Topic 8 · page 6 of 29

Building half-equations

A half-equation shows one species gaining or losing electrons. Balance the atoms (O with water, H with H⁺), then add electrons to balance the charge, and check against the change in oxidation number.

14 min read8.10

Read first: Oxidising and reducing agents

What a half-equation is 8.10

Simple cases need only the electrons:

Zn→ZnX2++2 eX−ClX2+2 eX−→2 ClX−FeX3++eX−→FeX2+ \ce{Zn -> Zn^2+ + 2e-} \qquad \ce{Cl2 + 2e- -> 2Cl-} \qquad \ce{Fe^3+ + e- -> Fe^2+}

Zinc loses electrons, so they are on the right: oxidation. Chlorine and iron(III) gain them, so they are on the left: reduction.

Many half-equations involve oxygen-containing ions such as MnO₄⁻, Cr₂O₇²⁻ and SO₄²⁻. In acid solution, the oxygen ends up as water, and H⁺ ions from the acid appear to balance the hydrogen.

The four-step method 8.10

MethodBalancing a half-equation in acid

  1. Balance the atoms of the element that changes (not O or H yet).
  2. Balance oxygen by adding H₂O to the side that needs oxygen.
  3. Balance hydrogen by adding H⁺ to the side that needs hydrogen.
  4. Balance the charge by adding e⁻ to the more positive side.

Then check: the number of electrons should equal the change in oxidation number, summed over all the atoms that change.

Worked exampleManganate(VII) to manganese(II)

  1. Atoms of Mn are already balanced: MnOX4X−→MnX2+\ce{MnO4- -> Mn^2+}.
  2. Oxygen: add 4 H₂O to the right: MnOX4X−→MnX2++4 HX2O\ce{MnO4- -> Mn^2+ + 4H2O}.
  3. Hydrogen: the right has 8 H, so add 8 H⁺ to the left: MnOX4X−+8 HX+→MnX2++4 HX2O\ce{MnO4- + 8H+ -> Mn^2+ + 4H2O}.
  4. Charge: the left is −1 + 8 = +7 and the right is +2, so add 5 e⁻ to the left: MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}.
  5. Check: Mn goes +7 → +2, a gain of 5 electrons. ✓

Worked exampleDichromate(VI) to chromium(III)

  1. Two chromium atoms are needed on the right: CrX2OX7X2−→2 CrX3+\ce{Cr2O7^2- -> 2Cr^3+}.
  2. Oxygen: 7 H₂O on the right: CrX2OX7X2−→2 CrX3++7 HX2O\ce{Cr2O7^2- -> 2Cr^3+ + 7H2O}.
  3. Hydrogen: 14 H⁺ on the left: CrX2OX7X2−+14 HX+→2 CrX3++7 HX2O\ce{Cr2O7^2- + 14H+ -> 2Cr^3+ + 7H2O}.
  4. Charge: left −2 + 14 = +12; right 2 × (+3) = +6. Add 6 e⁻ to the left.
  5. CrX2OX7X2−+14 HX++6 eX−→2 CrX3++7 HX2O\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}. Check: each Cr +6 → +3, a gain of 3 × 2 = 6 electrons. ✓

Half-equations you will meet 8.10

Species Half-equation Change in oxidation number
Fe²⁺ oxidised FeX2+→FeX3++eX−\ce{Fe^2+ -> Fe^3+ + e-} Fe +2 → +3
I⁻ oxidised 2 IX−→IX2+2 eX−\ce{2I- -> I2 + 2e-} I −1 → 0
H₂O₂ as oxidising agent HX2OX2+2 HX++2 eX−→2 HX2O\ce{H2O2 + 2H+ + 2e- -> 2H2O} O −1 → −2
H₂O₂ as reducing agent HX2OX2→OX2+2 HX++2 eX−\ce{H2O2 -> O2 + 2H+ + 2e-} O −1 → 0
SO₂ oxidised SOX2+2 HX2O→SOX4X2−+4 HX++2 eX−\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-} S +4 → +6
Conc. H₂SO₄ to SO₂ HX2SOX4+2 HX++2 eX−→SOX2+2 HX2O\ce{H2SO4 + 2H+ + 2e- -> SO2 + 2H2O} S +6 → +4
Conc. H₂SO₄ to S HX2SOX4+6 HX++6 eX−→S+4 HX2O\ce{H2SO4 + 6H+ + 6e- -> S + 4H2O} S +6 → 0
Conc. H₂SO₄ to H₂S HX2SOX4+8 HX++8 eX−→HX2S+4 HX2O\ce{H2SO4 + 8H+ + 8e- -> H2S + 4H2O} S +6 → −2

The last three are used on page 27.

Worked exampleSulfur dioxide to sulfate

  1. S is balanced: SOX2→SOX4X2−\ce{SO2 -> SO4^2-}.
  2. The right has two more O, so add 2 H₂O to the left: SOX2+2 HX2O→SOX4X2−\ce{SO2 + 2H2O -> SO4^2-}.
  3. The left has 4 H, so add 4 H⁺ to the right: SOX2+2 HX2O→SOX4X2−+4 HX+\ce{SO2 + 2H2O -> SO4^2- + 4H+}.
  4. Charge: left 0; right −2 + 4 = +2. Add 2 e⁻ to the right.
  5. SOX2+2 HX2O→SOX4X2−+4 HX++2 eX−\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}. Check: S +4 → +6, a loss of 2 electrons. ✓

Check yourself

  1. Write the half-equation for the oxidation of zinc to Zn²⁺, and for the reduction of Ag⁺ to Ag.
  2. Balance MnOX4X−→MnX2+\ce{MnO4- -> Mn^2+} in acid solution, and check the electron count against the oxidation numbers.
  3. Balance CrX2OX7X2−→CrX3+\ce{Cr2O7^2- -> Cr^3+} in acid.
  4. Write the half-equation for the oxidation of Br⁻ to Br₂.
  5. Hydrogen peroxide can be oxidised to oxygen. Write the half-equation and the change in oxidation number of oxygen.
  6. Write the half-equation for concentrated sulfuric acid, H₂SO₄, being reduced to H₂S in the presence of acid.

Answers cover these until you have tried the questions

  1. Zn→ZnX2++2 eX−\ce{Zn -> Zn^2+ + 2e-} (oxidation) and AgX++eX−→Ag\ce{Ag+ + e- -> Ag} (reduction).
  2. MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}. Mn +7 → +2 is a gain of 5 electrons, matching 5 e⁻. ✓
  3. CrX2OX7X2−+14 HX++6 eX−→2 CrX3++7 HX2O\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}. Balance charge: left −2 + 14 − 6 = +6, right +6. ✓
  4. 2 BrX−→BrX2+2 eX−\ce{2Br- -> Br2 + 2e-}. Br −1 → 0 for each atom: 2 electrons in total.
  5. HX2OX2→OX2+2 HX++2 eX−\ce{H2O2 -> O2 + 2H+ + 2e-}. Oxygen goes from −1 to 0, an increase of 1 per atom, so 2 electrons for the 2 oxygen atoms.
  6. S is balanced. O: 4 on the left, so 4 H₂O on the right: HX2SOX4→HX2S+4 HX2O\ce{H2SO4 -> H2S + 4H2O}. H: right has 10, left has 2, so add 8 H⁺ to the left. Charge: left +8, right 0, so add 8 e⁻ to the left. HX2SOX4+8 HX++8 eX−→HX2S+4 HX2O\ce{H2SO4 + 8H+ + 8e- -> H2S + 4H2O}. Check: S +6 → −2 is a gain of 8 electrons. ✓