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Developing Metals

Balancing redox equations involving acid-base reactions

  • 4 things to know
  • Components 01, 02 and 03

Overview

Recall questions on the manganate(VII) and dichromate(VI) half-equations, why the electron counts are what they are, and combining halves into a full equation.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Numbers

What is the half-equation for manganate(VII) reduced to manganese(II), and why are exactly five electrons needed?

Numbers

How are the manganate(VII) and iron(II) half-equations combined, and what is the dichromate(VI) half-equation?

Numbers

How do the charges balance in the dichromate(VI) half-equation, what is its full ionic equation with iron(II), and what is the half-equation for the oxidation of ethanedioate?

Numbers

Write the full ionic equation for the oxidation of hydrogen peroxide by manganate(VII) in acid.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Which two species are used to balance a redox half-equation taking place in acidic solution?

    Model answer

    Hydrogen ions H⁺ and water molecules H₂O are used.

  2. 2

    What two things must balance in a correctly written half-equation?

    Model answer

    Every type of atom must balance and the total electrical charge must be the same on both sides.

  3. 3

    Outline the steps for constructing a half-equation for a species that contains oxygen.

    Model answer

    Balance the element that changes oxidation state, add H₂O to balance the oxygen, add H⁺ to balance the hydrogen, then add electrons until the charges match.

  4. 4

    Write the half-equation for the reduction of manganate(VII) to manganese(II) in acid.

    Model answer

    The half-equation is MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

  5. 5

    Show that the charges balance in the half-equation MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

    Model answer

    The left side is (−1) + (+8) + (−5) = +2 and the right side is +2, so the charges are equal.

  6. 6

    Why are exactly five electrons needed in the manganate(VII) half-equation?

    Model answer

    Manganese falls from oxidation state +7 to +2, a change of five, so five electrons are gained.

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A3 ⁴₂He → ¹²₆C.
B¹²₆C + ⁴₂He → ¹⁶₈O.
C¹⁶₈O + ⁴₂He → ²⁰₁₀Ne.

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Nitrogen is +2 in NO and +4 in NO₂.
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