Reactions of the Halide Ions

AS · 12 min

The halide ions, $\ce{Cl-}$ , $\ce{Br-}$ and $\ce{I-}$ , mirror the halogens: where the halogens get weaker as oxidising agents down the group, the halide ions get stronger as reducing agents. This note covers the two sets of reactions the syllabus specifies: the test with aqueous silver nitrate followed by aqueous ammonia, used to identify halide ions in Paper 3, and the reactions of solid halides with concentrated sulfuric acid, where the different reducing power of the ions produces strikingly different products. Both are examined with equations and observations, and both need explaining.

Halide ions as reducing agents

A halide ion acts as a reducing agent by losing an electron (being oxidised to the halogen):

\ce{2X- -> X2 + 2e-}

Key result

Reducing power of the halide ions increases down the group: $\ce{Cl-} < \ce{Br-} < \ce{I-}$ .

Down the group the ion is larger: its outer electrons are further from the nucleus and more shielded by inner shells. They are held less strongly, so they are lost more easily. Iodide is the strongest reducing agent of the three.

This is the mirror image of the halogens' oxidising power. Chlorine is the strongest oxidising agent of the three halogens because chlorine atoms hold on to electrons most strongly; for exactly the same reason chloride is the weakest reducing agent.

Testing for halide ions with silver nitrate

The method

Method

Test for halide ions in solution

Add dilute nitric acid to a few $\text{cm}^3$ of the solution. This removes ions such as carbonate, which would also form a precipitate with silver ions. (Do not use hydrochloric acid, which adds chloride ions, or sulfuric acid, since silver sulfate can precipitate.)
Add a few drops of aqueous silver nitrate. A precipitate of the silver halide forms: $\ce{Ag+(aq) + X-(aq) -> AgX(s)}$ .
Note the colour of the precipitate.
Add aqueous ammonia to the precipitate and record whether it dissolves.

The results

halide ion	precipitate with $\ce{Ag+(aq)}$	effect of adding $\ce{NH3(aq)}$
chloride, $\ce{Cl-}$	white precipitate, $\ce{AgCl}$	soluble in $\ce{NH3(aq)}$ (dissolves to give a colourless solution)
bromide, $\ce{Br-}$	cream (off-white) precipitate, $\ce{AgBr}$	partially soluble in $\ce{NH3(aq)}$ (dissolves in concentrated ammonia)
iodide, $\ce{I-}$	pale yellow precipitate, $\ce{AgI}$	insoluble in $\ce{NH3(aq)}$ (even concentrated)

The ionic equations:

\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}

\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}

\ce{Ag+(aq) + I-(aq) -> AgI(s)}

Fluoride gives no precipitate, because silver fluoride is soluble.

Why ammonia distinguishes them

The colours (white, cream, pale yellow) are close, so ammonia is used to confirm. The explanation depends on how insoluble each silver halide is.

The solubility of the silver halides decreases from AgCl to AgI. Even a "precipitate" is in equilibrium with a very small concentration of ions: $\ce{AgX(s) <=> Ag+(aq) + X-(aq)}$ .
Ammonia molecules bond to $\ce{Ag+}$ ions in solution, forming a soluble complex ion and lowering the concentration of free $\ce{Ag+(aq)}$ . By Le Chatelier's principle, the equilibrium shifts to the right and more solid dissolves.
For AgCl, the most soluble, dilute ammonia removes enough $\ce{Ag+}$ for the precipitate to dissolve completely.
For AgBr, dilute ammonia dissolves only part of it; concentrated ammonia is needed.
For AgI, the least soluble, the concentration of $\ce{Ag+}$ in equilibrium with the solid is already so low that even concentrated ammonia cannot dissolve it.

Tip

The complex ion formed is the diamminesilver(I) ion, $\ce{[Ag(NH3)2]+}$ . The syllabus states that its formation and formula are not required at AS, but you must explain the solubility pattern. You will meet $\ce{[Ag(NH3)2]+}$ again as Tollens' reagent in organic chemistry.

Tip

Silver halide precipitates darken in sunlight, as light decomposes them to silver: $\ce{2AgX -> 2Ag + X2}$ . This was the basis of photographic film. Do the test away from strong sunlight and record colours promptly.

