Topic 16 · page 2 of 19

Building redox equations

Write each half-equation in four steps, multiply so the electrons match, add, and cancel. The same method gives every redox equation in this topic.

14 min read16.1

Read first: Oxidation and reduction

Half-equations in acid 16.1

A half-equation shows one species gaining or losing electrons. A redox equation is two half-equations added together so that the electrons cancel. Most reactions in this topic happen in acid, so you need a reliable way to write half-equations in acid.

MethodBalancing a half-equation in acid

  1. Write the species before and after, and balance the atom that changes.
  2. Add H₂O to balance oxygen.
  3. Add H⁺ to balance hydrogen.
  4. Add electrons to balance the charge.

Worked exampleManganate(VII) to manganese(II)

  1. MnO₄⁻ → Mn²⁺. One Mn on each side already.
  2. Four O on the left: add 4H₂O to the right. MnO₄⁻ → Mn²⁺ + 4H₂O
  3. Eight H on the right: add 8H⁺ to the left. MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
  4. Charge: left −1 + 8 = +7, right +2. Add 5 electrons to the left.
MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O \ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}

The number of electrons equals the fall in oxidation number (+7 to +2 is a fall of 5). Use that as a check on step 4.

Worked exampleDichromate(VI) to chromium(III)

  1. Cr₂O₇²⁻ → 2Cr³⁺. The 2 in front of Cr³⁺ is needed to balance chromium.
  2. Seven O on the left: add 7H₂O to the right.
  3. Fourteen H on the right: add 14H⁺ to the left.
  4. Charge: left −2 + 14 = +12, right 2 × 3 = +6. Add 6 electrons to the left.
CrX2OX7X2−+14 HX++6 eX−→2 CrX3++7 HX2O \small \ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}

Combining half-equations 16.1

To join an oxidation and a reduction, multiply one or both half-equations so that the electrons match, then add them and cancel the electrons.

Worked exampleManganate(VII) with iron(II)

The reduction needs 5 electrons. The oxidation is Fe²⁺ → Fe³⁺ + e⁻, which gives 1.

  1. Multiply the iron half-equation by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
  2. Add the two half-equations. The 5e⁻ cancel.
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}
  1. Check the charge: left −1 + 8 + 10 = +17, right +2 + 15 = +17.

Worked exampleDichromate(VI) with iron(II)

The reduction needs 6 electrons, so multiply the iron half-equation by 6.

CrX2OX7X2−+14 HX++6 FeX2+→2 CrX3++7 HX2O+6 FeX3+ \begin{aligned} &\ce{Cr2O7^2- + 14H+ + 6Fe^2+} \\ &\ce{-> 2Cr^3+ + 7H2O + 6Fe^3+} \end{aligned}

Charge: left −2 + 14 + 12 = +24, right 6 + 18 = +24.

Iodine and thiosulfate 16.1

The second equation of the titrations is built the same way. Iodine is reduced:

IX2+2 eX−→2 IX− \ce{I2 + 2e- -> 2I-}

Thiosulfate is oxidised to tetrathionate. Balance S first (2S₂O₃²⁻ → S₄O₆²⁻), then the charge: left −4, right −2, so the right side needs 2 electrons.

2 SX2OX3X2−→SX4OX6X2−+2 eX− \ce{2S2O3^2- -> S4O6^2- + 2e-}

Both half-equations involve 2 electrons, so add them directly:

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

The ratio of iodine to thiosulfate is 1 : 2. You will use it on page 13.

Check yourself

  1. Write the half-equation for MnO₄⁻ → Mn²⁺ in acid, and say how many electrons you expect from the oxidation number change.
  2. Write the half-equation for Cr₂O₇²⁻ → Cr³⁺ in acid and check that the charge balances.
  3. Combine your answer to question 2 with Fe²⁺ → Fe³⁺ + e⁻ to give the overall equation.
  4. Write the overall equation for MnO₄⁻ with Fe²⁺ in acid, then check the charge on each side.
  5. Write the half-equation for the oxidation of thiosulfate to tetrathionate, and the half-equation for I₂ → I⁻. Combine them.
  6. Hydrogen peroxide is reduced to water in acid: H₂O₂ → H₂O. Write the half-equation.

Answers cover these until you have tried the questions

  1. MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}. Mn falls from +7 to +2, a decrease of 5, so 5 electrons are gained.
  2. CrX2OX7X2−+14 HX++6 eX−→2 CrX3++7 HX2O\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}. Left: −2 + 14 − 6 = +6. Right: 2 × (+3) = +6. Balanced.
  3. Multiply the iron half-equation by 6 and add: CrX2OX7X2−+14 HX++6 FeX2+→2 CrX3++7 HX2O+6 FeX3+\ce{Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 7H2O + 6Fe^3+}. Charge: −2 + 14 + 12 = +24 on the left; 6 + 18 = +24 on the right.
  4. MnOX4X−+8 HX++5 FeX2+→MnX2++4 HX2O+5 FeX3+\ce{MnO4- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+}. Left: −1 + 8 + 10 = +17. Right: +2 + 15 = +17.
  5. Oxidation: 2 SX2OX3X2−→SX4OX6X2−+2 eX−\ce{2S2O3^2- -> S4O6^2- + 2e-}. Reduction: IX2+2 eX−→2 IX−\ce{I2 + 2e- -> 2I-}. Overall: IX2+2 SX2OX3X2−→2 IX−+SX4OX6X2−\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}.
  6. Balance O: H₂O₂ → 2H₂O. Balance H by adding 2H⁺ to the left. Charge: left +2, right 0, so add 2 electrons to the left. HX2OX2+2 HX++2 eX−→2 HX2O\ce{H2O2 + 2H+ + 2e- -> 2H2O}