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:

Oxidation-Reduction Reactions

AP Chemistry-Concise Summary Notes- All Topics

4.9.A.1 Constructing Balanced Chemical Equations from Half-Reactions:

1. Redox Reactions Overview:

Redox reactions, which stand for reduction-oxidation reactions, involve the transfer of electrons between different substances.

Oxidation is the process of losing electrons, leading to an increase in oxidation state.
Reduction involves gaining electrons, which results in a decrease in oxidation state.

Identifying Oxidation States:
a. The oxidation state of pure elements is 0.
b. For monatomic ions, the oxidation state corresponds to their charge (e.g., Na⁺ = +1, Cl⁻ = -1).
c. Oxygen usually has an oxidation state of -2, except in peroxides where it is -1.
d. Hydrogen has an oxidation state of +1 when bonded to nonmetals and -1 when bonded to metals.
e. In a neutral compound, the sum of oxidation states equals 0; for a polyatomic ion, it matches the ion’s charge.

Example:
Take a look at the reaction:

Hydrogen is oxidized from 0 to +1.
Oxygen is reduced from 0 to -2.

In redox reactions, oxidation and reduction occur simultaneously.

2. Writing and Balancing Half-Reactions:

i. Balancing Half-Reactions:

a. Write Half-Reactions:
Oxidation involves the loss of electrons.
Reduction involves the gain of electrons.

b. Balance Atoms (excluding O and H).

c. Balance Oxygen:
Introduce H₂O molecules to achieve oxygen balance.

d. Balance Hydrogen:
Use H⁺ ions (for acidic solutions) or OH⁻ (for basic solutions) to balance hydrogen.

e. Balance Charges:
Add **electrons (e⁻)** to ensure charge balance.

f. Combine Half-Reactions:
Adjust coefficients to equalize electrons, then sum the half-reactions.

Example:

Oxidation (Mg):

M g \to M g^{2 +} + 2 e^{-}

Reduction (O₂):

O_{2} + 4 e^{-} \to 2 O^{2 -}

Multiply oxidation by 2:

2 M g \to 2 M g^{2 +} + 4 e^{-}

Combine and cancel electrons:

2 M g + O_{2} \to 2 M g O

Balanced Redox Reaction:

2 M g + O_{2} \to 2 M g O

3. Combining Half-Reactions:

i. Combining Half-Reactions:

a. Equalize Electrons:
Adjust each half-reaction by multiplying it by a factor so that the number of electrons in both the oxidation and reduction half-reactions is equal.

b. Add Half-Reactions:
Merge the two half-reactions, eliminating the electrons.

c. Balance Atoms and Charges:
– Confirm that both atoms and charges are balanced on each side of the equation.

Example:

Step 1: Write Half-Reactions

Oxidation (Magnesium):
$M g \to M g^{2 +} + 2 e^{-}$
Reduction (Oxygen):
$O_{2} + 4 e^{-} \to 2 O^{2 -}$

Step 2: Equalize Electrons

Multiply the oxidation half-reaction by 2 to match the 4 electrons from the reduction half-reaction:
$2 M g \to 2 M g^{2 +} + 4 e^{-}$

Step 3: Combine Half-Reactions

Add the two half-reactions:
$2 M g + O_{2} + 4 e^{-} \to 2 M g^{2 +} + 4 e^{-} + 2 O^{2 -}$

Step 4: Cancel Electrons

The electrons cancel out:
$2 M g + O_{2} \to 2 M g O$

Final Balanced Equation:

2 M g + O_{2} \to 2 M g O

This is the complete balanced redox equation.

4. Balancing in Acidic and Basic Solutions:

The presence of H⁺ (acidic) and OH⁻ (basic) ions causes a little difference in the process of balancing redox reactions between acidic and basic solutions.

i. Acidic Solution Procedures:

a.Write Half-Reactions: – Distinguish between the half-reactions for oxidation and reduction.

Balance Atoms (apart from hydrogen and oxygen).

b. Balance Oxygen: To balance the oxygen atoms, add H₂O molecules. as for acidic solutions. Balance Hydrogen: To balance hydrogen atoms, add H⁺ ions.

ic. Balance Charges: To balance the charges, add electrons (e⁻).

c. Combine Half-Reactions: – To equalize the electrons, multiply as needed.
The electrons are cancelled out when the half-reactions are added together.

