Halide Ions: Reducing Ability & Sulfuric Acid Reactions Revision Notes | A Level AQA Chemistry

1. The Core Question: Why Do Halide Ions Differ as Reducing Agents?

Picture four test tubes, each containing a different sodium halide solid. You add concentrated sulphuric acid to each one. In the first tube, almost nothing interesting happens. In the last tube, you get a foul-smelling gas and a black solid forming before your eyes. Same acid, same type of reaction — yet completely different outcomes. The reason lies in a trend that runs straight down Group 7.

The key variable is reducing ability: how readily each halide ion donates electrons. As you move from fluoride ($F^-$) to iodide ($I^-$), the halide ions become progressively stronger reducing agents. Understanding why — and what it looks like in practice — is the heart of this topic.

2. The Trend in Reducing Ability: From Fluoride to Iodide

The reducing ability of halide ions increases down the group: $F^- < Cl^- < Br^- < I^-$. To understand why, think about what a reducing agent actually does — it donates electrons. The question is: how tightly does each ion hold onto its outer electrons?

Fluoride is a tiny ion with only two electron shells. Its outer electrons are close to the nucleus and strongly attracted to it — they are very difficult to remove. Iodide, by contrast, has five electron shells. Its outermost electrons are far from the nucleus, well-shielded, and comparatively straightforward to remove. This is why iodide is the strongest reducing agent of the four halide ions.

Notice the pattern: the stronger the reducing agent, the more it can reduce the sulphuric acid. Chloride barely manages to reduce it at all; iodide reduces it all the way to hydrogen sulfide. This graduated response is exactly what the exam expects you to explain and use as evidence.

3. Reactions of Sodium Halides with Concentrated Sulphuric Acid

Now the trend becomes concrete. When each solid sodium halide reacts with concentrated sulphuric acid, two things can happen. First, there is always an acid–base reaction producing the hydrogen halide gas. Second — and this is where the reducing ability matters — the hydrogen halide (or the halide ion itself) may then go on to reduce the sulphuric acid further.

Work through each halide in turn. The pattern builds logically, so follow the oxidation states carefully — they are your evidence for what is being oxidised and what is being reduced.

3a. Sodium Fluoride and Sodium Chloride

With sodium fluoride, only a simple acid–base reaction occurs. Concentrated sulphuric acid protonates the fluoride ion to produce hydrogen fluoride gas. No redox takes place because fluoride is such a weak reducing agent that it cannot reduce the sulphuric acid.

Sodium chloride behaves similarly. The reaction produces steamy fumes of hydrogen chloride gas, and that is essentially all that happens under typical conditions. Chloride is a weak enough reducing agent that concentrated sulphuric acid is not reduced to any significant extent.

The observation for both reactions is misty/steamy fumes of the hydrogen halide gas. For HCl, these fumes turn damp litmus paper red and produce white smoke with ammonia — useful confirmatory tests. Crucially, the sulphur in $H_2SO_4$ remains at oxidation state +6 throughout, confirming no reduction has occurred.

3b. Sodium Bromide — Moderate Reducing Ability

Bromide is a stronger reducing agent than chloride, and this shows up clearly in the reaction. The initial acid–base step still occurs, producing HBr gas. However, bromide ions are polarisable enough to then reduce the sulphuric acid. The sulphur is reduced from oxidation state +6 in $H_2SO_4$ down to +4 in $SO_2$.

The observations here are distinctive: misty HBr fumes initially, followed by orange-brown fumes of bromine vapour and the choking smell of sulphur dioxide. The appearance of bromine is your key visual evidence that a redox reaction has occurred — bromide has been oxidised.

3c. Sodium Iodide — Strong Reducing Ability

Iodide is the strongest reducing agent of the halide ions, and the reaction with concentrated sulphuric acid reflects this dramatically. Not only is $SO_2$ produced, but the iodide reduces the sulphuric acid further — all the way to elemental sulphur ($S$) and even hydrogen sulfide ($H_2S$). Multiple redox products form simultaneously.

