2.1 Redox Reactions and Half-Cell Reactions

Redox reactions involve electron transfer between species. Oxidation means losing electrons, while reduction means gaining them. These processes are key to understanding how chemicals interact and change during electrochemical reactions.

Half-cell reactions break down redox reactions into separate oxidation and reduction steps. Balancing these reactions is crucial for analyzing electrochemical systems and predicting the overall reaction outcomes in various applications.

Redox Reactions

Oxidation and reduction definitions

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• Oxidation
  • Involves the loss of electrons from a species
  • Results in an increase in the oxidation state of the species ($Fe^{2+} \rightarrow Fe^{3+}$)
  • The species undergoing oxidation acts as the reducing agent by donating electrons
• Reduction
  • Involves the gain of electrons by a species
  • Results in a decrease in the oxidation state of the species ($Cl_2 \rightarrow 2Cl^-$)
  • The species undergoing reduction acts as the oxidizing agent by accepting electrons

Oxidizing and reducing agents

• Oxidizing agent
  • Accepts electrons from another species causing its oxidation
  • Is reduced in the process as it gains electrons ($MnO_4^- \rightarrow Mn^{2+}$)
  • Examples include $O_2$, $Cl_2$, and $MnO_4^-$
• Reducing agent
  • Donates electrons to another species causing its reduction
  • Is oxidized in the process as it loses electrons ($2I^- \rightarrow I_2$)
  • Examples include $H_2$, $Na$, and $Fe^{2+}$

Half-Cell Reactions

Balancing half-cell reactions

• Oxidation half-reaction
  • Represents the loss of electrons by a species
  • The reactant loses electrons to form the oxidized product
  • Electrons appear as products on the right side ($Mg \rightarrow Mg^{2+} + 2e^-$)
• Reduction half-reaction
  • Represents the gain of electrons by a species
  • The reactant gains electrons to form the reduced product
  • Electrons appear as reactants on the left side ($Ag^+ + e^- \rightarrow Ag$)
• Steps for balancing half-cell reactions in acidic solution:
  1. Balance all atoms except H and O ($Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$)
  2. Balance O atoms by adding $H_2O$ ($Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O$)
  3. Balance H atoms by adding $H^+$ ($Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$)
  4. Balance charge by adding electrons ($MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$)

Combining half-cell reactions

• Identify the oxidation and reduction half-reactions
• Multiply each half-reaction by a factor to equalize the electrons transferred
• Add the half-reactions together and cancel out the electrons
• Simplify the overall reaction by canceling common terms on both sides
• Check that the final equation is balanced in terms of atoms and charge
• Example:
  • Oxidation: $2Al \rightarrow 2Al^{3+} + 6e^-$
  • Reduction: $3Cu^{2+} + 6e^- \rightarrow 3Cu$
  • Overall: $2Al + 3Cu^{2+} \rightarrow 2Al^{3+} + 3Cu$