Syllabus Edition

First teaching 2023

First exams 2025

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Half Equations (HL IB Chemistry)

Revision Note

Philippa

Author

Philippa

Expertise

Chemistry

Half Equations

How to balance a redox equation

  • Oxidation numbers can be used to balance chemical equations
  • Go through these steps to balance a redox equation:
    1. Write the unbalanced equation and identify the atoms which change in ox. no.
    2. Deduce the oxidation number changes
    3. Balance the oxidation number changes
    4. Balance the charges
    5. Balance the atoms

Worked example

Manganate(VII) ions (MnO4-) react with Fe2+ ions in the presence of acid (H+) to form Mn2+ ions, Fe3+ ions and water. Write the overall redox equation for the reaction

 

Answer:

  1. Write the unbalanced equation and identify the atoms which change in oxidation number 
    • Step 1 in balancing redox equation
  2. Deduce the oxidation number changes
    • Step 2 in balancing redox equation
  3. Balance the oxidation number changes
    • Step 3 in balancing redox equation
  4. Balance the charges
    • Step 4 in balancing redox equation
  5. Balance the atoms
    • Step 5 in balancing redox equation

Redox titrations

  • In titrations, the concentration of a solution is determined by titrating with a solution of known concentration.
  • In redox titrations, an oxidising agent is titrated against a reducing agent
    • Electrons are transferred from one species to the other
  • Indicators are sometimes used to show the endpoint of the titration
  • However, most transition metal ions naturally change colour when changing the oxidation state
  • There are two common redox titrations you should know about:
    • Manganate(VII) titrations
    • Iodine-thiosulfate titrations

Manganate(VII) titrations

  • A redox reaction occurs between acidified manganate(VII) ions and iron(II) ions:

MnO4 (aq) +  8H+ (aq) +  5Fe2+ (aq)  → Mn2+ (aq) +  5Fe3+ (aq) + 4H2O (l)

  • This reaction needs no indicator as the manganate(VII) is a strong purple colour which disappears at the endpoint, so the titration is self-indicating
  • This reaction is often used for the analysis of iron for example in iron tablets (health supplements)

Iodine-thiosulfate titrations

  • A redox reaction occurs between iodine and thiosulfate ions:

2S2O32– (aq) + I2 (aq) → 2I(aq) + S4O62– (aq)

  • The light brown/yellow colour of the iodine turns paler as it is converted to colourless iodide ions
  • When the solution is a straw colour, starch is added to clarify the endpoint
  • The solution turns blue/black until all the iodine reacts, at which point the colour disappears.
  • This titration can be used to determine the concentration of an oxidising agent, which oxidises iodide ions to iodine molecules
  • The amount of iodine is determined from titration against a known quantity of sodium thiosulfate solution
  • This reaction can be used for the analysis of chlorine in bleach

Worked example

A health supplement tablet containing iron(II)sulfate was analysed by titration. A tablet weighing 2.25 g was dissolved in dilute sulfuric acid and titrated against 0.100 mol dm-3 KMnO4.The titration required 26.50 cm3 for a complete reaction. Calculate the percentage by mass of iron in the table.

 

Answer:

  1. Write the balanced equation for the reaction
    • oxidation: Fe2+ (aq)  →   Fe3+ (aq)  + e-
    • reduction: MnO4- (aq) + 8H+ (aq)  + 5e-   →   Mn2+ (aq)  + 4H2O (l)
    • overall: MnO4- (aq) + 8H+ (aq)  + 5Fe2+ (aq)   →   Mn2+ (aq)  + 4H2O (l) + 5Fe3+ (aq)
       
  2. Determine the amount of MnO4- used in the titration
    • moles of MnO4 = 0.0265 dm3  x  0.100 mol dm-= 0.00265 mol
       
  3. Determine the amount of iron in the reaction
    • From the equation for the reaction, we know the reacting ratio  MnO4- : Fe2+ = 1: 5
    • ∴ moles of Fe2+ = 0.00265 mol MnO4- x 5 = 0.01325 mol
       
  4. Convert moles into the mass of iron
    • Mass of iron = 0.01325 mol x 55.85 gmol-1 = 0.740 g
       
  5. Find the percentage of iron in the tablet
    • ∴ % Fe in the tablet = (0.740/ 2.25)  x 100 = 32.9%

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