Edexcel IAL Chemistry 16.1 oxidation and reduction

Track the element that changes oxidation number, then state whether electrons are gained or lost and use that change to justify the redox decision.

Exam points

Use oxidation-number changes to justify whether a reaction is redox or not.
Link reduction to electron gain and a decrease in oxidation number in MCQ contexts.
Identify which species is oxidised or reduced from balanced equations or half-cells.

[Maximum number: 2]

This question is about mercury, Hg, and its compounds.
Mercury is a liquid element in the same group of the Periodic Table as zinc.
The electronic configuration of mercury is $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^{2}$ .

Mercury reacts with nitric acid to form an aqueous solution of $\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$ and nitrogen monoxide gas.

The unbalanced equation is shown.

\mathrm{Hg}(\mathrm{l})+\mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{NO}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})

Explain, using oxidation numbers, why this is a redox reaction.
(2)

An explanation that makes reference to the following
points:

Allow oxidation numbers from annotation to the equation
Ignore any reference to electron loss/gain

(2)

Expert

- Hg/mercury oxidised and
from 0 (in Hg ) to +2 (in $\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}$ )

(1)

Do not award reference to oxidation of any other element

- N / nitrogen is reduced
and
from +5 (in $\mathrm{HNO}_{3}$ ) to +2 (in NO)

(1)

Do not award $\mathrm{HNO}_{3}$ is reduced
Do not award reference to reduction of any other element

If no other mark awarded,
Hg/mercury oxidised and N/nitrogen reduced
OR
Hg / mercury from 0 to +2 and N / nitrogen from +5 to +2
scores (1)

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16.1—The terms ‘oxidation’ and ‘reduction’ in terms of electron transfer and changes in oxidation number, applied to s-, p- question 1