Chemistry (0620) | Redox

Understanding Redox Reactions: The Chemistry of Give and Take

Hello! Welcome to the exciting world of Redox reactions. This topic is absolutely fundamental to Chemistry (and lots of everyday life, like batteries and rust!). Redox stands for Reduction and Oxidation.

Don't worry if these terms sound tricky at first. We will break down three main ways chemists define these reactions, starting with the simplest one, so you can master them all!

Section 1: The Basic Definitions (Oxygen Transfer)

The most straightforward way to identify a redox reaction is by looking at what happens to the oxygen atoms in a chemical equation.

1.1 Definitions Based on Oxygen (Core Content)

A reaction is a redox reaction if oxidation and reduction happen simultaneously. They are like a chemical dance partner pair; one cannot occur without the other.

Oxidation

Oxidation is defined as the gain of oxygen by a substance.

• Example: When magnesium burns in air:
  \( 2\text{Mg} + \text{O}_2 \to 2\text{MgO} \)
  Magnesium (\(\text{Mg}\)) gains oxygen to become magnesium oxide (\(\text{MgO}\)). Therefore, \(\text{Mg}\) is oxidised.

Reduction

Reduction is defined as the loss of oxygen by a substance.

• Example: Reducing copper(II) oxide using hydrogen gas:
  \( \text{CuO} + \text{H}_2 \to \text{Cu} + \text{H}_2\text{O} \)
  Copper(II) oxide (\(\text{CuO}\)) loses oxygen to become copper (\(\text{Cu}\)). Therefore, \(\text{CuO}\) is reduced.

1.2 Identifying Redox by Oxygen Transfer (Core Content)

In the reaction above (\( \text{CuO} + \text{H}_2 \to \text{Cu} + \text{H}_2\text{O} \)):

• \(\text{CuO}\) is reduced (it lost oxygen).
• \(\text{H}_2\) is oxidised (it gained oxygen, becoming \(\text{H}_2\text{O}\)).

Since both reduction and oxidation occurred in the same equation, this is a redox reaction.

Did you know? The name 'oxidation' originally came from the discovery of oxygen by Antoine Lavoisier!

Section 2: The Electron Definitions (Extended Content)

The oxygen definition is useful but limited. What about reactions that don't involve oxygen (like displacement reactions)? For these, we use the definition based on electron movement.

This definition is often easier to apply, especially when dealing with ions.

2.1 Definitions Based on Electrons (Supplement Content)

When atoms react, they either lose or gain electrons. This movement determines if the substance is oxidised or reduced.

Memory Aid: LEO the lion says GER

This is the most common mnemonic you will use for redox!

• LEO: Loss of Electrons is Oxidation.
• GER: Gain of Electrons is Reduction.

The key idea here:

• Oxidation: A substance loses electrons (often becoming a positive ion, or becoming a more positive ion).
• Reduction: A substance gains electrons (often becoming a negative ion, or becoming a less positive ion).

Example: The reaction of Zinc and Copper(II) ions

When a zinc metal bar is placed in copper(II) sulfate solution, the zinc dissolves and copper metal precipitates.

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

We can look at the half-equations (syllabus 4.1.11):

• Zinc atoms: \( \text{Zn} \to \text{Zn}^{2+} + 2\text{e}^- \)
  Zinc loses electrons (LEO). Zinc is oxidised.
• Copper ions: \( \text{Cu}^{2+} + 2\text{e}^- \to \text{Cu} \)
  Copper ions gain electrons (GER). Copper ions are reduced.

Connection to Electrochemistry: In electrolysis (Syllabus Section 4), oxidation always occurs at the anode (where electrons are lost), and reduction always occurs at the cathode (where electrons are gained).

Section 3: Oxidation Numbers (The Professional Method)

The most precise way chemists track redox is using Oxidation Numbers (O.N.) or oxidation states. This method works for every type of reaction, including those with covalent compounds where electrons aren't fully lost or gained, but merely shared unequally.

3.1 Using Roman Numerals (Core Content)

For simple compounds, especially transition metal salts, we use Roman numerals in the name to show the oxidation number (or valency) of the metal.

• Example: Iron(III) oxide ($\text{Fe}_2\text{O}_3$). The Roman numeral (III) tells us the O.N. of Iron is +3.
• Example: Copper(I) chloride ($\text{CuCl}$). The Roman numeral (I) tells us the O.N. of Copper is +1.

3.2 Definitions Based on Oxidation Number (Supplement Content)

If the oxidation number changes during a reaction, it is a redox reaction:

• Oxidation: An increase in oxidation number (becoming more positive).
• Reduction: A decrease in oxidation number (becoming less positive/more negative).

Don't worry if this seems tricky at first. Practice the rules below!

