Topic 8 · page 7 of 29

Full ionic equations

Multiply the two half-equations until the electrons match, add them, and cancel anything that appears on both sides. Then check atoms and charge, and make sure no electrons are left.

13 min read8.10

Read first: Building half-equations

Combining two half-equations 8.10

Electrons are not left over in a real reaction: the number lost in the oxidation equals the number gained in the reduction. So a full equation is built from one oxidation half-equation and one reduction half-equation, with the electrons made equal.

MethodFrom half-equations to a full ionic equation

  1. Write the reduction half-equation and the oxidation half-equation (electrons on opposite sides).
  2. Multiply one or both by whole numbers so that the electrons are equal. Multiply every term, including the H₂O and H⁺.
  3. Add the two equations together.
  4. Cancel the electrons, and any H₂O or H⁺ that appear on both sides.
  5. Check: atoms balance, charge balances, and no electrons appear.

Worked exampleManganate(VII) and iron(II)

Acidified MnO₄⁻ oxidises Fe²⁺ to Fe³⁺. Construct the ionic equation.

  1. Reduction: MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}
  2. Oxidation: FeX2+→FeX3++eX−\ce{Fe^2+ -> Fe^3+ + e-}
  3. The oxidation gives 1 electron, the reduction takes 5. Multiply the oxidation by 5: 5 FeX2+→5 FeX3++5 eX−\ce{5Fe^2+ -> 5Fe^3+ + 5e-}
  4. Add and cancel the 5 e⁻:
MnOX4X−+8 HX++5 FeX2+→MnX2++4 HX2O+5 FeX3+ \ce{MnO4- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+}
  1. Check charge: left −1 + 8 + 10 = +17; right +2 + 15 = +17. ✓ Atoms: Mn 1, O 4, H 8, Fe 5 on each side. ✓

Worked exampleDichromate(VI) and iron(II)

Acidified Cr₂O₇²⁻ oxidises Fe²⁺.

  1. Reduction: CrX2OX7X2−+14 HX++6 eX−→2 CrX3++7 HX2O\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}
  2. Oxidation, × 6: 6 FeX2+→6 FeX3++6 eX−\ce{6Fe^2+ -> 6Fe^3+ + 6e-}
  3. Add:
CrX2OX7X2−+14 HX++6 FeX2+→2 CrX3++7 HX2O+6 FeX3+ \ce{Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 7H2O + 6Fe^3+}
  1. Check charge: left −2 + 14 + 12 = +24; right 6 + 18 = +24. ✓

When the numbers of electrons do not divide, use the lowest common multiple. For 2 electrons in one equation and 5 in the other, multiply by 5 and by 2 to get 10 each.

Worked exampleManganate(VII) and ethanedioate

Acidified MnO₄⁻ oxidises C₂O₄²⁻ (ethanedioate) to CO₂. This pair is used in titrations in the practical on page 21.

  1. Reduction: MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}
  2. Oxidation: CX2OX4X2−→2 COX2+2 eX−\ce{C2O4^2- -> 2CO2 + 2e-}
  3. Lowest common multiple of 5 and 2 is 10. Multiply the reduction by 2 and the oxidation by 5.
  4. Add:
2 MnOX4X−+16 HX++5 CX2OX4X2−→2 MnX2++8 HX2O+10 COX2 \ce{2MnO4- + 16H+ + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2}
  1. Check charge: left −2 + 16 − 10 = +4; right +4. ✓ Oxygen: left 8 + 20 = 28; right 8 + 20 = 28. ✓

Simple redox equations 8.10

Metals reacting with acid or with ions follow the same method, and often the half-equations are short enough to combine by eye.

Worked exampleMagnesium and silver ions

  1. Oxidation: Mg→MgX2++2 eX−\ce{Mg -> Mg^2+ + 2e-}
  2. Reduction: AgX++eX−→Ag\ce{Ag+ + e- -> Ag}. Multiply by 2: 2 AgX++2 eX−→2 Ag\ce{2Ag+ + 2e- -> 2Ag}
  3. Add: Mg+2 AgX+→MgX2++2 Ag\ce{Mg + 2Ag+ -> Mg^2+ + 2Ag}

Worked exampleHydrogen peroxide and iodide

Hydrogen peroxide oxidises iodide ions in acid to iodine.

