Welcome to Redox I!
Welcome to one of the most important topics in your AS Chemistry journey! Redox might sound like a complex word, but it is just a "mash-up" of two words: Reduction and Oxidation. These reactions are happening around you every second—from the battery powering your phone to the way your body turns food into energy. In this chapter, we are going to learn how to track where electrons go during a reaction and how to balance equations like a pro. Don't worry if it feels like "chemical accountancy" at first; once you learn the rules, it becomes a very satisfying puzzle!
1. The Basics: What is Redox?
At its heart, a redox reaction is all about the movement of electrons. If one atom loses an electron, another atom must be there to catch it. We use a famous mnemonic to remember which is which:
The "OIL RIG" Mnemonic
Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)
Wait, why is gaining electrons called "reduction"?
This is a common point of confusion! It’s called reduction because electrons have a negative charge. When an atom gains an electron, its overall charge (or "oxidation number") is reduced (it goes down). Think of it like gaining debt—your bank balance goes down!
Quick Review: Metals vs. Non-Metals
• Metals generally like to lose electrons to become stable. This means they are oxidised and form positive ions (cations). Their oxidation number increases.
• Non-metals generally like to gain electrons. This means they are reduced and form negative ions (anions). Their oxidation number decreases.
Key Takeaway: You cannot have oxidation without reduction. They always happen together as a pair!
2. Oxidising and Reducing Agents
In chemistry, an "agent" is something that makes something else happen. This can be a bit of a brain-twister, so let’s use an analogy.
Analogy: A Travel Agent
A travel agent doesn't go on holiday; they help you go on holiday. Similarly:
• An Oxidising Agent oxidises something else. To do that, it must take electrons away from the other substance. Therefore, the oxidising agent gains electrons and is itself reduced.
• A Reducing Agent reduces something else. It gives its electrons away to the other substance. Therefore, the reducing agent loses electrons and is itself oxidised.
Common Mistake Alert!
Students often think the "oxidising agent" is the one being oxidised. It's the opposite! Always remember: The agent does the work to the other person/atom.
Key Takeaway: Oxidising agents gain electrons; Reducing agents lose electrons.
3. Oxidation Numbers: Chemical Accountancy
An oxidation number is a number assigned to an element in a compound that represents the number of electrons lost or gained by an atom of that element. We write them with the sign before the number (e.g., \( +2 \) or \( -1 \)).
The Golden Rules for Assigning Oxidation Numbers
To solve redox puzzles, you must learn these rules. They are not optional!
1. Uncombined elements (e.g., \( \text{He}, \text{Cl}_2, \text{Mg}, \text{S}_8 \)) always have an oxidation number of 0.
2. Simple ions have an oxidation number equal to their charge (e.g., \( \text{Na}^+ \) is \( +1 \), \( \text{Mg}^{2+} \) is \( +2 \), \( \text{Cl}^- \) is \( -1 \)).
3. The sum of oxidation numbers in a neutral compound is 0.
4. The sum of oxidation numbers in a polyatomic ion (like \( \text{SO}_4^{2-} \)) equals the overall charge of the ion.
5. Fluorine is always -1 in compounds (it's the most "greedy" element!).
6. Oxygen is usually -2, except in peroxides (like \( \text{H}_2\text{O}_2 \)) where it is -1.
7. Hydrogen is usually +1, except in metal hydrides (like \( \text{NaH} \)) where it is -1.
Using Roman Numerals
We use Roman numerals in the names of compounds to show the oxidation number of an element that can have more than one state. For example:
• Iron(II) chloride contains \( \text{Fe} \) with an oxidation number of +2 (\( \text{FeCl}_2 \)).
• Iron(III) chloride contains \( \text{Fe} \) with an oxidation number of +3 (\( \text{FeCl}_3 \)).
Did you know?
The "Old Names" for iron(II) and iron(III) were ferrous and ferric. Using oxidation numbers and Roman numerals is much clearer for scientists everywhere!
Key Takeaway: Use the rules as a checklist to find the "missing" oxidation number in any molecule.
4. How to Calculate Oxidation Numbers (Step-by-Step)
Let's find the oxidation number of Sulfur (\( \text{S} \)) in Sulfuric Acid (\( \text{H}_2\text{SO}_4 \)).
Step 1: Identify the knowns. Hydrogen is \( +1 \). Oxygen is \( -2 \).
Step 2: Multiply by the number of atoms. We have two \( \text{H} \) atoms (\( 2 \times +1 = +2 \)) and four \( \text{O} \) atoms (\( 4 \times -2 = -8 \)).
Step 3: Set up a simple math equation. The total must be \( 0 \) because it's a neutral compound.
\( (+2) + (\text{S}) + (-8) = 0 \)
Step 4: Solve for \( \text{S} \).
\( \text{S} - 6 = 0 \)
\( \text{S} = +6 \)
The oxidation number of Sulfur in \( \text{H}_2\text{SO}_4 \) is +6.
5. Disproportionation: The "Self-Redox" Reaction
Usually, one thing is oxidised and another is reduced. But sometimes, the same element in a single species is simultaneously oxidised and reduced. This is called disproportionation.
Example: Decomposition of Hydrogen Peroxide
\( 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \)
In \( \text{H}_2\text{O}_2 \), Oxygen is -1.
In \( \text{H}_2\text{O} \), Oxygen is -2 (Reduced: \( -1 \rightarrow -2 \)).
In \( \text{O}_2 \), Oxygen is 0 (Oxidised: \( -1 \rightarrow 0 \)).
Key Takeaway: If you see one element's oxidation number go both up and down from the same starting point, it's disproportionation!
6. Writing Redox Equations
To see the electron transfer clearly, we split a full equation into two half-equations. One shows oxidation (electrons on the right), and one shows reduction (electrons on the left).
Step-by-Step: Constructing a Full Ionic Equation
Let's react \( \text{Mg} \) with \( \text{Cu}^{2+} \) ions.
1. Write the oxidation half-equation: \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2\text{e}^- \)
2. Write the reduction half-equation: \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \)
3. Check the electrons: Both have \( 2\text{e}^- \). If they were different (e.g., 1 and 2), you would have to multiply the whole equation to make them equal.
4. Combine them: Add the left sides together and the right sides together.
\( \text{Mg} + \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Mg}^{2+} + 2\text{e}^- + \text{Cu} \)
5. Cancel the electrons: Since they appear on both sides, they cancel out.
Final Equation: \( \text{Mg(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{Cu(s)} \)
Quick Tip:
In a final full ionic equation, you should never see any electrons (\( \text{e}^- \)). If you do, you haven't finished balancing them yet!
Key Takeaway: Balance the electrons in your half-equations before you join them together.
Final Summary Review
• Oxidation: Loss of electrons / Increase in oxidation number.
• Reduction: Gain of electrons / Decrease in oxidation number.
• Oxidising Agent: Takes electrons (is reduced).
• Reducing Agent: Gives electrons (is oxidised).
• Oxidation Numbers: A way to track electrons. Know your "Golden Rules"!
• Disproportionation: When one element is both oxidised and reduced in the same reaction.
Don't worry if this seems tricky at first—keep practicing the oxidation number rules, and the rest will fall into place!