Welcome to Redox Reactions!

Hello! Get ready to explore one of the most important types of chemical reactions: redox reactions. You might not have heard the name, but you see them happening all around you – from a battery powering your phone, to a slice of apple turning brown, to the way our bodies get energy from food. Pretty important, right?

In this chapter, we're going to demystify redox. We'll learn how to spot these reactions, understand the special 'book-keeping' tool called oxidation numbers, and master the skill of balancing redox equations. Don't worry if it seems tricky at first, we'll break it all down step-by-step!


Section 1: What are Oxidation and Reduction?

A redox reaction is simply a reaction where oxidation and reduction happen at the same time. You can't have one without the other! They are like two sides of the same coin. There are a few ways to define them.

1. The Classic Definition: In terms of Oxygen and Hydrogen

This is the original way scientists thought about redox. It's simple and works for many common reactions.

Oxidation is the gain of oxygen or loss of hydrogen.
Reduction is the loss of oxygen or gain of hydrogen.

Example: The reaction of copper(II) oxide with hydrogen.

$$ CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l) $$

• Copper(II) oxide ($$CuO$$) loses oxygen to become copper ($$Cu$$). So, $$CuO$$ is reduced.
• Hydrogen ($$H_2$$) gains oxygen to become water ($$H_2O$$). So, $$H_2$$ is oxidised.

2. The Modern Definition: In terms of Electrons (The Most Important One!)

The real story behind redox reactions is the transfer of electrons. This definition is more powerful because it works for all redox reactions, not just those involving oxygen or hydrogen.

Here's a super useful mnemonic to help you remember:

OIL RIG

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

Example: Placing a zinc strip in copper(II) sulphate solution.

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

Let's break this down into two half-equations:

• Oxidation Half-Equation: The zinc atom loses two electrons to become a zinc ion.
$$ Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- $$ (Loss of electrons = Oxidation)

• Reduction Half-Equation: The copper(II) ion gains two electrons to become a copper atom.
$$ Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) $$ (Gain of electrons = Reduction)

Agents of Change: Oxidising and Reducing Agents

In a redox reaction, one substance causes another to change. We call these 'agents'.

• An Oxidising Agent (or oxidant) is a substance that causes oxidation in another substance. In the process, the oxidising agent gets reduced itself (it gains electrons).

• A Reducing Agent (or reductant) is a substance that causes reduction in another substance. In the process, the reducing agent gets oxidised itself (it loses electrons).

Analogy Time! Think of a travel agent. A travel agent doesn't travel themselves; they make other people travel. Similarly, a reducing agent doesn't get reduced; it makes another substance get reduced!

In our example, $$ Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s) $$:

- $$Zn$$ is oxidised, so it is the reducing agent.
- $$Cu^{2+}$$ is reduced, so it is the oxidising agent.

Key Takeaway for Section 1

Redox is all about electron transfer. Remember OIL RIG: Oxidation Is Loss, Reduction Is Gain. The substance that gets oxidised is the reducing agent, and the substance that gets reduced is the oxidising agent.


Section 2: Oxidation Numbers - The Ultimate Redox Tool

So, electron transfer is easy to see with ions. But what about reactions with covalent compounds, where electrons are shared, not fully transferred? For this, we use a concept called oxidation numbers (O.N.), sometimes called oxidation states.

The oxidation number is the hypothetical charge an atom would have if all its bonds were 100% ionic. It's a way of keeping track of where electrons are, even when they're being shared.

How to Assign Oxidation Numbers: The Rules

Follow these rules in order. The rule higher up the list takes priority!

1. An atom in its elemental form: O.N. is 0. (e.g., $$Na$$, $$Cl_2$$, $$O_2$$, $$P_4$$ all have O.N. = 0)
2. A simple ion: O.N. is equal to its charge. (e.g., in $$Na^+$$, O.N. of Na is +1; in $$S^{2-}$$, O.N. of S is -2)
3. Group 1 and Group 2 metals in compounds: Group 1 is always +1. Group 2 is always +2.
4. Fluorine in compounds: Always -1.
5. Hydrogen in compounds: Almost always +1. (The rare exception is in metal hydrides like $$NaH$$, where H is -1).
6. Oxygen in compounds: Almost always -2. (Exceptions include peroxides like $$H_2O_2$$, where O is -1).
7. For a neutral compound: The sum of all oxidation numbers must be 0.
8. For a polyatomic ion: The sum of all oxidation numbers must equal the charge of the ion.

Step-by-Step Examples:

Example 1: Find the oxidation number of sulphur (S) in sulphuric acid, $$H_2SO_4$$.