Basic Solutions Steps:

After balancing, add OH⁻ ions to both sides to neutralize, but otherwise, follow the same steps as for acidic solutions.

Balancing Redox Mechanisms in Acidic and Basic Substances

The method of balancing redox reactions between acidic and basic solutions is slightly different when H⁺ (acidic) and OH⁻ (basic) ions are present.

Methods for Acidic Solutions:

Write Half-Reactions: – Write the half-reactions for reduction and oxidation separately.

Balance Atoms — except for oxygen and hydrogen.

Balance Oxygen: Add H₂O molecules to bring the oxygen atoms into balance.

4. Balance Hydrogen: Add H⁺ ions to balance hydrogen atoms.

Balance Charges: Add **electrons (e⁻) to balance the charges.

The sixth step is to combine half-reactions: When necessary, multiply to equalize the electrons.
When the half-reactions are put together, the electrons are cancelled out.

Steps for Basic Solutions:

Add OH⁻ ions to both sides after balancing to neutralize; otherwise, follow the same steps.

Balancing Redox Mechanisms in Acidic and Basic Substances

The method of balancing redox reactions between acidic and basic solutions is slightly different when H⁺ (acidic) and OH⁻ (basic) ions are present.

i. Methods for Acidic Solutions:

1. Write Half-Reactions: – Write the half-reactions for reduction and oxidation separately.

Balance Atoms — except for oxygen and hydrogen.

3. Balance Oxygen: Add H₂O molecules to bring the oxygen atoms into balance.

4. Balance Hydrogen: Add H⁺ ions to balance hydrogen atoms.

5. Balance Charges: Add electrons (e⁻) to balance the charges.

The sixth step is to combine half-reactions: When necessary, multiply to equalize the electrons.
When the half-reactions are put together, the electrons are cancelled out.

Steps for Basic Solutions:

Add OH⁻ ions to both sides after balancing to neutralize; otherwise, follow the same steps.as for acidic solutions.

OLD Content

Oxidation-Reduction Rates

Know that it is a redox reaction by a change in oxidation states

Oxidation States

Oxidation Numbers: signifies the number of charges an atom would have in a molecule or ionic compound if electrons were completely transferred
Know which atom has been reduced/oxidized based on the changes in their oxidation states
- Reduced = charge/oxidation state decreases (gains electrons); Oxidizing agent: electron acceptor
- Oxidized = charge/oxidation state increases (loses electrons); Reducing agent: electron donor

- Na is oxidized = reducing agent; chlorine is reduced = oxidizing agent
Note: actual charges are written n+ or n-; oxidation states are written +n or –n

Oxidation State Rules

Any element by itself: 0
Monatomic ion = charge of ion
Oxygen is usually -2 in its compounds

- Exception: peroxide (O₂²⁻) which is -1

Hydrogen: +1
Fluorine and the rest of the halogens are -1 (most of the time)
Sulfur in SO4: +6

Sum of oxidation states = 0 in compounds; sum of oxidation states = charge of the ion
When don’t have rule for one of atoms/polyatomic ions, use the atom that does have a rule to find out oxidation state
- Ex:

Balancing Oxidation-Reduction Equations

Using Oxidation Numbers

Assign oxidation numbers to each atom
Determine which atoms are being reduced and oxidized
Write each half-reactions
Balance elements
For each half-reaction, balance charge using electrons

- Electrons must be on opposite sides and MUST have the same coefficients

If necessary, multiply by integer to equalize electron count
Add up half-reactions and write overall equation
Balance remaining elements/compounds

Redox Reactions in Acidic Solutions vs Basic Solutions

Acidic Solutions

Reaction involves H+ ions → For each half-reaction
- Balance all elements except H and O
- Balance oxygen using H2O
- Balance H using H+
- Balance the charge using electrons

Basic Solutions

Reaction involves OH- ions → For each half-reaction
- Use the above method to obtain the final balanced equation as if H+ were present
- Add a number of OH- that is equal to the number of H+ ions to both sides of the equation
  - We want to eliminate H+ by forming H2O
- Eliminate the number of H2O molecules that appear on both sides of the equation
- Check that the elements and charges are balanced

AP Chemistry 4.9 Oxidation-Reduction Rates Study Notes

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