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Recall the definitions of oxidation and reduction.

Chemistry

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  1. Recall the definitions of oxidation and reduction.

    • Oxidation is the loss of electrons, resulting in an increase in oxidation state.
    • Reduction is the gain of electrons, resulting in a decrease in oxidation state.
  2. Analyze each half-reaction to determine if electrons are gained or lost, and how the oxidation state changes.

    • (i) Fe3++3e−⟶Fe\text{Fe}^{3+} + 3\text{e}^- \longrightarrow \text{Fe}

      • Fe3+\text{Fe}^{3+} gains 3 electrons to become Fe\text{Fe}. The oxidation state changes from +3 to 0.
      • This is a gain of electrons, so it is reduction.
    • (ii) Cr⟶Cr3++3e−\text{Cr} \longrightarrow \text{Cr}^{3+} + 3\text{e}^-

      • Cr\text{Cr} loses 3 electrons to become Cr3+\text{Cr}^{3+}. The oxidation state changes from 0 to +3.
      • This is a loss of electrons, so it is oxidation.
    • (iii) MnO42−⟶MnO4−+e−\text{MnO}_4^{2-} \longrightarrow \text{MnO}_4^- + \text{e}^-

      • In MnO42−\text{MnO}_4^{2-}, the oxidation state of Mn is +6 (x+4(−2)=−2  ⟹  x=+6x + 4(-2) = -2 \implies x = +6).
      • In MnO4−\text{MnO}_4^-, the oxidation state of Mn is +7 (y+4(−2)=−1  ⟹  y=+7y + 4(-2) = -1 \implies y = +7).
      • MnO42−\text{MnO}_4^{2-} loses 1 electron to become MnO4−\text{MnO}_4^-. The oxidation state of Mn changes from +6 to +7.
      • This is a loss of electrons, so it is oxidation.
    • (iv) Li++e−⟶Li\text{Li}^+ + \text{e}^- \longrightarrow \text{Li}

      • Li+\text{Li}^+ gains 1 electron to become Li\text{Li}. The oxidation state changes from +1 to 0.
      • This is a gain of electrons, so it is reduction.

    Here are the classifications:

    • (i) Fe3++3e−⟶Fe\text{Fe}^{3+} + 3\text{e}^- \longrightarrow \text{Fe}: Reduction
    • (ii) Cr⟶Cr3++3e−\text{Cr} \longrightarrow \text{Cr}^{3+} + 3\text{e}^-: Oxidation
    • (iii) MnO42−⟶MnO4−+e−\text{MnO}_4^{2-} \longrightarrow \text{MnO}_4^- + \text{e}^-: Oxidation
    • (iv) Li++e−⟶Li\text{Li}^+ + \text{e}^- \longrightarrow \text{Li}: Reduction
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