Oxidation and reduction are key processes in electrochemistry. They involve the transfer of electrons between species, changing their oxidation states. Understanding these concepts is crucial for grasping how batteries work and how chemical reactions occur.

Oxidizing and reducing agents play vital roles in these electron transfers. Oxidizing agents accept electrons, while reducing agents donate them. Knowing how to identify and balance these reactions is essential for predicting and controlling chemical processes in various applications.

Oxidation and Reduction Fundamentals

Oxidation and reduction definitions

• Oxidation involves the loss of electrons from a species resulting in an increase in its oxidation number
  • Occurs at the anode in an electrochemical cell (battery)
  • Examples: $Fe^{2+} \rightarrow Fe^{3+} + e^-$, $2Cl^- \rightarrow Cl_2 + 2e^-$
• Reduction involves the gain of electrons by a species resulting in a decrease in its oxidation number
  • Occurs at the cathode in an electrochemical cell
  • Examples: $Cu^{2+} + 2e^- \rightarrow Cu$, $2H^+ + 2e^- \rightarrow H_2$
• Mnemonic device OIL RIG helps remember Oxidation Is Loss of electrons and Reduction Is Gain of electrons

Oxidizing and reducing agents

• Oxidizing agent (oxidant) is a species that accepts electrons causing the oxidation of another species and is reduced in the process
  • Examples: $H_2O_2$ in the reaction $2Fe^{2+} + H_2O_2 + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O$, $MnO_4^-$ in the reaction $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$
• Reducing agent (reductant) is a species that donates electrons causing the reduction of another species and is oxidized in the process
  • Examples: $Na$ in the reaction $2Na + Cl_2 \rightarrow 2NaCl$, $Sn^{2+}$ in the reaction $Sn^{2+} + 2Fe^{3+} \rightarrow Sn^{4+} + 2Fe^{2+}$

Oxidation numbers in compounds

• Rules for assigning oxidation numbers to elements in compounds:
  1. Free elements have an oxidation number of 0 ($Na$, $H_2$, $O_2$)
  2. Monatomic ions have an oxidation number equal to their charge ($Na^+$ is +1, $Cl^-$ is -1)
  3. Hydrogen has an oxidation number of +1 except in metal hydrides where it is -1 ($NaH$)
  4. Oxygen has an oxidation number of -2 except in peroxides where it is -1 ($H_2O_2$) and in compounds with fluorine ($OF_2$)
  5. Fluorine always has an oxidation number of -1 ($HF$, $SF_6$)
  6. In neutral compounds, the sum of oxidation numbers must equal 0 ($KMnO_4$: +1 + +7 + 4(-2) = 0)
  7. In polyatomic ions, the sum of oxidation numbers must equal the charge of the ion ($SO_4^{2-}$: +6 + 4(-2) = -2)

Balancing half-reactions

• Half-reactions separate oxidation and reduction processes
  • Oxidation half-reaction shows the species losing electrons ($Zn \rightarrow Zn^{2+} + 2e^-$)
  • Reduction half-reaction shows the species gaining electrons ($Cu^{2+} + 2e^- \rightarrow Cu$)
• Steps to balance half-reactions:
  1. Balance atoms other than H and O ($Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$)
  2. Balance oxygen atoms by adding $H_2O$ ($Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O$)
  3. Balance hydrogen atoms by adding $H^+$ ions (already balanced in the previous step)
  4. Balance charge by adding electrons ($Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$)
• Combining half-reactions:
  1. Multiply half-reactions to equalize electrons transferred
  2. Add half-reactions together
  3. Cancel out common terms
  4. Verify that atoms and charges are balanced