Reactions of Halides

Question 1

Explain the trend in reducing power of the halides

Answer 1

The reducing power of the halides increase down the group.
-This is because as you go down the group the IONS get bigger
-The number of shells increase so the electrons are further away from the positive nucleus
-Therefore the attraction between the nucleus and the outer shell electron is weaker. Making it easier for the halide to lose an electron

Question 2

Explain the reaction of KF and KCL with sulphuric acid

Answer 2

KF(s) + H2SO4(l) → KHSO4(s) + HF(g)
Produces misty fumes because of the Hydrogen fluoride

KCl(s) + H2SO4(l) → KHSO4(s) + HCl(g)
Produces misty fumes because of the hydrogen chloride

Both the fluoride ions and the chloride ions aren’t strong enough to reduce again, so the reaction stops there. It is not a redox reaction

Question 3

Write the equation for concentrated Sulfuric acid partial ionisation

Answer 3

H2SO4 ⇌ H+ + HSO4-
Sulfuric acid, especially when concentrated, can act as an oxidising agent, it itself is being reduced. But the extent of its reduction and the products formed depends on the species being oxidised
The three possible reduction products are
-Sulfur dioxide
-Sulfur
-Hydrogen sulfide

Question 4

Write the three different hand equations repenting its oxidising action

Answer 4

1) H2SO4 + 2H+ + 2e- → 2H2O + SO2
+6 +4
2)H2SO4 + 6H+ + 6e- → 4H2O + S
+6 0
3) H2SO4 + 8H+ + 8e- → 4H2O + H2S
+6 -2

Question 5

Stat the observation and products of the sodium halide NaCl with concentrated sulfuric acid

Answer 5

Observation: Misty fumes
Products: Hydrogen chloride (HCl)

Question 5

Stat the observation and products of the sodium halide NaBr with concentrated sulfuric acid

Answer 5

Observation: -Misty fumes
-Brown fumes
-Colourless gas with choking smell
Products: -Hydrogen bromide (HBr)
-Bromine (Br)
-Sulfur dioxide (SO2)

Question 6

Stat the observation and products of the sodium halide Nal with concentrated sulfuric acid

Answer 6

Observation: -Misty fumes
-Purple fumes or black solid
-Colourless gas with choking smell
-Yellow solid
-Colourless gas with rotten egg smell
Products: -Hydrogen iodide (HI)
-Iodine (I2)
-Sulfur dioxide (SO2)
-Sulfur (S)
-Hydrogen sulfide (H2S)

Question 7

Explain the the sulfuric acid acts as an oxidising agent for the three halides

Answer 7

-With sodium chloride, the sulfuric acid behaves only as an acid and not as an oxidising agent. This because the chloride ions have low reducing power
-With sodium bromide, the greater reducing power of bromide ions cause the sulfuric acid to be reduced
-With sodium iodide, the much greater reducing power of iodide ions cause the sulfuric acid to be reduced

Question 8

State the equation of sodium chloride with concentrated sulfuric acid

Answer 8

NaCl + H2SO4 → NaHSO4 + HCl
Only one equation because no redox reaction are occurring

Question 9

State the equation of sodium bromide with concentrated sulfuric acid

Answer 9

NaBr + H2S04 → NaHSO4 + HBr

The half equations:
2Br- → Br2 + 2e-
H2SO4 + 2H+ + 2e-→ 2H2O + SO2

Add them together
2Br- + H2SO4 + 2H+ → 2H2O+ SO2 + Br2
2HBr + H2SO4 → 2H2) + SO2 + Br2
This equation represents the oxidation of the misty fumes and hydrogen bromide

Question 10

State the equation of sodium Iodide with concentrated sulfuric acid

Answer 10

NaI + H2S04 → NaSO4 + HI
This equation represents the formation of misty fumes and hydrogen iodide

The half equations:
2I- → I2 + 2e- (Need to x3 to get same electrons)
H2SO4 + 6H+ + 6e- → 4H2O + S

Add them together
6I- + H2SO4 +6H+ → 4H2SO4 + S + 3I2
6HI + H2SO4 → 4H2SO4 + S + 3I2
This equation represents the oxidation of the misty fumes and hydrogen iodide

Question 11

What is the test for halide ions

Answer 11

-Add dilute nitric acid(To make sure that any other anions like carbonate ions are removed because they would also form a precipitate
-Then add dilute silver nitrate solution
-If a precipitate is obtained then add some ammonia solution can be dilute of concentrated

Question 12

The result obtained for the halide ions

Answer 12

Add silver nitrate solution:
Chloride ions = White precipitate
Bromide ions = Cream precipitate
Iodine ions = Yellow precipitate

Add dilute aqueous ammonia:
Chloride ions = Soluble (precipitate dissolves to give a colourless solution)
Bromide ions = Insoluble ( Precipitate remains unchanged) Iodide ions = insoluble ( Precipitate remains unchanged)

Add concentrated aqueous ammonia:
Chloride ions = Soluble (precipitate dissolves to give a colourless solution)
Bromide ions = Soluble (Precipitate dissolves to give a colourless solution)
iodide Ions = Insoluble ( Precipitate remains unchanged)

Question 13

State the feral ions equation for the formation of the precipitate

Answer 13

Ag+(aq) + X-(aq) → AgX(s)

Question 14

Write the equation for Sodium chloride with the silver nitrate solution

Answer 14

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

Question 15

Why do we use aqueous ammonia for testing halide ions

Answer 15

Because the precipitates from the halides have different solubilities
-Silver chloride dissolves readily in both dilute and concentrated aqueous ammonia
-Silver bromide dissolves really in concentrated aqueous ammonia, but not in dilute aqueous ammonia
-Silver nitrate dissolves in neither

Dissolving occur because of the formation of a complex ion. In the case of silver chloride
AgCl(s) + 2NH3(aq) → (Ag(NH3)2)+(aq) + Cl-(Aq)

Question 16

Are the hydrogen halides polar

Answer 16

All the hydrogen halides are colourless gasses and exists a polar diatomic molecules

Question 17

Test for carbonates and hydrogencarbonates Ions

Answer 17

Add dilute hydrochloric acid - Carbonates and hydro carbonates will fiz because they give of CO2

Question 18

Test for sulfate Ions

Answer 18

• Add dilute acid (Can be HCL or Nitric acid)( This is added to Gert rid of carbonate ions as they also form precipitate)
  -Add barium chloride solution
  -If a white precipitate forms of barium sulfate forms that indicates the compound contains a sulfate

The precipitate forming:
Ba2+ + SO42- → BaSO4

Question 19

Test for ammonium ions

Answer 19

-Add some sodium hydroxide, to the mystery substance in a test tube
-Gently heat the mixture
-Hold over a piece of damp red litmus paper
-If there’s ammonia present, it’ll dissolve the water on the damp litmus paper, Turing it blue
NH4+ + OH- → NH3(g) + H20(I)