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AP Chemistry 4.9 Oxidation-Reduction Rates Study Notes - New Syllabus Effective fall 2024

AP Chemistry 4.9 Oxidation-Reduction Rates Study Notes- New syllabus

AP Chemistry 4.9 Oxidation-Reduction Rates Study Notes – AP Chemistry –  per latest AP Chemistry Syllabus.

LEARNING OBJECTIVE

Represent a balanced redox reaction equation using half-reactions.

Key Concepts: 

  • Representing Redox Reactions Using Half-Reactions

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AP Chemistry Exam Questions MCQs and FRQs

Representing Redox Reactions Using Half-Reactions

A redox (oxidation–reduction) reaction involves the transfer of electrons between chemical species. Balanced redox equations can be systematically constructed using half-reactions, which separately represent the oxidation (loss of electrons) and reduction (gain of electrons) processes. The overall balanced equation is obtained by combining the two half-reactions so that the total number of electrons lost equals the number of electrons gained.

1. Key Concepts of Redox Processes:

  • Oxidation: Loss of electrons; increase in oxidation number.
  • Reduction: Gain of electrons; decrease in oxidation number.
  • Oxidizing agent: The species that is reduced (gains electrons).
  • Reducing agent: The species that is oxidized (loses electrons).

Example Concept: In the reaction \(\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}\):

  • Zn → Zn²⁺ + 2e⁻ (oxidation)
  • Cu²⁺ + 2e⁻ → Cu (reduction)

Net Reaction: \(\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}\)

2. Purpose of Using Half-Reactions:

  • Clarifies electron transfer explicitly.
  • Ensures conservation of mass and charge.
  • Facilitates systematic balancing of complex redox reactions in acidic or basic media.

3. Rules for Balancing Redox Equations Using Half-Reactions:

(Applicable for reactions in acidic or basic solutions)

  1. Write separate half-reactions for oxidation and reduction.
  2. Balance all elements except hydrogen and oxygen.
  3. Balance oxygen atoms by adding \(\mathrm{H_2O}\).
  4. Balance hydrogen atoms by adding \(\mathrm{H^+}\) (for acidic medium) or \(\mathrm{OH^-}\) (for basic medium).
  5. Balance charge by adding electrons (\(\mathrm{e^-}\)) to the more positive side.
  6. Equalize the number of electrons lost and gained between the two half-reactions.
  7. Add the half-reactions together and cancel identical species on both sides.

Example 

Balance the following redox reaction in acidic solution: \(\mathrm{MnO_4^-(aq) + Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + Fe^{3+}(aq)}\)

▶️Answer / Explanation

Step 1: Write the two half-reactions:

Oxidation: \(\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}\)
Reduction: \(\mathrm{MnO_4^- \rightarrow Mn^{2+}}\)

Step 2: Balance oxygen by adding water:

\(\mathrm{MnO_4^- \rightarrow Mn^{2+} + 4H_2O}\)

Step 3: Balance hydrogen by adding \(\mathrm{H^+}\):

\(\mathrm{8H^+ + MnO_4^- \rightarrow Mn^{2+} + 4H_2O}\)

Step 4: Balance charge by adding electrons:

\(\mathrm{8H^+ + MnO_4^- + 5e^- \rightarrow Mn^{2+} + 4H_2O}\)

Step 5: Multiply the oxidation half-reaction by 5 to equalize electrons:

\(\mathrm{5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-}\)

Step 6: Add the two half-reactions and cancel electrons:

\(\mathrm{8H^+ + MnO_4^- + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}}\)

Step 7: Verify atom and charge balance:

  • Atoms: Mn, Fe, H, and O all balanced.
  • Charge: +13 on both sides 

Final Balanced Equation: \(\mathrm{MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}\)

Example

Balance the following reaction in basic solution: \(\mathrm{Cr(OH)_3 + ClO^- \rightarrow CrO_4^{2-} + Cl^-}\)

▶️Answer / Explanation

Step 1: Write separate half-reactions:

Oxidation: \(\mathrm{Cr(OH)_3 \rightarrow CrO_4^{2-}}\)
Reduction: \(\mathrm{ClO^- \rightarrow Cl^-}\)

Step 2: Balance O and H using water and hydroxide.

For oxidation half-reaction (add \(\mathrm{H_2O}\) and \(\mathrm{OH^-}\)): \(\mathrm{Cr(OH)_3 + 5OH^- \rightarrow CrO_4^{2-} + 4H_2O + 3e^-}\)

For reduction half-reaction: \(\mathrm{ClO^- + H_2O + 2e^- \rightarrow Cl^- + 2OH^-}\)

Step 3: Equalize electrons (LCM of 3 and 2 = 6):

Multiply first half by 2, second by 3.

Step 4: Add both half-reactions and cancel identical species.

\(\mathrm{2Cr(OH)_3 + 3ClO^- + 4OH^- \rightarrow 2CrO_4^{2-} + 3Cl^- + 5H_2O}\)

Final Balanced Equation: \(\mathrm{2Cr(OH)_3 + 3ClO^- + 4OH^- \rightarrow 2CrO_4^{2-} + 3Cl^- + 5H_2O}\)

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