Welcome to the World of Redox Reactions!

Hello future chemist! This chapter is one of the most fundamental and important concepts in all of chemistry. Redox reactions are everywhere—they power your phone battery, they cause iron to rust, and they help your body process food.

We are moving beyond simple electron transfer definitions (like in ionic bonding) to a powerful system called Oxidation Numbers. Mastering this will allow you to analyze *any* chemical reaction, even those involving complex covalent molecules, and identify exactly what is happening. Don't worry if assigning these numbers seems like learning a new language at first; with practice, it becomes second nature!

Section 1: The Basics of Electron Transfer

1.1 Defining Oxidation and Reduction (The Electron View)

When you first learned about ionic reactions, you learned that atoms gain or lose electrons. Redox reactions are simply reactions where this electron movement takes place.

We use a famous mnemonic to remember which process involves losing electrons and which involves gaining them:

🧠 Memory Aid: OIL RIG

Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)

  • Oxidation: The loss of one or more electrons.
    Example: \(\text{Mg} \rightarrow \text{Mg}^{2+} + 2\text{e}^-\)
  • Reduction: The gain of one or more electrons.
    Example: \(\text{Cl}_2 + 2\text{e}^- \rightarrow 2\text{Cl}^-\)

Important Note: Oxidation and Reduction always occur together! If one substance loses electrons (is oxidized), another substance must gain those electrons (be reduced). This is why we call them Redox Processes (Reduction + Oxidation).

Key Takeaway: Redox involves electron movement. Oxidation = loss (OIL). Reduction = gain (RIG).

Section 2: Oxidation Numbers (Oxidation State)

2.1 Why We Need Oxidation Numbers (ON)

For simple ionic compounds, it's easy to track electrons. But what about covalent compounds like \(\text{SO}_2\) or \(\text{HNO}_3\)? Electrons are shared, not fully transferred.

The Oxidation Number (or Oxidation State) is a theoretical charge assigned to an atom in a molecule or ion, assuming all bonds are 100% ionic (meaning electrons are fully transferred to the more electronegative atom). This system allows us to track electron movement even in covalent reactions.

2.2 Rules for Assigning Oxidation Numbers (ON)

To calculate ON, follow these rules in order of priority. (It’s like an electronic game of chess—some elements have higher priority than others!)

  1. Elements: The ON of any uncombined element is zero (0).
    Example: \(\text{O}_2\), \(\text{Na}\), \(\text{Fe}\). ON for all atoms = 0.
  2. Ions: For a simple ion (like \(\text{Mg}^{2+}\)), the ON is equal to the charge of the ion.
    Example: \(\text{Fe}^{3+}\) has an ON of +3.
  3. Group 1 Metals: Always +1 (e.g., \(\text{Na}, \text{K}\)).
  4. Group 2 Metals: Always +2 (e.g., \(\text{Mg}, \text{Ca}\)).
  5. Fluorine (F): Always –1 (it is the most electronegative element).
  6. Hydrogen (H): Usually +1, except in metal hydrides (like \(\text{NaH}\)), where it is –1.
  7. Oxygen (O): Usually –2, except in peroxides (like \(\text{H}_2\text{O}_2\)), where it is –1, and when bonded to Fluorine (\(\text{OF}_2\)), where it is +2.
  8. Halogens (Cl, Br, I): Usually –1, unless they are bonded to a more electronegative atom (like Oxygen or Fluorine).
Rule 9: The Summation Rule

The sum of all oxidation numbers in a:
(a) Neutral Compound must equal zero (0).
(b) Polyatomic Ion must equal the charge of the ion.

2.3 Step-by-Step Calculation Example

Let's calculate the oxidation number of Sulfur (\(\text{S}\)) in the sulfate ion, \(\text{SO}_4^{2-}\).

  1. Identify the Target: We want the ON of S (let's call it \(x\)).
  2. Assign Known ONs: Oxygen is usually –2 (Rule 7).
  3. Apply the Summation Rule: The total charge of the ion is –2 (Rule 9b).
    \((\text{ON of S}) + 4 \times (\text{ON of O}) = -2\)
  4. Calculate:
    \(x + 4(-2) = -2\)
    \(x - 8 = -2\)
    \(x = +6\)

The oxidation number of sulfur in \(\text{SO}_4^{2-}\) is +6.

