9701 · 6.1

Redox processes: electron transfer and changes in oxidation number (oxidation state) — practice questions

Practice and worked examples for 9701 Redox processes: electron transfer and changes in oxidation number (oxidation state). Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Determine the oxidation number of chromium in the dichromate(VI) ion, $\text{Cr}_2\text{O}_7^{2-}$ .

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The overall charge of the ion is -2, so the sum of oxidation numbers must be -2.
Assign the known oxidation number: Oxygen is -2 (as it's not a peroxide or bonded to fluorine).
Let the oxidation number of Cr be $x$ . There are two Cr atoms and seven O atoms.
Set up the equation: $(2 \times x) + (7 \times -2) = -2$ .
Solve for $x$ : $2x - 14 = -2 \implies 2x = +12 \implies x = +6$ .
The oxidation number of chromium in $\text{Cr}_2\text{O}_7^{2-}$ is +6.

Worked example 2

Consider the reaction: $2\text{FeCl}_3(\text{aq}) + \text{SnCl}_2(\text{aq}) \rightarrow 2\text{FeCl}_2(\text{aq}) + \text{SnCl}_4(\text{aq})$ . Identify the oxidising agent and the reducing agent, justifying your answer with oxidation numbers.

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Assign oxidation numbers to Fe:
- In $\text{FeCl}_3$ , Cl is -1, so Fe must be +3.
- In $\text{FeCl}_2$ , Cl is -1, so Fe must be +2.
- The oxidation number of Fe decreases from +3 to +2. This is reduction.
Assign oxidation numbers to Sn:
- In $\text{SnCl}_2$ , Cl is -1, so Sn must be +2.
- In $\text{SnCl}_4$ , Cl is -1, so Sn must be +4.
- The oxidation number of Sn increases from +2 to +4. This is oxidation.
Identify Agents:
- $\text{FeCl}_3$ contains the Fe that is reduced, so $\text{FeCl}_3$ is the oxidising agent.
- $\text{SnCl}_2$ contains the Sn that is oxidised, so $\text{SnCl}_2$ is the reducing agent.

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