Redox reactions and oxidation numbers (KCSE Chemistry Form 3)

TL;DR

Redox reactions involve the transfer of electrons, where one substance is oxidized (loses electrons) and another is reduced (gains electrons). Oxidation numbers help us track these electron transfers and identify what's being oxidized or reduced. Understanding these concepts is crucial for explaining many chemical processes.

1. The Mental Model

Think of redox reactions like a chemical "give and take" of electrons. One atom gives electrons away, becoming more positive, while another atom takes those electrons, becoming more negative. Oxidation numbers are like a score-keeping system for this electron exchange.

2. The Core Material

What is a Redox Reaction?

"Redox" is short for Reduction-Oxidation. These are reactions where there's a change in the oxidation state (or oxidation number) of atoms. This change happens because electrons are transferred between reactants.

• Oxidation: This is the loss of electrons. When an atom loses electrons, its oxidation number increases. A good way to remember this is OIL RIG: Oxidation Is Loss.
• Reduction: This is the gain of electrons. When an atom gains electrons, its oxidation number decreases. Remember: Reduction Is Gain.

It's important to remember that oxidation and reduction always happen together. You can't have one without the other! The substance that gets oxidized is called the reducing agent (because it causes the other substance to be reduced). The substance that gets reduced is called the oxidizing agent (because it causes the other substance to be oxidized).

What are Oxidation Numbers?

An oxidation number (or oxidation state) is a hypothetical charge an atom would have if all bonds were purely ionic. It's a way to keep track of electron distribution in compounds and identify electron transfer in redox reactions.

Rules for Assigning Oxidation Numbers

These rules are super important and you need to know them well. Always apply them in order from 1 to 7.

1. Elements: The oxidation number of an atom in its elemental form (e.g., Fe, O₂, Cl₂) is 0.
2. Monatomic Ions: The oxidation number of a monatomic ion (e.g., Na⁺, Cl⁻, Mg²⁺) is equal to its charge.
3. Group 1 Metals: In compounds, Group 1 metals (Li, Na, K, etc.) always have an oxidation number of +1.
4. Group 2 Metals: In compounds, Group 2 metals (Mg, Ca, Ba, etc.) always have an oxidation number of +2.
5. Hydrogen: In most compounds, hydrogen has an oxidation number of +1. The exception is in metal hydrides (e.g., NaH), where it's -1.
6. Oxygen: In most compounds, oxygen has an oxidation number of -2. Exceptions include peroxides (e.g., H₂O₂, where it's -1) and when bonded to fluorine (e.g., OF₂, where it's +2).
7. Halogens: In most compounds, halogens (F, Cl, Br, I) have an oxidation number of -1. Fluorine always has -1. Other halogens can have positive oxidation numbers when bonded to more electronegative elements (like oxygen).
8. Sum of Oxidation Numbers:
   • For a neutral compound, the sum of the oxidation numbers of all atoms is 0.
   • For a polyatomic ion, the sum of the oxidation numbers of all atoms is equal to the charge of the ion.

Identifying Redox Reactions Using Oxidation Numbers

Here's how you can use oxidation numbers to identify what's oxidized and reduced:

graph TD
    A[Start: Write balanced chemical equation] --> B{Assign oxidation numbers to ALL atoms in reactants and products};
    B --> C{Compare oxidation numbers of each element from reactants to products};
    C --> D{Did an element's oxidation number increase?};
    D -- Yes --> E[That element was OXIDIZED];
    D -- No --> F{Did an element's oxidation number decrease?};
    F -- Yes --> G[That element was REDUCED];
    F -- No --> H[No change in oxidation numbers for any element];
    H --> I[This is NOT a redox reaction];
    E --> J[Identify Oxidizing and Reducing Agents];
    G --> J;
    J --> K[End];

Example:
Consider the reaction: 2Na(s) + Cl₂(g) → 2NaCl(s)

1. Na: In Na(s), its oxidation number is 0 (elemental form). In NaCl, it's Na⁺, so its oxidation number is +1.
   • Oxidation number increased from 0 to +1. So, Na is oxidized.
2. Cl: In Cl₂(g), its oxidation number is 0 (elemental form). In NaCl, it's Cl⁻, so its oxidation number is -1.
   • Oxidation number decreased from 0 to -1. So, Cl is reduced.

Since Na is oxidized, it is the reducing agent. Since Cl₂ is reduced, it is the oxidizing agent.

3. Worked Example

Problem: For the reaction CuO(s) + H₂(g) → Cu(s) + H₂O(l), identify the substance oxidized, the substance reduced, the oxidizing agent, and the reducing agent.

Solution:

1. Assign oxidation numbers to all atoms in the reactants:

   • In CuO: Oxygen is usually -2. Since the compound is neutral, Cu must be +2 (to balance -2).
   • In H₂: This is an elemental form, so H is 0.

2. Assign oxidation numbers to all atoms in the products:

   • In Cu: This is an elemental form, so Cu is 0.
   • In H₂O: Oxygen is -2. Hydrogen is usually +1. (2 x +1) + (-2) = 0, so it balances.

3. Compare oxidation numbers from reactants to products:

   • Cu: Changed from +2 (in CuO) to 0 (in Cu).
     • The oxidation number decreased. Therefore, Cu is reduced.
   • H: Changed from 0 (in H₂) to +1 (in H₂O).
     • The oxidation number increased. Therefore, H is oxidized.
   • O: Remained -2 (in CuO and H₂O). Oxygen was neither oxidized nor reduced.

4. Identify agents:

   • Since Cu (in CuO) was reduced, CuO is the oxidizing agent.
   • Since H (in H₂) was oxidized, H₂ is the reducing agent.

4. Key Takeaways

• Oxidation is the loss of electrons, resulting in an increase in oxidation number.
• Reduction is the gain of electrons, resulting in a decrease in oxidation number.
• Oxidation and reduction always occur simultaneously in a redox reaction.
• The substance that gets oxidized is the reducing agent; the substance that gets reduced is the oxidizing agent.
• Oxidation numbers are a tool to track electron transfer and identify redox processes.
• Always follow the rules for assigning oxidation numbers in the correct order.

Common Mistakes to Avoid:
- Confusing oxidation (loss) with reduction (gain) – remember OIL RIG!
- Forgetting that elemental forms have an oxidation number of 0.
- Incorrectly calculating the sum of oxidation numbers for polyatomic ions (it must equal the ion's charge).
- Assuming oxygen is always -2 or hydrogen is always +1 (remember the exceptions!).

5. Now Try It

Consider the reaction: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

Your task is to:
1. Assign oxidation numbers to all atoms in all reactants and products.
2. Identify which substance is oxidized.
3. Identify which substance is reduced.
4. Identify the oxidizing agent.
5. Identify the reducing agent.

You should be able to clearly show the change in oxidation numbers for the elements that are oxidized and reduced. Success looks like correctly identifying all five points, with clear reasoning based on oxidation number changes.