and justify your answer using oxidation numbers.
Solution
  1. In H2O2 oxygen is in the −1 state (peroxide).
  2. In H2O oxygen is −2 (reduced); in O2 oxygen is 0 (oxidised).
  3. The same element (O) is simultaneously oxidised and reduced from one intermediate state.

Answer: It is a disproportionation reaction.

6
Worked Example
Find the oxidation number of chromium in the dichromate ion Cr2O72-.
Solution
  1. Oxygen is −2 each; there are seven O atoms.
  2. Let each Cr be x. Ion charge is −2: 2x + 7(−2) = −2.
  3. 2x − 14 = −2 ⇒ 2x = 12.
  4. Solve: x = +6.

Answer: Each chromium is in the +6 state.

Key Points

  • Oxidation = loss of electrons (or addition of O / loss of H); reduction = gain of electrons (or loss of O / addition of H) — remember OIL RIG.
  • Oxidation number is the imaginary ionic charge; assign by rules (free element 0, O = -2, H = +1, sum = charge) and solve for the unknown.
  • The reducing agent is oxidised (its oxidation number rises); the oxidising agent is reduced (its oxidation number falls).
  • Redox reactions come in five families: combination, decomposition, displacement, disproportionation and comproportionation.
  • Disproportionation: one intermediate oxidation state of an element is both oxidised and reduced; comproportionation is its reverse.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.According to the electronic concept, oxidation is:
Explanation: OIL RIG: Oxidation Is Loss of electrons; reduction is gain.
Q2.The oxidation number of sulphur in H2SO4 is:
Explanation: 2(+1) + x + 4(-2) = 0 gives x = +6.
Q3.In Zn + Cu2+ → Zn2+ + Cu, the reducing agent is:
Explanation: Zn loses electrons (0 to +2), so it is oxidised and acts as the reducing agent.
Q4.The reaction 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O is an example of:
Explanation: Chlorine (0) is both reduced to Cl⁻ (-1) and oxidised to ClO₃⁻ (+5) — disproportionation.
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