Using Roman Numerals

We often use Roman numerals to show the oxidation number of an element, especially metals that have variable oxidation states (like transition metals).
Example: \(\text{Iron(III) chloride}\) means iron has an ON of +3 (\(\text{FeCl}_3\)).
Example: Copper(I) oxide (\(\text{Cu}_2\text{O}\)) contains copper with an ON of +1.

Quick Review: Oxidation Number Checklist

When calculating ON, always prioritize:
1. Free Element (0)
2. Group 1/2 (+1 / +2)
3. Fluorine (–1)
4. Hydrogen (+1, unless hydride)
5. Oxygen (–2, unless peroxide)

Key Takeaway: Oxidation numbers follow strict rules based on charge and electronegativity. The ON helps track the "theoretical charge" of an atom.

Section 3: Defining Redox by Oxidation Number Change

This is the modern, more reliable way to define redox. It works for all reactions, ionic or covalent.

3.1 The Definitions (ON View)

  • Oxidation: An increase in oxidation number (ON).
  • Reduction: A decrease in oxidation number (ON).
Analogy: The Chemist’s Scoreboard

Imagine the ON is a scoreboard.

  • If the score goes Up (e.g., from +2 to +4), the atom has been Oxidized. (The chemical feels positive about its score!)
  • If the score goes Down (e.g., from +7 to +2), the atom has been Reduced.

3.2 Combining Definitions: Electron vs. ON Change

Don't forget the two definitions are linked!

  • Oxidation: Loss of electrons \((e^-)\) causes ON to become more positive (increase).
  • Reduction: Gain of electrons \((e^-)\) causes ON to become more negative (decrease).

Example: Reaction between Iron(II) and Chlorine

\(\text{2Fe}^{2+}(\text{aq}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{Cl}^-(\text{aq})\)


Iron: ON changes from +2 to +3. (Increase). Oxidized.
Chlorine: ON changes from 0 (\(\text{Cl}_2\)) to –1. (Decrease). Reduced.

Key Takeaway: Oxidation is an ON increase. Reduction is an ON decrease. Track the ON before and after the reaction to identify redox.

Section 4: Oxidising Agents and Reducing Agents

This is where terminology can be confusing, but the logic is simple: the agent is the species that causes the change in the *other* substance.

4.1 Defining the Agents

  • Oxidising Agent (OA):
    This substance causes oxidation in another species. To do this, the OA must take electrons. Therefore, the Oxidising Agent itself gets Reduced.
  • Reducing Agent (RA):
    This substance causes reduction in another species. To do this, the RA must give away electrons. Therefore, the Reducing Agent itself gets Oxidized.
🧠 Memory Aid: The "Agent" is Self-Sacrificing

An OA is a bad guy: it reduces *itself* so it can oxidize others.
A good OA wants electrons (it is easily reduced).
A good RA gives electrons easily (it is easily oxidized).

4.2 Identifying Agents in Practice

Let’s look at the reaction of zinc with copper ions:

\(\text{Zn}(\text{s}) + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu}(\text{s})\)

  • Zinc (\(\text{Zn}\)): ON goes from 0 to +2. (Oxidized). Since \(\text{Zn}\) was oxidized, it must be the Reducing Agent (RA).
  • Copper (\(\text{Cu}^{2+}\)): ON goes from +2 to 0. (Reduced). Since \(\text{Cu}^{2+}\) was reduced, it must be the Oxidising Agent (OA).
Did You Know?

Common industrial Oxidizing Agents often contain elements in their highest possible oxidation state, making them desperate to be reduced (e.g., \(\text{KMnO}_4\) (Mn is +7) or concentrated nitric acid (\(\text{HNO}_3\)) (N is +5)).

Key Takeaway: The OA is reduced, and the RA is oxidized. The reaction is defined by the change in the ON of the species involved.

Section 5: Disproportionation Reactions

5.1 Definition and Characteristics

Sometimes, a single element can play both roles in a reaction—it is oxidized and reduced simultaneously.

