Welcome to the World of Redox!

In this chapter, we are going to explore one of the most exciting parts of Physical Chemistry: Redox reactions. Don't let the name scare you—"Redox" is just a shorthand for Reduction and Oxidation. These reactions are happening everywhere around you, from the battery in your phone to the way your body turns food into energy!

Think of a redox reaction like a game of catch, but instead of a ball, the atoms are throwing and catching electrons. By the end of these notes, you’ll be able to track where those electrons are going and write the "playbook" for the whole game.


1. Oxidation and Reduction: The "OIL RIG" Mnemonic

In chemistry, oxidation and reduction always happen at the same time. If one atom loses an electron, another atom must be there to grab it.

To help you remember which is which, we use the world-famous mnemonic: OIL RIG.

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

What are Oxidising and Reducing Agents?

This is where students sometimes get a little confused, but here is a simple trick: an agent is someone who makes something happen for someone else.

  • An oxidising agent makes oxidation happen to another substance. To do this, it accepts electrons (it gets reduced itself).
  • A reducing agent makes reduction happen to another substance. To do this, it donates electrons (it gets oxidised itself).

Analogy: Think of a reducing agent like a "Pizza Delivery Driver." The driver gives you the pizza (the electron). The driver is the "agent" of delivery, but because they gave the pizza away, they now have "lost" it.

Quick Review Box:
Oxidation: Loss of electrons / Increase in oxidation state.
Reduction: Gain of electrons / Decrease in oxidation state.


2. Oxidation States (The Bookkeeping System)

How do we know if something has been oxidised or reduced if we can't see the electrons moving? We use oxidation states (or oxidation numbers). Think of this as a chemical "bank account" that tracks how many electrons an atom has.

The Rules for Assigning Oxidation States

You need to learn these rules by heart. Don't worry, they are quite logical!

  1. Uncombined elements: Any element on its own (like \(Na\), \(He\), or \(O_2\)) always has an oxidation state of 0.
  2. Simple ions: For a single atom ion, the oxidation state is the same as the charge. Example: \(Mg^{2+}\) is +2, \(Cl^-\) is -1.
  3. Fluorine: Always -1 in its compounds.
  4. Oxygen: Usually -2. (Except in peroxides like \(H_2O_2\) where it is -1, or when combined with Fluorine).
  5. Hydrogen: Usually +1. (Except in metal hydrides like \(NaH\) where it is -1).
  6. The "Sum" Rules:
    • In a neutral compound (like \(H_2O\)), the sum of all oxidation states must be 0.
    • In a complex ion (like \(SO_4^{2-}\)), the sum of all oxidation states must equal the charge of the ion (-2 in this case).

Example Walkthrough: What is the oxidation state of Sulfur in \(H_2SO_4\)?
1. We know Hydrogen is \(+1\). There are two: \(2 \times (+1) = +2\).
2. We know Oxygen is \(-2\). There are four: \(4 \times (-2) = -8\).
3. The total must be \(0\). So: \((+2) + S + (-8) = 0\).
4. Solving for \(S\): \(S - 6 = 0\), so \(S = \mathbf{+6}\).

Key Takeaway: If the oxidation state goes UP, the atom has been oxidised. If it goes DOWN, it has been reduced.


3. Writing Half-Equations

A half-equation shows us exactly what is happening to one specific species (either just the oxidation part or just the reduction part). These equations must include electrons, written as \(e^-\).

Step-by-Step: Balancing Half-Equations in Acidic Conditions

Don't worry if this seems tricky at first! Just follow these steps in order every single time:

  1. Balance the element that is being oxidised or reduced.
  2. Balance any Oxygen atoms by adding \(H_2O\) to the other side.
  3. Balance any Hydrogen atoms by adding \(H^+\) ions to the other side.
  4. Balance the charges by adding electrons (\(e^-\)) to the side that is more positive.

Example: Change \(MnO_4^-\) into \(Mn^{2+}\).
Step 1: \(Mn\) is already balanced.
Step 2: Add \(4H_2O\) to the right to balance the 4 Oxygens: \(MnO_4^- \rightarrow Mn^{2+} + 4H_2O\).
Step 3: Add \(8H^+\) to the left to balance the Hydrogens: \(MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O\).
Step 4: Check charges. Left side is \(( -1 + 8 ) = +7\). Right side is \(+2\). Add \(5e^-\) to the left to make both sides \(+2\).
Final: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)


4. Combining Half-Equations

To get the overall redox equation, we combine the oxidation half-equation and the reduction half-equation. There is one golden rule here: The number of electrons lost must equal the number of electrons gained.

How to combine them:

  1. Find the oxidation half-equation and the reduction half-equation.
  2. Multiply one or both equations by a number so that the number of electrons is the same in both.
  3. Add the two equations together.
  4. Cancel out anything that appears on both sides (usually the electrons, and sometimes \(H^+\) or \(H_2O\)).

Example: Combine the \(MnO_4^-\) equation (gains \(5e^-\)) with an Iron equation: \(Fe^{2+} \rightarrow Fe^{3+} + e^-\).
1. The Iron equation only has \(1e^-\), but the Manganese needs \(5e^-\).
2. Multiply the whole Iron equation by 5: \(5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-\).
3. Add them together: \(MnO_4^- + 8H^+ + 5e^- + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} + 5e^-\).
4. Cancel the \(5e^-\) from both sides.
Result: \(MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}\)

Did you know? This specific reaction is used in titrations to find out how much iron is in a sample of ore or even in iron supplement tablets!


Common Mistakes to Avoid

  • Mistaking the charge for the oxidation state: In \(O_2\), the molecule has no charge, but the oxidation state is 0. In \(SO_4^{2-}\), the charge is -2, but the oxidation state of Oxygen is still -2 (it's the sum that matters).
  • Forgetting to balance Oxygen/Hydrogen: Always check for Oxygen first, then Hydrogen. If you skip this, your charges will never balance.
  • Electrons on the wrong side: In oxidation, electrons are lost (placed on the right/product side). In reduction, they are gained (placed on the left/reactant side).

Summary Checklist

- Can I define oxidation and reduction in terms of electrons? (Remember OIL RIG!)
- Can I assign oxidation states to any atom in a formula? (Use the rules!)
- Can I identify the oxidising and reducing agents? (Who took the electrons?)
- Can I write and balance a half-equation? (Element, \(H_2O\), \(H^+\), then \(e^-\))
- Can I combine two half-equations into one overall equation? (Make the electrons match!)