1. This is a neutral compound, so the total O.N. must be 0. (Rule 7)
2. We know O.N. of H is +1 (Rule 5) and O.N. of O is -2 (Rule 6).
3. Let the O.N. of S be 'x'.
4. Set up the equation: $$ (2 \times H) + (1 \times S) + (4 \times O) = 0 $$
$$ (2 \times (+1)) + (x) + (4 \times (-2)) = 0 $$
$$ +2 + x - 8 = 0 $$
$$ x = +6 $$
So, the oxidation number of S in $$H_2SO_4$$ is +6.

Example 2: Find the oxidation number of chromium (Cr) in the dichromate ion, $$Cr_2O_7^{2-}$$.

1. This is a polyatomic ion with a charge of 2-, so the total O.N. must be -2. (Rule 8)
2. We know O.N. of O is -2 (Rule 6).
3. Let the O.N. of Cr be 'y'. There are two Cr atoms, so we have '2y'.
4. Set up the equation: $$ (2 \times Cr) + (7 \times O) = -2 $$
$$ (2y) + (7 \times (-2)) = -2 $$
$$ 2y - 14 = -2 $$
$$ 2y = +12 $$
$$ y = +6 $$
So, the oxidation number of each Cr atom in $$Cr_2O_7^{2-}$$ is +6.

3. The Final Definition: In terms of Oxidation Numbers

Using oxidation numbers gives us the most powerful and universal definition of redox:

Oxidation is an increase in oxidation number.
Reduction is a decrease in oxidation number.

Think of a number line! Moving to the right (e.g., -1 to 0, +2 to +4) is oxidation. Moving to the left (e.g., 0 to -2, +7 to +2) is reduction.

Key Takeaway for Section 2

Oxidation numbers are our 'book-keeping' tool for electrons. Learn the rules for assigning them! Oxidation means the O.N. goes up; reduction means the O.N. goes down. This definition works for EVERY reaction.


Section 3: Common Oxidising and Reducing Agents

In the lab, you'll meet a cast of common chemical agents. It's very useful to know who they are and what they do!

Common Oxidising Agents (They love to gain electrons and get reduced!)

Agent: Acidified potassium manganate(VII), $$MnO_4^-(aq)/H^+(aq)$$
What happens: The Mn atom goes from O.N. +7 to +2.
Colour change: Purple to colourless
Half-equation: $$ MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l) $$

Agent: Acidified potassium dichromate(VI), $$Cr_2O_7^{2-}(aq)/H^+(aq)$$
What happens: The Cr atoms go from O.N. +6 to +3.
Colour change: Orange to green
Half-equation: $$ Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l) $$

Agent: Iron(III) ion, $$Fe^{3+}(aq)$$
What happens: The Fe ion goes from O.N. +3 to +2.
Colour change: Yellow/brown to pale green
Half-equation: $$ Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq) $$

Other important oxidising agents include: Chlorine ($$Cl_2$$), Nitric acid ($$HNO_3$$), and concentrated Sulphuric acid (conc. $$H_2SO_4$$).

Common Reducing Agents (They love to lose electrons and get oxidised!)

Agent: Iron(II) ion, $$Fe^{2+}(aq)$$
What happens: The Fe ion goes from O.N. +2 to +3.
Colour change: Pale green to yellow/brown
Half-equation: $$ Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^- $$

Agent: Iodide ion, $$I^-(aq)$$
What happens: The I ion goes from O.N. -1 to 0 (in $$I_2$$).
Observation: A colourless solution turns brown.
Half-equation: $$ 2I^-(aq) \rightarrow I_2(aq) + 2e^- $$

Agent: Sulphite ion, $$SO_3^{2-}(aq)$$
What happens: The S atom goes from O.N. +4 to +6 (in $$SO_4^{2-}$$).
Half-equation: $$ SO_3^{2-}(aq) + H_2O(l) \rightarrow SO_4^{2-}(aq) + 2H^+(aq) + 2e^- $$

Other important reducing agents include: Reactive metals like Zinc ($$Zn(s)$$).

Did you know?

The orange-to-green colour change of dichromate(VI) ions was used in old police breathalyser tests. The ethanol from the driver's breath acted as a reducing agent, reducing the orange dichromate to green chromium(III). The more green it turned, the more alcohol was present!


Section 4: The Electrochemical Series (ECS)

How can we predict if a redox reaction will happen? We use the Electrochemical Series (ECS). Think of it as a "league table" for redox, ranking species on their strength as oxidising or reducing agents.

You'll usually be given a section of the ECS in an exam. It's often written as a list of reduction half-equations.

How to Read the ECS

• Strongest Oxidising Agents are on the LEFT side, towards the BOTTOM of the series.
• Strongest Reducing Agents are on the RIGHT side, towards the TOP of the series.

How to Predict Reactions

A spontaneous reaction can occur between a reducing agent (from the top-right of the ECS) and an oxidising agent (from the bottom-left of the ECS).

