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Redox Reactions

Building ion–electron half-equations

  • 3 things to know
  • A2 Unit 3

Overview

Nineteen recall questions on Building ion–electron half-equations.

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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?

Questions from this subtopic

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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