Unit 6: Constructing Half-Equations

Learning to represent oxidation and reduction processes as balanced chemical equations.

6.4 Constructing Half-Equations (Methods 1 & 2)

A half-equation shows either the oxidation or the reduction part of a redox reaction, including the electrons lost or gained. For simple ions, this is straightforward (e.g., $Na \rightarrow Na^+ + e^-$). For more complex species, especially those involving oxygen in acidic solutions, a systematic approach is needed.

Method 1: Using Oxidation Numbers
  1. Balance the main atom: Balance the atom being oxidised or reduced.
  2. Add electrons: Calculate the change in oxidation number. Add the corresponding number of electrons ($e^-$) to the right for oxidation (loss) or to the left for reduction (gain).
  3. Balance oxygen: Add water molecules ($H_2O$) to the side that is deficient in oxygen.
  4. Balance hydrogen: Add hydrogen ions ($H^+$) to the side that is deficient in hydrogen.

Example: Convert permanganate ($MnO_4^-$) to manganese(II) ions ($Mn^{2+}$).

  1. Mn is already balanced.
  2. Oxidation number of Mn in $MnO_4^-$ is +7; in $Mn^{2+}$ it is +2. The change is -5 (a gain of 5 electrons). Add $5e^-$ to the left:
    $MnO_4^- + 5e^- \rightarrow Mn^{2+}$
  3. There are 4 oxygen atoms on the left. Add 4 $H_2O$ to the right:
    $MnO_4^- + 5e^- \rightarrow Mn^{2+} + 4H_2O$
  4. There are now 8 hydrogen atoms on the right. Add 8 $H^+$ to the left:
    $MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l)$
Method 2: Balancing Atoms and Charge (No Oxidation Numbers Needed)
  1. Balance the main atom: Balance the atom being oxidised or reduced.
  2. Balance oxygen: Add water molecules ($H_2O$) to balance the oxygen atoms.
  3. Balance hydrogen: Add hydrogen ions ($H^+$) to balance the hydrogen atoms.
  4. Balance charge: Add electrons ($e^-$) to the more positive side to make the total charge on both sides equal.

Example: Convert dichromate ($Cr_2O_7^{2-}$) to chromium(III) ions ($Cr^{3+}$).