Example: Let's look at a small part of the ECS.

$$ Zn^{2+}(aq) + 2e^- \rightleftharpoons Zn(s) $$
$$ Fe^{2+}(aq) + 2e^- \rightleftharpoons Fe(s) $$
$$ Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu(s) $$

Question: Will zinc metal, $$Zn(s)$$, react with iron(II) ions, $$Fe^{2+}(aq)$$?
Answer: 1. Find our potential reactants. $$Zn(s)$$ is on the top-right (a reducing agent). $$Fe^{2+}(aq)$$ is on the bottom-left (an oxidising agent).
2. The reducing agent ($$Zn$$) is above the oxidising agent ($$Fe^{2+}$$). A reaction is possible!
3. The reaction would be: $$ Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) $$.

Question: Will copper metal, $$Cu(s)$$, react with zinc ions, $$Zn^{2+}(aq)$$?
Answer: 1. Find our reactants. $$Cu(s)$$ is a reducing agent. $$Zn^{2+}(aq)$$ is an oxidising agent.
2. The reducing agent ($$Cu$$) is below the oxidising agent ($$Zn^{2+}$$). A reaction is NOT spontaneous. Nothing will happen.

Key Takeaway for Section 4

The ECS is a ranking of oxidising and reducing power. A substance on the top-right (stronger reducing agent) can react with a substance on the bottom-left (stronger oxidising agent).


Section 5: Balancing Redox Equations - The Master Skill

Okay, this is the big one. Balancing redox equations can seem complex, but if you follow a set of steps, you can solve any of them! We'll use the ion-electron half-equation method.

Let's balance the reaction between manganate(VII) ions and iron(II) ions in acidic solution.

Unbalanced equation: $$ MnO_4^-(aq) + Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + Fe^{3+}(aq) $$

Step-by-Step Guide (for Acidic Solution)

Step 1: Write the two half-equations.
Identify what gets oxidised and what gets reduced.
Reduction: $$ MnO_4^- \rightarrow Mn^{2+} $$
Oxidation: $$ Fe^{2+} \rightarrow Fe^{3+} $$

Step 2: Balance each half-equation separately.
Let's start with the oxidation one – it's easier!
$$ Fe^{2+} \rightarrow Fe^{3+} $$
a. Atoms are already balanced (1 Fe on each side).
b. Balance the charge by adding electrons ($$e^-$$). The left is +2, the right is +3. We need to add one electron to the right side.
Balanced Oxidation Half-Equation: $$ Fe^{2+} \rightarrow Fe^{3+} + e^- $$

Now for the reduction half-equation:
$$ MnO_4^- \rightarrow Mn^{2+} $$
a. Balance atoms other than O and H. (Mn is already balanced).
b. Balance Oxygen atoms by adding $$H_2O$$. There are 4 O's on the left, so add 4 $$H_2O$$ to the right.
$$ MnO_4^- \rightarrow Mn^{2+} + 4H_2O $$
c. Balance Hydrogen atoms by adding $$H^+$$. There are $$4 \times 2 = 8$$ H's on the right, so add 8 $$H^+$$ to the left.
$$ MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O $$
d. Balance the charge by adding electrons ($$e^-$$).
- Total charge on left: $$(-1) + (+8) = +7$$
- Total charge on right: $$(+2) + (0) = +2$$
To get from +7 to +2, we must add 5 negative charges. Add 5 $$e^-$$ to the left.
Balanced Reduction Half-Equation: $$ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O $$

Step 3: Equalise the electrons.
The oxidation half-equation produces 1 $$e^-$$.
The reduction half-equation consumes 5 $$e^-$$.
We need to multiply the oxidation half-equation by 5 so that the electrons lost equal the electrons gained.
$$ 5(Fe^{2+} \rightarrow Fe^{3+} + e^-) $$ gives $$ 5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^- $$

Step 4: Add the two half-equations together.
$$ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O $$
$$ 5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^- $$
------------------------------------------------------------------
$$ MnO_4^-(aq) + 8H^+(aq) + 5Fe^{2+}(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l) + 5Fe^{3+}(aq) + 5e^- $$

Step 5: Cancel anything that's the same on both sides.
The 5$$e^-$$ on each side cancel out.

Step 6: Write the final balanced equation and do a final check!
$$ MnO_4^-(aq) + 8H^+(aq) + 5Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l) + 5Fe^{3+}(aq) $$
Atom check: 1 Mn, 8 H, 5 Fe, 4 O on both sides. Perfect!
Charge check: Left = (-1) + (+8) + 5(+2) = +17. Right = (+2) + 0 + 5(+3) = +17. Perfect!

Key Takeaway for Section 5

Balancing redox equations is a systematic process. Follow the steps carefully: split into half-equations, balance atoms (O with H₂O, H with H⁺), balance charge with e⁻, equalise e⁻, and add them back together. Practice makes perfect!