IB DP Chemistry - R3.2.1 Oxidation and reduction definitions - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R3.2.1 Oxidation and reduction definitions – Study Notes – New Syllabus
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Reactivity 3.2.1 — Oxidation and Reduction
Reactivity 3.2.1 — Oxidation and Reduction
Oxidation and reduction can be described in four different but complementary ways, depending on the context:
| Process | Oxidation | Reduction |
|---|---|---|
| Electron transfer | Loss of electrons | Gain of electrons |
| Oxidation state | Increase in oxidation state | Decrease in oxidation state |
| Oxygen | Gain of oxygen | Loss of oxygen |
| Hydrogen | Loss of hydrogen | Gain of hydrogen |
Oxidation States (Oxidation Numbers):
The oxidation state (or oxidation number) of an atom in a compound indicates the number of electrons lost or gained compared to the uncombined atom.
- Oxidation states are assigned using a set of formal rules (see below).
- They are useful for identifying redox changes and balancing redox equations.
Rules for Determining Oxidation States:
- Atoms in their elemental form: \( 0 \) (e.g. \( \text{Cl}_2, \text{O}_2, \text{Na} \)).
- Monoatomic ions: Equal to their ionic charge (e.g. \( \text{Na}^+ = +1 \), \( \text{O}^{2-} = -2 \)).
- Oxygen: Usually \( -2 \), except in peroxides (\( -1 \)) and OF\(_2\) (\( +2 \)).
- Hydrogen: Usually \( +1 \), except in metal hydrides like \( \text{NaH} \) where it is \( -1 \).
- Fluorine: Always \( -1 \) in compounds.
- The sum of oxidation numbers in:
- a neutral compound is \( 0 \)
- a polyatomic ion is equal to the ion’s charge
Examples:
- In \( \text{H}_2\text{SO}_4 \): H = \( +1 \), O = \( -2 \), so S = \( +6 \)
- In \( \text{NO}_3^- \): O = \( -2 \), so N = \( +5 \)
- In \( \text{Fe}_2\text{O}_3 \): O = \( -2 \), Fe = \( +3 \)
Identifying Redox Species and Agents:
In a redox reaction:
- The substance that is oxidized loses electrons → called the reducing agent
- The substance that is reduced gains electrons → called the oxidizing agent
Example
In the reaction:
\( \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \)
▶️Answer/Explanation
– Zn is oxidized (0 to +2): it’s the reducing agent.
– Cu\( ^{2+} \) is reduced (+2 to 0): it’s the oxidizing agent.
This is a classic electron transfer redox reaction.
Variable Oxidation States in Transition Metals:
Transition metals often form multiple oxidation states, depending on the compound and bonding environment. Their ability to use d orbitals for bonding enables this flexibility.
| Transition Metal | Common Oxidation States | Examples |
|---|---|---|
| Fe | +2, +3 | \( \text{Fe}^{2+} \) in FeSO\(_4\), \( \text{Fe}^{3+} \) in FeCl\(_3\) |
| Mn | +2 to +7 | \( \text{MnO}_4^- \) (Mn = +7), \( \text{Mn}^{2+} \) |
| Cr | +3, +6 | \( \text{Cr}^{3+} \), \( \text{Cr}_2\text{O}_7^{2-} \) |
Variable Oxidation States in Non-Metals:
Non-metals also show variable oxidation numbers, especially in oxyanions:
- Nitrogen: −3 in \( \text{NH}_3 \), +3 in \( \text{HNO}_2 \), +5 in \( \text{HNO}_3 \)
- Chlorine: −1 in HCl, +1 in \( \text{ClO}^- \), +3 in \( \text{ClO}_2^- \), +5 in \( \text{ClO}_3^- \), +7 in \( \text{ClO}_4^- \)
- Sulfur: −2 in H\(_2\)S, +4 in SO\(_2\), +6 in H\(_2\)SO\(_4\)
Use of Oxidation Numbers in Naming Compounds:
When elements show more than one oxidation state, Roman numerals are used in compound names to indicate the oxidation number of the central atom:
- \( \text{FeCl}_2 \): iron(II) chloride → Fe is +2
- \( \text{FeCl}_3 \): iron(III) chloride → Fe is +3
- \( \text{MnO}_4^- \): manganate(VII) → Mn is +7
Example
Identify the oxidation state of Mn in \( \text{KMnO}_4 \), and name the compound.
▶️Answer/Explanation
Example
A student is given the redox reaction:
\( \text{Cl}_2(g) + 2\text{I}^-(aq) \rightarrow 2\text{Cl}^-(aq) + \text{I}_2(s) \)
Identify the species that is oxidized and the species that is reduced. State the oxidizing and reducing agents.
▶️Answer/Explanation
Iodide ions \( \text{I}^- \) are oxidized to \( \text{I}_2 \): oxidation (−1 to 0).
Chlorine \( \text{Cl}_2 \) is reduced to \( \text{Cl}^- \): reduction (0 to −1).
Reducing agent: \( \text{I}^- \) (donates electrons)
Oxidizing agent: \( \text{Cl}_2 \) (accepts electrons)
Example
Determine the oxidation state of chromium in \( \text{Cr}_2\text{O}_7^{2-} \).
Then, use this to write the IUPAC name of \( \text{K}_2\text{Cr}_2\text{O}_7 \).
▶️Answer/Explanation
Let the oxidation state of Cr be \( x \).
Oxygen = \( -2 \), total from O = \( 7 \times (-2) = -14 \)
\( 2x – 14 = -2 \Rightarrow 2x = +12 \Rightarrow x = +6 \)
Chromium is in the \( +6 \) oxidation state.
The compound is named: potassium dichromate(VI).