  1. Balance Cr: Add a 2 in front of $Cr^{3+}$:
    $Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$
  2. Balance O: There are 7 oxygen atoms on the left. Add 7 $H_2O$ to the right:
    $Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$
  3. Balance H: There are 14 hydrogen atoms on the right. Add 14 $H^+$ to the left:
    $14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$
  4. Balance Charge:
    • Left side charge: 14(+1) + (-2) = +12
    • Right side charge: 2(+3) + 0 = +6
    The left side is more positive. Add 6 electrons ($e^-$) to the left to balance the charge (+12 - 6 = +6).
    $14H^+(aq) + Cr_2O_7^{2-}(aq) + 6e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$
Solved Examples:
  1. Write a half-equation for the oxidation of chloride ions ($Cl^-$) to chlorine gas ($Cl_2$).
    Solution: Balance Cl: $2Cl^- \rightarrow Cl_2$. Balance charge: add $2e^-$ to the right. $2Cl^- \rightarrow Cl_2 + 2e^-$.
  2. Write a half-equation for the reduction of sulphur dioxide ($SO_2$) to elemental sulphur (S) in acidic solution.
    Solution: Balance S (done). Balance O: $SO_2 \rightarrow S + 2H_2O$. Balance H: $SO_2 + 4H^+ \rightarrow S + 2H_2O$. Balance charge: left is +4, right is 0. Add $4e^-$ to the left. $SO_2 + 4H^+ + 4e^- \rightarrow S + 2H_2O$.
  3. Construct the half-equation for the oxidation of sulphite ions ($SO_3^{2-}$) to sulphate ions ($SO_4^{2-}$).
    Solution: Balance S (done). Balance O: $SO_3^{2-} + H_2O \rightarrow SO_4^{2-}$. Balance H: $SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+$. Balance charge: left is -2, right is 0. Add $2e^-$ to the right. $SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-$.
  4. Is the conversion of $NO_3^-$ to $NO$ an oxidation or a reduction? Write the half-equation.
    Solution: N goes from +5 to +2, so it is a reduction. Balance N (done). Balance O: $NO_3^- \rightarrow NO + 2H_2O$. Balance H: $NO_3^- + 4H^+ \rightarrow NO + 2H_2O$. Balance charge: left is +3, right is 0. Add $3e^-$ to the left. $NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O$.
  5. Write the half-equation for the oxidation of iron(II) ions to iron(III) ions.
    Solution: $Fe^{2+} \rightarrow Fe^{3+}$. Balance charge by adding one electron to the right: $Fe^{2+} \rightarrow Fe^{3+} + e^-$.
  6. Balance the half-equation: $H_2O_2 \rightarrow O_2$ in acidic solution.
    Solution: Balance O (done). Balance H: $H_2O_2 \rightarrow O_2 + 2H^+$. Balance charge: left is 0, right is +2. Add $2e^-$ to the right. $H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$.
  7. What is the purpose of adding $H^+$ and $H_2O$ when balancing half-equations?
    Solution: To balance the hydrogen and oxygen atoms that are often involved in the reaction, especially in aqueous acidic solutions.
  8. Write the half-equation for the reduction of chlorine gas ($Cl_2$) to chloride ions ($Cl^-$).
    Solution: Balance Cl: $Cl_2 \rightarrow 2Cl^-$. Balance charge: add $2e^-$ to the left. $Cl_2 + 2e^- \rightarrow 2Cl^-$.
  9. Construct the half-equation for the oxidation of ethanol ($CH_3CH_2OH$) to ethanoic acid ($CH_3COOH$).
    Solution: Balance C (done). Balance O: $CH_3CH_2OH + H_2O \rightarrow CH_3COOH$. Balance H: $CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+$. Balance charge: left is 0, right is +4. Add $4e^-$ to the right. $CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+ + 4e^-$.
  10. Balance the half-equation for the reduction of $IO_3^-$ to $I_2$.
    Solution: Balance I: $2IO_3^- \rightarrow I_2$. Balance O: $2IO_3^- \rightarrow I_2 + 6H_2O$. Balance H: $2IO_3^- + 12H^+ \rightarrow I_2 + 6H_2O$. Balance charge: left is +10, right is 0. Add $10e^-$ to the left. $2IO_3^- + 12H^+ + 10e^- \rightarrow I_2 + 6H_2O$.

Knowledge Check (20 Questions)

Answer: Either the oxidation or reduction part of a redox reaction, showing the electrons transferred.

Answer: On the right side (products), as they are lost.

Answer: Water ($H_2O$).

Answer: Two electrons on the right side ($Zn \rightarrow Zn^{2+} + 2e^-$).

Answer: Hydrogen ions ($H^+$).

Answer: Reactants (on the left side).

Answer: 5 electrons.

Answer: Balancing the total charge by adding electrons.

Answer: $I_2 \rightarrow 2I^-$.

Answer: Yes, all atoms and the charge are balanced.

Answer: 14 $H^+$ on the left side.

Answer: Oxidation (N goes from +4 to +5).

Answer: 0 (assuming one H+ for now: -1 + 1 = 0). The final balanced equation will have a different charge.

Answer: $Ag^+ + e^- \rightarrow Ag$.

Answer: Three $H_2O$ molecules on the right side.

Answer: Two electrons on the left side ($I_2 + 2e^- \rightarrow 2I^-$).

Answer: Balance the main element (Chromium in this case).

Answer: False. There are also methods for balancing in basic/alkaline solution (which typically involve adding $OH^-$).

Answer: 8 electrons. ($S_2O_3^{2-} + 5H_2O \rightarrow 2SO_4^{2-} + 10H^+ + 8e^-$).

Answer: The number of atoms of each element AND the total electrical charge.
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