3.3 Rules for Assigning Oxidation Numbers (Supplement Content)

You need to know four essential rules to calculate oxidation numbers:

1. Uncombined Element: The oxidation number of any element in its uncombined state is zero (0).
   Examples: $\text{Zn}$, $\text{O}_2$, $\text{Cl}_2$ all have an O.N. of 0.
2. Monatomic Ion: The oxidation number of a monatomic ion (a single atom ion) is equal to the charge on the ion.
   Examples: $\text{Na}^+$ is +1. $\text{Cl}^-$ is -1. $\text{O}^{2-}$ is -2.
3. Neutral Compound: The sum of the oxidation numbers of all elements in a neutral compound must be zero (0).
   Example: In $\text{H}_2\text{O}$, $\text{H}$ is +1, $\text{O}$ is -2. Total charge: \( (2 \times +1) + (-2) = 0 \).
4. Polyatomic Ion: The sum of the oxidation numbers of all elements in a polyatomic ion (like $\text{SO}_4^{2-}$ or $\text{NO}_3^-$) is equal to the charge on the ion.
   Example: In $\text{SO}_4^{2-}$, the overall charge is -2.

Step-by-Step Example: Calculating Sulfur's O.N. in \(\text{H}_2\text{SO}_4\)

We know: $\text{H}$ is +1, $\text{O}$ is -2. The compound $\text{H}_2\text{SO}_4$ is neutral (sum = 0). Let $x$ be the O.N. of S.

\( (2 \times \text{O.N. of H}) + (1 \times \text{O.N. of S}) + (4 \times \text{O.N. of O}) = 0 \)
\( (2 \times +1) + x + (4 \times -2) = 0 \)
\( +2 + x - 8 = 0 \)
\( x - 6 = 0 \)
\( x = +6 \)

Therefore, the oxidation number of sulfur (S) in sulfuric acid is +6.

Section 4: Oxidising and Reducing Agents (The Causers)

In any redox reaction, the substance that causes the other reactant to be oxidized is called the oxidising agent, and the substance that causes the other reactant to be reduced is called the reducing agent.

4.1 Definitions of Agents (Supplement Content)

This is the trickiest part conceptually, but use the analogy: an agent does the job for someone else.

The Oxidising Agent (OA)

An Oxidising Agent is a substance that oxidises another substance and is itself reduced.

• It gains electrons (GER).
• Its oxidation number decreases.

The Reducing Agent (RA)

A Reducing Agent is a substance that reduces another substance and is itself oxidised.

• It loses electrons (LEO).
• Its oxidation number increases.

Example: Look again at \( \text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu} \)

• $\text{Zn}$ is oxidised (lost electrons). Since $\text{Zn}$ caused the $\text{Cu}^{2+}$ to be reduced, $\text{Zn}$ is the Reducing Agent.
• $\text{Cu}^{2+}$ is reduced (gained electrons). Since $\text{Cu}^{2+}$ caused the $\text{Zn}$ to be oxidised, $\text{Cu}^{2+}$ is the Oxidising Agent.

Section 5: Identifying Redox Reactions Using Colour Changes (Extended Content)

In the lab, we often use specific chemical agents to prove if a reaction involves oxidation or reduction, by looking for a characteristic colour change.

5.1 Testing for Reduction (Using an Oxidising Agent)

A common oxidising agent used in redox titrations is acidified aqueous potassium manganate(VII), \(\text{KMnO}_4\).

• $\text{KMnO}_4$ is a very strong oxidising agent.
• Because it is an oxidising agent, it must be reduced itself when it reacts.
• Colour Change: When $\text{KMnO}_4$ is reduced, the purple colour of the manganate(VII) ion ($\text{MnO}_4^-$) disappears, and the solution becomes colourless.
• If you add purple $\text{KMnO}_4$ solution to an unknown substance and the purple colour disappears, it means the unknown substance oxidised the $\text{KMnO}_4$, making the unknown substance a reducing agent.

5.2 Testing for Oxidation (Using a Reducing Agent)

A common reducing agent used in redox tests is aqueous potassium iodide, \(\text{KI}\).

• $\text{KI}$ is a reducing agent because the iodide ion ($\text{I}^-$) easily loses electrons.
• Because it is a reducing agent, it must be oxidised itself when it reacts.
• Colour Change: When iodide ions ($\text{I}^-$) are oxidised, they form iodine ($\text{I}_2$).
  \( 2\text{I}^- \to \text{I}_2 + 2\text{e}^- \)
• The solution changes from colourless (iodide ions) to brown/yellow (iodine).
• If you add colourless $\text{KI}$ solution to an unknown substance and it turns brown/yellow, it means the unknown substance reduced the $\text{KI}$, making the unknown substance an oxidising agent.