A Disproportionation Reaction is a reaction in which an element in a specific oxidation state is simultaneously oxidized to a higher oxidation state and reduced to a lower oxidation state.

5.2 Example: Chlorine in Cold Aqueous NaOH

A classic syllabus example is the reaction of chlorine gas with cold, dilute aqueous sodium hydroxide:

\(\text{Cl}_2(\text{g}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{NaOCl}(\text{aq}) + \text{H}_2\text{O}(\text{l})\)

Let's track the ON of Chlorine:

  1. Reactant \(\text{Cl}_2\): ON = 0 (Element).
  2. Product \(\text{NaCl}\): Cl is more electronegative than Na. ON = –1.
  3. Product \(\text{NaOCl}\) (Sodium chlorate(I)): Na is +1, O is –2. To make the total charge 0, Cl must be +1.
  • Chlorine is Reduced (0 to –1).
  • Chlorine is Oxidized (0 to +1).

Since the same element (\(\text{Cl}\)) is both oxidized and reduced, this is a disproportionation reaction.

Key Takeaway: Disproportionation is simultaneous oxidation and reduction of the same element.

Section 6: Using Oxidation Numbers to Balance Equations

A major application of ON is helping to balance complex redox equations, especially in acidic or alkaline solutions. The key idea is that the total increase in ON must equal the total decrease in ON (conservation of electrons).

6.1 Step-by-Step Guide to Balancing (Simple Method)

This method ensures the electron transfers balance, giving you the correct stoichiometry (mole ratios).

Consider the reaction: \(\text{HNO}_3 + \text{Cu} \rightarrow \text{NO}_2 + \text{Cu}(\text{NO}_3)_2 + \text{H}_2\text{O}\)

  1. Assign ONs and Identify Changes:
    • In \(\text{HNO}_3\), Nitrogen (N) = +5.
    • In \(\text{Cu}\), Copper (Cu) = 0.
    • In \(\text{NO}_2\), Nitrogen (N) = +4. (Change: Decrease by 1)
    • In \(\text{Cu}(\text{NO}_3)_2\), Copper (Cu) = +2. (Change: Increase by 2)
  2. Determine Total Change Ratio:
    • N change (Reduction) = 1 (from +5 to +4).
    • Cu change (Oxidation) = 2 (from 0 to +2).
  3. Find Stoichiometric Ratio: To balance the electrons, the total loss must equal the total gain. We need 2 atoms of N reacting for every 1 atom of Cu to make the electron changes equal (2 x 1 decrease = 1 x 2 increase).
  4. Insert Coefficients: Set the coefficient of \(\text{NO}_2\) (containing N(+4)) to 2, and the coefficient of \(\text{Cu}(\text{NO}_3)_2\) (containing Cu(+2)) to 1.

    \(\text{HNO}_3 + \text{Cu} \rightarrow 2\text{NO}_2 + \text{Cu}(\text{NO}_3)_2 + \text{H}_2\text{O}\)

  5. Balance Remaining Atoms (by inspection): Now balance H and O.
    • The final equation must account for all N atoms: 2 N atoms went to \(\text{NO}_2\) and 2 N atoms are in \(\text{Cu}(\text{NO}_3)_2\). Total N on the product side is 4.
    • Balance N by changing the \(\text{HNO}_3\) coefficient to 4.
    • \(\text{4HNO}_3 + \text{Cu} \rightarrow 2\text{NO}_2 + \text{Cu}(\text{NO}_3)_2 + \text{2H}_2\text{O}\) (Balance H/O with 2 \(\text{H}_2\text{O}\))

The final balanced equation is: \(\text{4HNO}_3 + \text{Cu} \rightarrow 2\text{NO}_2 + \text{Cu}(\text{NO}_3)_2 + 2\text{H}_2\text{O}\)

⚠️ Common Mistake Alert

Students often forget that the total oxidation number must be balanced, not just the single atom change. Always use the Least Common Multiple (LCM) of the change values to get the correct mole ratio.

Key Takeaway: Balancing redox equations relies on ensuring the total increase in Oxidation Number equals the total decrease in Oxidation Number.