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Redox Reactions

  • 5 subtopics
  • 15 things to know
  • A2 Unit 3

What Redox Reactions covers

Redox Reactions is examined on A2 Unit 3 of WJEC Chemistry, Physical and Inorganic Chemistry. 81 recall questions across 5 subtopics.

Where it is assessed

A2 Unit 3

Physical and Inorganic Chemistry.

Check the WJEC specification ↗

The 5 subtopics

Each one has its own page.

What you need to know

Answer each one before you reveal it.

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Key term

How do you construct a half-equation for a species reacting in acid solution?

Key term

What is the half-equation for the reduction of the manganate(VII) ion to Mn²⁺ in acid solution?

Key term

What is the half-equation for the reduction of VO₂⁺ to VO²⁺ in acid solution?

Numbers

Combine the dichromate(VI) half-equation with the oxidation of iron(II).

Key term

What is the overall equation for the reaction of acidified dichromate(VI) with iodide ions?

Key term

What is the overall equation for the reaction of acidified manganate(VII) with iodide ions?

Key term

What is seen when an excess of potassium iodide solution is added to a solution of a copper(II) salt?

Key term

Use oxidation numbers to show that the reaction of copper(II) with iodide is a redox reaction.

Numbers

Why is an excess of potassium iodide added when copper(II) ions are treated with iodide?

Key term

Why is no indicator needed in a titration with potassium manganate(VII)?

Key term

Which acid is used to acidify the flask in a manganate(VII) titration, and why not hydrochloric acid?

Key term

Why is nitric acid not used to acidify a manganate(VII) titration?

Numbers

25.0 cm³ of an iron(II) solution requires 22.40 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII). Calculate the concentration of the iron(II) solution.

Numbers

Calculate the volume of 0.0200 mol dm⁻³ potassium manganate(VII) needed to react with 25.0 cm³ of 0.0500 mol dm⁻³ sodium ethanedioate in acid solution.

Numbers

Calculate the volume of 0.100 mol dm⁻³ iron(II) sulfate solution that would react with 25.0 cm³ of 0.0250 mol dm⁻³ potassium dichromate(VI) in acid solution.

Take Redox Reactions further

The same content, four other ways.

Commonly confused

Pairs that lose marks when they are blurred together.

  • An oxidising agent vs A reducing agent

    An oxidising agent: a species that takes electrons from another and is itself reduced. A reducing agent: a species that gives electrons to another and is itself oxidised.

  • The standard electrode potential vs The cell EMF

    The standard electrode potential: the voltage of a half-cell measured against the standard hydrogen electrode. The cell EMF: the difference between the two electrode potentials in a complete cell.

Questions from this topic

Each one with the model answer underneath.

  1. 1

    How do you construct a half-equation for a species reacting in acid solution?

    Model answer

    Write the species that changes on each side and balance the atoms of the element being oxidised or reduced. Add water molecules to balance the oxygen atoms, then add hydrogen ions to balance the hydrogen atoms. Finally add electrons to whichever side is needed so that the total charge is the same on both sides.

  2. 2

    What is the half-equation for the reduction of the manganate(VII) ion to Mn²⁺ in acid solution?

    Model answer

    MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Four water molecules balance the four oxygen atoms, eight hydrogen ions balance the hydrogen, and five electrons make the charge +2 on both sides, since −1 + 8 − 5 = +2.

  3. 3

    Construct the half-equation for the reduction of the chlorate(I) ion, ClO⁻, to chloride ions in acid solution.

    Model answer

    The half-equation is ClO⁻ + 2H⁺ + 2e⁻ → Cl⁻ + H₂O. The single oxygen atom leaves as one water molecule, which needs two hydrogen ions to supply its hydrogen. Two electrons are then added on the left so that the charge is −1 + 2 − 2 = −1 on each side, matching chlorine falling from +1 in ClO⁻ to −1 in Cl⁻.

  4. 4

    What is the half-equation for the oxidation of thiosulfate ions to tetrathionate ions?

    Model answer

    2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻. Two thiosulfate ions are needed to supply the four sulfur atoms, and two electrons balance the charge, since −4 on the left equals −2 − 2 on the right.

  5. 5

    Why do hydrogen ions appear in the half-equation for the reduction of manganate(VII)?

    Model answer

    The manganate(VII) ion carries four oxygen atoms which are not present in the Mn²⁺ product. Those oxygen atoms leave as water, and the hydrogen needed to form that water is supplied by the acid, so hydrogen ions must appear on the left of the equation.

  6. 6

    Why must a solution of dichromate(VI) or manganate(VII) be acidified before it can act as an oxidising agent in these ways?

    Model answer

    Both half-equations consume a large number of hydrogen ions, eight for each manganate(VII) ion and fourteen for each dichromate(VI) ion. Without a plentiful supply of acid the reduction to Mn²⁺ or Cr³⁺ cannot take place.

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Redox Reactions revision questions

What does Redox Reactions cover in WJEC A-Level Chemistry?
5 subtopics: building ion–electron half-equations, combining half-equations into full equations, copper, iodide and thiosulfate, carrying out and calculating a redox titration: procedure for a manganate titration and carrying out and calculating a redox titration: redox titration calculations. Each one has its own page with questions and model answers.
How is Redox Reactions organised?
By the WJEC 1410QS specification, subtopic by subtopic, so it matches the order you are taught it in.

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