Notes 13.2 - Half-Reactions and Balancing Equations

Half-Reactions and Balancing Equations

Aim

• What is a "half-reaction"? How can half-reactions be used to help balance an equation?

Do Now

• Is it possible to have oxidation without reduction? Reduction without oxidation?
• Attempt to balance the following equations:
  • $KIO<em>3 + H</em>2SO<em>3 \rightarrow KI + H</em>2SO_4$
  • $S \rightarrow SO_4$
  • $KOH + Cl<em>2 \rightarrow KClO</em>3 + KCl + H_2O$
  • $CrO<em>2 \rightarrow Cr</em>2O_7$

Half-Reactions

• Assign oxidation numbers to each element in the following reaction:
  • $KClO<em>3 \rightarrow KCl + O</em>2$
• Oxidation half-reaction:
  • $O^{-2} \rightarrow O_2^0$
• Reduction half-reaction:
  • $Cl^{+5} \rightarrow Cl^{-1}$
• How many electrons are "lost" during oxidation? 12
• How many electrons are "gained" during reduction? 6
• Is this a problem? What must be done?
  • Problem: We must have the same # of electrons lost as gained.
• Oxidation half-reaction (with electron adjustment):
  • $O^{-2} \times 2$
• Reduction half-reaction (with electron adjustment):
  • $Cl^{+5} + 6e^- \rightarrow Cl^{-1}$
• Use your coefficients from the "adjusted" half-reactions to deduce the coefficients of the overall equation:
  • $2KClO<em>3 \rightarrow 2KCl + 3O</em>2$

Balancing Equations Using the Half-Reaction Method

• Use the half-reaction method to balance the following equations:

  • $Sn + Ag^+ \rightarrow Sn^{2+} + Ag$

    • Ox: $Sn^0 \rightarrow Sn^{+2} + 2e^-$
    • Red: $(Ag^+ + 1e^-) \times 2 \rightarrow Ag^0$

  • $Cr + Pb^{2+} \rightarrow Cr^{3+} + Pb$

    • Ox: $Cr \rightarrow Cr^{+3} + 3e^-$
    • Red: $(Pb^{+2} + 2e^- \rightarrow Pb) \times 3$

  • $KIO<em>3 + H</em>2SO<em>3 \rightarrow KI + H</em>2SO_4$

    • $+5 \rightarrow -1$
    • Ox: $(S^{+4} \rightarrow S^{+6} + 2e^-) \times 3$
    • Red: $(I^{+5} + 6e^- \rightarrow I^{-1})$

  • $S \rightarrow SO_4$

  • $KOH + Cl<em>2 \rightarrow KClO</em>3 + KCl + H_2O$

    • $+0 \rightarrow +5$
    • Ox: $(Cl \rightarrow Cl^{+5} + 5e^-)$
    • Red: $(Cl + 1e^- \rightarrow Cl^-)$
    • Final Balanced Equation: $6KOH + 3Cl<em>2 \rightarrow KClO</em>3 + 5KCl + 3H_2O$

  • $NH<em>3 + O</em>2 \rightarrow NO + H_2O$

    • Ox: $(N^{-3} \rightarrow N^{+2} + 5e^-) \times 4$
    • Red: $(O_2 + 4e^- \rightarrow 2O^{-2}) \times 5$

  • $PbS + H<em>2O</em>2 \rightarrow PbSO<em>4 + H</em>2O$

    • Ox: $(S^{-2} \rightarrow S^{+6} + 8e^-) \times 1$
    • Red: $(O + 2e^- \rightarrow O^{-2}) \times 4$

  • $MnO<em>2 + H</em>2SO<em>4 \rightarrow MnSO</em>4 + H<em>2O + O</em>2$

    • Ox: $(2O \rightarrow O_2 + 2e^-)$
    • Red: $Mn \rightarrow Mn + 2e^- \rightarrow Mn2$

  • $KClO<em>3 + HCl \rightarrow Cl</em>2 + KCl + H_2O$

    • Ox: $(2Cl^- \rightarrow Cl_2 + 2e^-) \times 3$
    • Red: $(Cl^{+5} + 6e^- \rightarrow Cl^-)$

  • $CrO<em>2 \rightarrow Cr</em>2O_7$