  1. Reduction: HX2OX2+2 HX++2 eX−→2 HX2O\ce{H2O2 + 2H+ + 2e- -> 2H2O}
  2. Oxidation: 2 IX−→IX2+2 eX−\ce{2I- -> I2 + 2e-}
  3. Electrons already match (2 each). Add:
HX2OX2+2 HX++2 IX−→2 HX2O+IX2 \ce{H2O2 + 2H+ + 2I- -> 2H2O + I2}

An ionic equation can be turned into a full equation by adding the spectator ions. In the manganate(VII) and iron(II) reaction in sulfuric acid, for example, K⁺ from KMnO₄ and SO₄²⁻ from FeSO₄ and H₂SO₄ are unchanged, and they are not shown in the ionic equation.

Check yourself

  1. Combine Zn→ZnX2++2 eX−\ce{Zn -> Zn^2+ + 2e-} and CuX2++2 eX−→Cu\ce{Cu^2+ + 2e- -> Cu} into an ionic equation.
  2. Combine FeX3++eX−→FeX2+\ce{Fe^3+ + e- -> Fe^2+} and 2 IX−→IX2+2 eX−\ce{2I- -> I2 + 2e-} into an ionic equation.
  3. Construct the ionic equation for the reaction of acidified manganate(VII) with SO₂, which is oxidised to sulfate (use the half-equation SOX2+2 HX2O→SOX4X2−+4 HX++2 eX−\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}).
  4. Check that the equation CrX2OX7X2−+14 HX++6 FeX2+→2 CrX3++7 HX2O+6 FeX3+\ce{Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 7H2O + 6Fe^3+} is balanced for charge.
  5. Magnesium reacts with dilute hydrochloric acid. Write the two half-equations and the ionic equation.
  6. A student writes MnOX4X−+8 HX++5 eX−+FeX2+→MnX2++4 HX2O+FeX3+\ce{MnO4- + 8H+ + 5e- + Fe^2+ -> Mn^2+ + 4H2O + Fe^3+} as the full equation. Give two things wrong with it.

Answers cover these until you have tried the questions

  1. The electrons already match: Zn+CuX2+→ZnX2++Cu\ce{Zn + Cu^2+ -> Zn^2+ + Cu}.
  2. Multiply the first by 2: 2 FeX3++2 eX−→2 FeX2+\ce{2Fe^3+ + 2e- -> 2Fe^2+}. Add: 2 FeX3++2 IX−→2 FeX2++IX2\ce{2Fe^3+ + 2I- -> 2Fe^2+ + I2}.
  3. Reduction × 2: 2 MnOX4X−+16 HX++10 eX−→2 MnX2++8 HX2O\ce{2MnO4- + 16H+ + 10e- -> 2Mn^2+ + 8H2O}. Oxidation × 5: 5 SOX2+10 HX2O→5 SOX4X2−+20 HX++10 eX−\ce{5SO2 + 10H2O -> 5SO4^2- + 20H+ + 10e-}. Add and cancel the electrons, 10 H₂O (from 8 on the right and 10 on the left leaves 2 on the left) and 16 H⁺ (20 − 16 leaves 4 on the right): 2 MnOX4X−+5 SOX2+2 HX2O→2 MnX2++5 SOX4X2−+4 HX+\ce{2MnO4- + 5SO2 + 2H2O -> 2Mn^2+ + 5SO4^2- + 4H+}. Charge: left −2; right +4 − 10 + 4 = −2. ✓
  4. Left: −2 + 14 + 6 × (+2) = −2 + 14 + 12 = +24. Right: 2 × (+3) + 6 × (+3) = 6 + 18 = +24. ✓ Charge balances.
  5. Oxidation: Mg→MgX2++2 eX−\ce{Mg -> Mg^2+ + 2e-}. Reduction: 2 HX++2 eX−→HX2\ce{2H+ + 2e- -> H2}. Add: Mg+2 HX+→MgX2++HX2\ce{Mg + 2H+ -> Mg^2+ + H2}.
  6. (1) Electrons must not appear in the full equation. (2) The electrons do not match: the oxidation half-equation must be multiplied by 5 so that 5 Fe²⁺ lose the 5 electrons gained by one MnO₄⁻. The charge is also unbalanced as written (left +4, right +5).