Reactions with concentrated sulfuric acid

When concentrated sulfuric acid is added to a solid sodium or potassium halide, two kinds of reaction can happen:

Acid–base: sulfuric acid protonates the halide ion, releasing the hydrogen halide as steamy fumes. This happens for all three.
Redox: the hydrogen halide (or the halide ion) reduces the sulfuric acid. This happens only if the halide is a strong enough reducing agent, so it happens for bromide and iodide but not chloride.

Sodium chloride

\ce{NaCl(s) + H2SO4(l) -> NaHSO4(s) + HCl(g)}

Observation: steamy white fumes of hydrogen chloride. No further reaction: chloride ions are too weak a reducing agent to reduce sulfuric acid. Sulfur stays at $+6$ .

Sodium bromide

First, the acid–base reaction:

\ce{NaBr(s) + H2SO4(l) -> NaHSO4(s) + HBr(g)}

Then some of the hydrogen bromide reduces the concentrated sulfuric acid to sulfur dioxide:

\ce{2HBr(g) + H2SO4(l) -> Br2(g) + SO2(g) + 2H2O(l)}

Observations: steamy fumes (HBr) and orange-brown fumes (bromine); sulfur dioxide is colourless with a choking smell. Oxidation numbers: Br from $-1$ to 0 (oxidised); S from $+6$ to $+4$ (reduced).

Sodium iodide

\ce{NaI(s) + H2SO4(l) -> NaHSO4(s) + HI(g)}

Iodide is the strongest reducing agent, so it reduces sulfur much further, all the way to $-2$ . Several reactions happen together:

\ce{2HI(g) + H2SO4(l) -> I2(s) + SO2(g) + 2H2O(l)}

\ce{6HI(g) + H2SO4(l) -> 3I2(s) + S(s) + 4H2O(l)}

\ce{8HI(g) + H2SO4(l) -> 4I2(s) + H2S(g) + 4H2O(l)}

Observations: only a little steamy fume (most HI is oxidised); purple fumes of iodine vapour and a black solid (iodine); a yellow solid (sulfur); a smell of bad eggs (hydrogen sulfide).

Key result

halide	products	most reduced sulfur product	oxidation number of S in it	observations
$\ce{NaCl}$	$\ce{HCl}$	none (no redox)	$+6$ (unchanged)	steamy fumes
$\ce{NaBr}$	$\ce{HBr}$ , $\ce{Br2}$ , $\ce{SO2}$	$\ce{SO2}$	$+4$	steamy fumes, orange-brown fumes
$\ce{NaI}$	$\ce{HI}$ , $\ce{I2}$ , $\ce{SO2}$ , $\ce{S}$ , $\ce{H2S}$	$\ce{H2S}$	$-2$	purple fumes, black solid, yellow solid, bad-egg smell

The further the sulfur is reduced, the stronger the reducing agent: this is direct evidence that reducing power increases $\ce{Cl-} < \ce{Br-} < \ce{I-}$ .

Building the redox equations from half-equations

You may be asked to construct these equations. Use half-equations (see Formulas and equations for balancing).

Oxidation: $\ce{2Br- -> Br2 + 2e-}$

Reduction: $\ce{H2SO4 + 2H+ + 2e- -> SO2 + 2H2O}$

Add them (electrons cancel):

\ce{2Br- + H2SO4 + 2H+ -> Br2 + SO2 + 2H2O}

Since $\ce{2Br- + 2H+}$ is the same as $\ce{2HBr}$ , this matches $\ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O}$ .

For hydrogen sulfide: $\ce{H2SO4 + 8H+ + 8e- -> H2S + 4H2O}$ , combined with $\ce{8I- -> 4I2 + 8e-}$ , gives $\ce{8I- + H2SO4 + 8H+ -> 4I2 + H2S + 4H2O}$ .

Worked examples

Identifying a halide ion

A solution of an unknown sodium halide is acidified with dilute nitric acid, and aqueous silver nitrate is added. A cream precipitate forms. When dilute aqueous ammonia is added, the precipitate partially dissolves. Identify the halide ion and write the ionic equation.

Solution

Cream precipitate, partially soluble in dilute ammonia: bromide.

