Topic 16 · page 1 of 19

Oxidation and reduction

Oxidation is loss of electrons and an increase in oxidation number; reduction is the opposite. Oxidation numbers let you follow the electrons in any compound, in the s-, p- and d-block alike.

14 min read16.1

Two ways to describe the same change 16.1

Every redox reaction moves electrons from one species to another. You can describe the move in two ways. The first follows the electrons. The second follows oxidation numbers, which is the only way that works when no free ions are involved, as in MnO₄⁻.

The two always happen together. Electrons that one species loses must be gained by another. The species doing the job gets its own name:

Take zinc in blue copper(II) sulfate solution. Copper coats the zinc and the blue colour fades.

Zn(s)+CuX2+(aq)→ZnX2+(aq)+Cu(s) \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}

Zinc loses two electrons, so zinc is oxidised and is the reducing agent. Each copper(II) ion gains them, so copper(II) is reduced and is the oxidising agent.

Assigning oxidation numbers 16.1

An oxidation number is the charge an atom would have if every bond were ionic. It is a bookkeeping device. Use these rules in order:

Worked exampleOxidation numbers in three ions

  1. Cr in Cr₂O₇²⁻: 2x + 7(−2) = −2, so 2x = +12 and x = +6.
  2. S in S₂O₃²⁻: 2x + 3(−2) = −2, so x = +2.
  3. S in S₄O₆²⁻: 4x + 6(−2) = −2, so x = +2.5. This is an average. The four sulfur atoms share a total of +10, so oxidation numbers can be fractions.

A Roman numeral in a name is the oxidation number. Manganate(VII) contains manganese at +7, and iron(III) is Fe at +3.

One example from each block 16.1

The spec asks you to apply these ideas to s-, p- and d-block elements. One reaction from each block is enough to show the pattern.

s-block: sodium. In sodium burning in chlorine, sodium goes from 0 to +1.

2 Na(s)+ClX2(g)→2 NaCl(s) \ce{2Na(s) + Cl2(g) -> 2NaCl(s)}

Sodium loses an electron, so it is oxidised and is the reducing agent: Na→NaX++eX−\ce{Na -> Na+ + e-}.

p-block: chlorine, iodine and sulfur. In the same reaction, chlorine goes from 0 to −1. It gains electrons, so it is reduced: ClX2+2 eX−→2 ClX−\ce{Cl2 + 2e- -> 2Cl-}. You will meet a second p-block reaction in the titrations:

IX2+2 SX2OX3X2−→2 IX−+SX4OX6X2− \ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}

Iodine goes from 0 to −1 (reduced). Sulfur goes from +2 to +2.5 (oxidised). Each of the two thiosulfate ions gives up half an electron per sulfur atom, which is 2 electrons in total. Iodine takes exactly those 2 electrons.

d-block: manganese and iron. This is the reaction of Core Practical 13a:

MnOX4X−+8 HX++5 FeX2+→MnX2++4 HX2O+5 FeX3+ \begin{aligned} &\ce{MnO4- + 8H+ + 5Fe^2+} \\ &\ce{-> Mn^2+ + 4H2O + 5Fe^3+} \end{aligned}

Manganese goes from +7 to +2, a decrease of 5, so it gains 5 electrons. Each iron goes from +2 to +3, so it loses 1 electron. It takes 5 Fe²⁺ for each MnO₄⁻.

Block Element Change Type
s Na in NaCl formation 0 → +1 oxidised
p Cl in NaCl formation 0 → −1 reduced
p I in I₂ with thiosulfate 0 → −1 reduced
p S in S₂O₃²⁻ to S₄O₆²⁻ +2 → +2.5 oxidised
d Mn in MnO₄⁻ to Mn²⁺ +7 → +2 reduced
d Fe in Fe²⁺ to Fe³⁺ +2 → +3 oxidised

Check yourself

  1. Give the oxidation number of manganese in MnO₄⁻, of chlorine in ClO₃⁻ and of oxygen in H₂O₂.
  2. In Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g), say which species is oxidised, which is reduced, the oxidising agent and the reducing agent. Give the oxidation number changes.
  3. Write the half-equation for the oxidation of sodium and for the reduction of chlorine in 2Na + Cl₂ → 2NaCl. Show that the electrons balance.
  4. Use oxidation numbers to explain why 5 Fe²⁺ ions react with each MnO₄⁻ ion.
  5. For I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, state the change in oxidation number of iodine and of sulfur and name the reducing agent.
  6. A student writes "zinc is the oxidising agent in Zn + Cu²⁺ → Zn²⁺ + Cu". Correct the statement.

Answers cover these until you have tried the questions

  1. Mn: x + 4(−2) = −1, so +7. Cl: x + 3(−2) = −1, so +5. O in H₂O₂: 2(+1) + 2x = 0, so −1 (a peroxide).
  2. Mg goes 0 → +2, so Mg is oxidised and is the reducing agent. H goes +1 → 0, so H⁺ is reduced and is the oxidising agent.
  3. Oxidation: Na→NaX++eX−\ce{Na -> Na+ + e-} (×2 gives 2 electrons lost). Reduction: ClX2+2 eX−→2 ClX−\ce{Cl2 + 2e- -> 2Cl-} (2 electrons gained). 2 lost = 2 gained, so the electrons balance.
  4. Mn falls from +7 to +2, a decrease of 5. Each Fe rises from +2 to +3, an increase of 1. For the total increase to equal the total decrease, 5 Fe are needed per Mn: 5 Fe²⁺ : 1 MnO₄⁻.
  5. Iodine: 0 → −1 (reduced). Sulfur: +2 → +2.5 (oxidised). Thiosulfate donates the electrons, so S₂O₃²⁻ is the reducing agent.
  6. Zinc goes 0 → +2, so zinc is oxidised and is the reducing agent. The oxidising agent is Cu²⁺, which is reduced to copper.