Group 17: Question 8

Worked solution

Part (a): Sodium chloride with concentrated sulfuric acid

Concentrated sulfuric acid protonates the chloride ion, displacing hydrogen chloride gas:

$\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(aq)} + \text{HCl(g)}$

Observation: steamy/misty white fumes of $\text{HCl}$ gas are seen escaping (they fume in moist air as $\text{HCl}$ dissolves in atmospheric water vapour).

Why no redox occurs: in this reaction, chlorine stays as $\text{Cl}^-$ throughout (it is not oxidised to $\text{Cl}_2$, and sulfur stays as $+6$) it is not reduced. Chloride is too weak a reducing agent (its outer electrons are held tightly, being a small ion) to reduce sulfur from $+6$ in $\text{H}_2\text{SO}_4$. The reaction is therefore only a proton-transfer (acid–base) reaction, not a redox reaction.

Part (b): Sodium bromide with concentrated sulfuric acid

Bromide is a stronger reducing agent than chloride. The initially formed $\text{HBr}$ is further oxidised by concentrated sulfuric acid, which is itself reduced to sulfur dioxide. Combining the initial acid–base step with this redox step gives the overall balanced equation:

$2\text{NaBr(s)} + 3\text{H}_2\text{SO}_4\text{(l)} \rightarrow 2\text{NaHSO}_4\text{(aq)} + \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$

Checking the balance: Na: $2=2$; Br: $2=2$; S: $3=2+1$; H: $6=2+4$; O: $12=8+2+2$. All atoms balance.

Oxidation number of sulfur: in $\text{H}_2\text{SO}_4$, sulfur is $\boxed{+6}$; in $\text{SO}_2$, oxygen is $-2$ (two of them, total $-4$) and the molecule is neutral, so sulfur is $\boxed{+4}$, a reduction of $2$. This is balanced by two bromide ions each losing one electron ($\text{Br}^- \rightarrow \tfrac{1}{2}\text{Br}_2 + \text{e}^-$), giving the $2:1$ ratio of $\text{Br}^-$ to $\text{S}$ reduced seen in the equation.

Observation: steamy fumes are again seen (from the acid–base step), together with orange/red-brown fumes as bromine gas is released, and the sharp, choking (pungent, acrid) smell of sulfur dioxide gas.

Part (c): Explaining the trend in reducing power

A halide ion acts as a reducing agent by losing an electron (being oxidised) to whatever it reduces. Down Group 17, from $\text{Cl}^-$ to $\text{Br}^-$ to $\text{I}^-$, ionic radius increases, because each successive ion has one more occupied electron shell. The outer electrons of a larger ion are further from the nucleus and are shielded by more inner shells of electrons, so they experience a weaker net attraction to the nucleus and are held less tightly.

This means larger halide ions lose their outer electron more readily, so reducing power increases down the group: $\text{I}^-$ is the strongest reducing agent (reducing sulfur all the way to $\text{H}_2\text{S}$), $\text{Br}^-$ is intermediate (reducing sulfur only as far as $\text{SO}_2$), and $\text{Cl}^-$ is too weak to reduce sulfuric acid at all.

Final answers

• (a) $\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(aq)} + \text{HCl(g)}$; steamy white fumes; no redox as chloride is too weak a reducing agent.
• (b) $2\text{NaBr(s)} + 3\text{H}_2\text{SO}_4\text{(l)} \rightarrow 2\text{NaHSO}_4\text{(aq)} + \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$; sulfur goes from $+6$ to $+4$; steamy fumes plus orange/red-brown bromine fumes and a choking smell of $\text{SO}_2$.
• (c) Reducing power increases $\text{Cl}^- < \text{Br}^- < \text{I}^-$ because ionic radius increases down the group, so the outer electron is held less tightly and is lost more easily.