\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}

Explaining the ammonia test

Explain why silver chloride dissolves in dilute aqueous ammonia but silver iodide does not.

Solution

Each precipitate is in equilibrium with its ions: $\ce{AgX(s) <=> Ag+(aq) + X-(aq)}$ . Ammonia forms a complex with $\ce{Ag+(aq)}$ , lowering $\ce{[Ag+(aq)]}$ , so the position of equilibrium shifts to the right and solid dissolves.

Silver chloride is more soluble than silver iodide, so this shift is enough to dissolve all the AgCl. Silver iodide is so insoluble that $\ce{[Ag+(aq)]}$ is already extremely low; ammonia cannot lower it enough, so AgI does not dissolve.

Concentrated sulfuric acid and oxidation numbers

Concentrated sulfuric acid is added to solid sodium bromide. Write equations for the two reactions that occur, state two observations, and use oxidation numbers to show which reaction is redox.

Solution

$\ce{NaBr(s) + H2SO4(l) -> NaHSO4(s) + HBr(g)}$ : not redox (Br stays $-1$ , S stays $+6$ ). This is an acid–base reaction.

$\ce{2HBr(g) + H2SO4(l) -> Br2(g) + SO2(g) + 2H2O(l)}$ : redox. Br: $-1 \to 0$ , oxidised. S: $+6 \to +4$ , reduced.

Observations: steamy fumes (HBr) and orange-brown fumes (bromine).

Exam-style: identifying a halide by mass of precipitate

$0.500\ \text{g}$ of a sodium halide, NaX, is dissolved in water and excess acidified silver nitrate is added. The dried precipitate of AgX has a mass of $0.912\ \text{g}$ . Identify X and predict the colour of the precipitate and its behaviour with aqueous ammonia. ( $A_r$ : Na 23.0, Ag 107.9)

Solution

$n(\ce{NaX}) = n(\ce{AgX})$ , so with $A_r(\ce{X}) = x$ :

\frac{0.500}{23.0 + x} = \frac{0.912}{107.9 + x}

$0.500(107.9 + x) = 0.912(23.0 + x)$

$53.95 + 0.500x = 20.98 + 0.912x$

$32.97 = 0.412x$ , so $x = 80.0$ : bromine, X is bromide.

The precipitate AgBr is cream and is partially soluble in dilute aqueous ammonia (soluble in concentrated ammonia).

Exam-hard: sodium iodide with concentrated sulfuric acid

(a) Write half-equations for the oxidation of iodide ions and for the reduction of sulfuric acid to sulfur, and combine them into an overall ionic equation.

(b) Explain why sodium iodide produces sulfur and hydrogen sulfide with concentrated sulfuric acid, but sodium chloride produces neither.

(c) $1.66\ \text{g}$ of potassium iodide is treated with excess concentrated sulfuric acid. Assuming all the iodide is converted to iodine by the reaction producing sulfur dioxide only, calculate the maximum mass of iodine formed and the volume of $\ce{SO2}$ at room conditions. ( $A_r$ : K 39.1, I 126.9)

Solution

(a) Oxidation: $\ce{2I- -> I2 + 2e-}$ (multiply by 3: $\ce{6I- -> 3I2 + 6e-}$ ).

Reduction: sulfur goes from $+6$ to 0, gaining 6 electrons: $\ce{H2SO4 + 6H+ + 6e- -> S + 4H2O}$ .

Overall: $\ce{6I- + H2SO4 + 6H+ -> 3I2 + S + 4H2O}$ . Check charge: left $-6 + 6 = 0$ ; right 0.

(b) Iodide ions are large, so their outer electrons are far from the nucleus and well shielded, and are lost easily: iodide is a strong reducing agent, strong enough to reduce sulfur from $+6$ to 0 (S) and $-2$ ( $\ce{H2S}$ ). Chloride ions are small, hold their electrons strongly and are too weak a reducing agent to reduce sulfuric acid at all, so only the acid–base reaction producing HCl occurs.

$n(\ce{I2}) = 0.00500\ \text{mol}$ ; mass $= 0.00500 \times 253.8 = 1.27\ \text{g}$ .