The observations for sodium iodide are the most dramatic: misty HI fumes, purple/violet vapour and grey-black solid of iodine, yellow solid sulphur, and the unmistakable smell of rotten eggs from hydrogen sulfide. This combination of products is direct evidence that iodide is a far stronger reducing agent than bromide or chloride.

4. Tracking Oxidation States: The Redox Evidence

Oxidation state changes are the language of redox, and examiners expect you to use them as evidence. In every reaction above, the halide ion starts at oxidation state -1. When it acts as a reducing agent, it is oxidised — its oxidation state increases towards 0 (forming the halogen molecule). The sulphur in sulphuric acid starts at +6 and is reduced to lower values depending on how strong the reducing agent is.

Worked example: Tracking Oxidation State Changes in the NaI Reaction

For the reaction $8NaI_{(s)} + 5H_2SO_4_{(l)} \rightarrow 4Na_2SO_4_{(s)} + H_2S_{(g)} + 4I_2_{(s)} + 4H_2O_{(l)}$, identify what is oxidised and what is reduced, and state the change in oxidation state for each.

1. Na is +1, overall charge is 0, so I = -1 — Iodide ion carries a charge of -1 in the ionic compound.
2. $I_2$ is an element in its standard state, so oxidation state = 0 — All elements in their standard state have an oxidation state of zero.
3. Iodine: $-1 \rightarrow 0$, an increase of +1 per iodine atom — An increase in oxidation state = oxidation. Iodide is the reducing agent — it donates electrons.
4. H = +1 (×2 = +2), O = -2 (×4 = -8), overall = 0: so S = +6 — Using the rule that the sum of oxidation states in a neutral compound equals zero: +2 + S + (-8) = 0, giving S = +6.
5. H = +1 (×2 = +2), overall = 0: so S = -2 — Sum of oxidation states: +2 + S = 0, giving S = -2.
6. Sulphur: $+6 \rightarrow -2$, a decrease of 8 — A decrease in oxidation state = reduction. Sulphuric acid is the oxidising agent — it accepts electrons from iodide.

Answer: Iodide is oxidised (-1 to 0); sulphur in $H_2SO_4$ is reduced (+6 to -2). Iodide is the reducing agent; $H_2SO_4$ is the oxidising agent.

Compare this with the bromide reaction, where sulphur only goes from +6 to +4. The greater the drop in oxidation state of sulphur, the stronger the reducing agent that caused it. This is the quantitative backbone of the trend.

5. Summary of Observations and Products

Pulling the four reactions together into a comparison is exactly the kind of task an examiner will set. The table below consolidates the key observations and products — this is the information you need to recall and apply under exam conditions.

The progression from no redox (fluoride, chloride) through moderate redox (bromide) to extensive redox (iodide) is the trend in reducing ability made visible. Each additional product formed by iodide represents a deeper reduction of sulphuric acid, which is only possible because iodide can donate electrons so readily.

6. Exam Technique: Scoring Full Marks on This Topic

This topic rewards precision. Examiners are looking for specific chemical language, correct oxidation states, and observations tied to named products. The following guidance targets the most common ways marks are dropped.

The reducing ability trend is one of the most reliably examined aspects of Group 7 chemistry. Master the three-step explanation (ionic radius → shielding → ease of electron donation), know your observations for each reaction, and track oxidation states carefully — those are the tools that convert understanding into marks.

Worked example: Progression of halide reactions with concentrated sulphuric acid

Compare what happens when fluoride, chloride, bromide, and iodide ions are treated with concentrated sulphuric acid.

1. F- and Cl- do not reduce concentrated sulphuric acid under normal conditions; they mainly give HF or HCl in an acid-base reaction. — They are weaker reducing agents than Br- and I-.
2. Br- reduces some H2SO4 to SO2 while being oxidised to Br2. — Bromide is a stronger reducing agent than chloride.
3. I- reduces H2SO4 more strongly, giving products such as SO2, S, and H2S while iodide is oxidised to I2. — Iodide is the strongest reducing agent of the common halide ions in this context.