$n(\ce{SO2}) = 0.00500\ \text{mol}$ ; volume $= 0.00500 \times 24.0 = 0.120\ \text{dm}^3$ ( $120\ \text{cm}^3$ ).

Watch out

Acidify with dilute nitric acid, not hydrochloric or sulfuric acid, before adding silver nitrate.
The colours are white, cream, pale yellow. Do not call AgI "yellow-orange" or AgBr "yellow".
Concentrated sulfuric acid reacts with solid halides. In dilute aqueous solution none of these redox reactions occur.
With NaCl, the only product gas is HCl. Writing $\ce{Cl2}$ as a product is a common error: chloride cannot reduce sulfuric acid.
The halide ion is the reducing agent; it is oxidised. Sulfuric acid is the oxidising agent; it is reduced.

Exam tip

The silver nitrate test answer needs three things: the reagent sequence (dilute $\ce{HNO3}$ , then $\ce{AgNO3(aq)}$ , then $\ce{NH3(aq)}$ ), the precipitate colour, and the solubility in ammonia.
For concentrated sulfuric acid questions, examiners often ask for the oxidation number of sulfur in each product: $\ce{SO2}$ $+4$ , $\ce{S}$ 0, $\ce{H2S}$ $-2$ .
"Explain the different reactions" questions want the link: increasing ionic radius and shielding, electrons lost more easily, increasing reducing power, so sulfur is reduced to a lower oxidation number.
Observations are separate marks: steamy fumes (HX), orange fumes ( $\ce{Br2}$ ), purple fumes or black solid ( $\ce{I2}$ ), yellow solid (S), bad-egg smell ( $\ce{H2S}$ ).

Practical skills

In Paper 3 the silver nitrate test appears in qualitative analysis. Use about $1\ \text{cm}^3$ of the unknown, add dilute nitric acid until acidic, then a few drops of $\ce{AgNO3(aq)}$ . Record the colour precisely, then add aqueous ammonia (several $\text{cm}^3$ , enough to be in excess) and shake. Record "precipitate dissolves to give a colourless solution", "precipitate partially dissolves" or "no change". Comparing with known chloride, bromide and iodide solutions side by side makes the cream/pale yellow distinction much easier. Reactions with concentrated sulfuric acid are teacher demonstrations in a fume cupboard and are not part of Paper 3 tests.

Summary

Reducing power: $\ce{Cl-} < \ce{Br-} < \ce{I-}$ (larger ion, outer electrons less strongly held).
Silver nitrate test (after dilute $\ce{HNO3}$ ): AgCl white, soluble in $\ce{NH3(aq)}$ ; AgBr cream, partially soluble; AgI pale yellow, insoluble.
Ammonia lowers $\ce{[Ag+]}$ by complexing; this dissolves AgCl (most soluble) but not AgI (least soluble).
Conc. $\ce{H2SO4}$ + NaCl: HCl only (steamy fumes).
- NaBr: HBr, then $\ce{Br2}$ and $\ce{SO2}$ (S $+6 \to +4$ ); orange fumes.
- NaI: HI, then $\ce{I2}$ with $\ce{SO2}$ , S and $\ce{H2S}$ (S down to $-2$ ); purple fumes, black and yellow solids, bad-egg smell.

Practice

Question

State the colour of the precipitate formed when acidified silver nitrate solution is added to solutions of chloride, bromide and iodide ions.
Explain why dilute nitric acid, rather than dilute hydrochloric acid, is added before the silver nitrate.
Write the equation for the reaction of concentrated sulfuric acid with solid sodium chloride, and explain why no further reaction occurs.
Give the oxidation numbers of sulfur in $\ce{H2SO4}$ , $\ce{SO2}$ , $\ce{S}$ and $\ce{H2S}$ .
Explain why the reducing power of halide ions increases from chloride to iodide.
Calculate the mass of silver chloride precipitated when excess silver nitrate is added to $25.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ sodium chloride. ( $A_r$ : Ag 107.9, Cl 35.5)
Construct the equation for the reaction of hydrogen iodide with concentrated sulfuric acid to form iodine, hydrogen sulfide and water, using half-equations.
$2.06\ \text{g}$ of sodium bromide reacts with excess concentrated sulfuric acid. Assuming all the bromide is converted to bromine, calculate the mass of bromine and the volume of sulfur dioxide (room conditions) formed. ( $A_r$ : Na 23.0, Br 79.9)
A solution may contain chloride ions, iodide ions, or both. Describe how silver nitrate and aqueous ammonia could be used to show whether both are present.
A $0.150\ \text{g}$ sample of impure sodium chloride is dissolved in water and treated with excess acidified silver nitrate. The precipitate is filtered, washed, dried and weighed: $0.358\ \text{g}$ . Calculate the percentage purity of the sodium chloride, and suggest one reason why the precipitate should be dried in the dark and one reason why it must be washed. ( $A_r$ : Na 23.0, Cl 35.5, Ag 107.9)

Answers

Chloride: white. Bromide: cream. Iodide: pale yellow.
The acid removes ions such as carbonate that would also form a precipitate with $\ce{Ag+}$ . Hydrochloric acid would add chloride ions, which give a white precipitate of AgCl whatever the unknown contains.
$\ce{NaCl(s) + H2SO4(l) -> NaHSO4(s) + HCl(g)}$ . Chloride ions are small and hold their electrons strongly, so they are too weak a reducing agent to reduce sulfuric acid; HCl is not oxidised to $\ce{Cl2}$ .
$\ce{H2SO4}$ $+6$ ; $\ce{SO2}$ $+4$ ; $\ce{S}$ 0; $\ce{H2S}$ $-2$ .
From $\ce{Cl-}$ to $\ce{I-}$ the ion is larger, with more shells; the outer electrons are further from the nucleus and more shielded, so they are attracted less strongly and are lost more easily. A species that loses electrons more easily is a stronger reducing agent.
$n(\ce{NaCl}) = 0.100 \times 25.0 / 1000 = 2.50 \times 10^{-3}\ \text{mol} = n(\ce{AgCl})$ . $M(\ce{AgCl}) = 143.4$ ; mass $= 0.359\ \text{g}$ (0.3585 g).
Oxidation: $\ce{8I- -> 4I2 + 8e-}$ . Reduction: $\ce{H2SO4 + 8H+ + 8e- -> H2S + 4H2O}$ . Overall: $\ce{8I- + H2SO4 + 8H+ -> 4I2 + H2S + 4H2O}$ , i.e. $\ce{8HI + H2SO4 -> 4I2 + H2S + 4H2O}$ .
$n(\ce{NaBr}) = 2.06 / 102.9 = 0.02002\ \text{mol}$ . $\ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O}$ : $n(\ce{Br2}) = n(\ce{SO2}) = 0.01001\ \text{mol}$ . Mass of $\ce{Br2}$ $= 0.01001 \times 159.8 = 1.60\ \text{g}$ . Volume of $\ce{SO2}$ $= 0.01001 \times 24.0 = 0.240\ \text{dm}^3$ .
Acidify with dilute nitric acid and add excess silver nitrate. Filter (or allow to settle) and add dilute aqueous ammonia to the precipitate. If only chloride is present, the white precipitate dissolves completely. If only iodide is present, the pale yellow precipitate does not dissolve at all. If both are present, part of the precipitate dissolves (the AgCl) and a pale yellow solid (AgI) remains; acidifying the ammonia filtrate with nitric acid re-forms a white precipitate of AgCl, confirming chloride.
$n(\ce{AgCl}) = 0.358 / 143.4 = 2.497 \times 10^{-3}\ \text{mol} = n(\ce{NaCl})$ . Mass of $\ce{NaCl} = 2.497 \times 10^{-3} \times 58.5 = 0.1461\ \text{g}$ . Purity $= 0.1461 / 0.150 \times 100 = 97.4\%$ . Dry in the dark because light decomposes silver chloride to silver and chlorine, changing the mass. Wash to remove soluble impurities (excess silver nitrate, sodium nitrate) that would otherwise add to the dried mass.

Halide ions as reducing agents

Testing for halide ions with silver nitrate

The method

The results

Why ammonia distinguishes them

Reactions with concentrated sulfuric acid

Sodium chloride

Sodium bromide

Sodium iodide

Building the redox equations from half-equations

Worked examples

Practice

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