StudyXY
RegisterLog in
Back to AI Flashcard MakerChemistry /A-level Chemistry: 3.2.3 Group 7, The Halogens Part 2

A-level Chemistry: 3.2.3 Group 7, The Halogens Part 2

Chemistry34 CardsCreated 2 months ago

These flashcards outline the qualitative test for halide ions using silver nitrate, including expected precipitate colours for each halide and the purpose of adding dilute nitric acid to avoid interference from other ions.

PrintEmbedImportReport

Test for Halides

What is the result for fluoride?

No precipitate

Tap or swipe ↕ to flip
Swipe ←→Navigate
1/34

Key Terms

Term
Definition

Test for Halides

What is the result for fluoride?

No precipitate

Test for Halides

What is the result for chloride?

White precipitate of silver chloride

Test for Halides

What is the result for bromide?

Cream precipitate of silver bromide

Test for Halides

What is the result for iodide?

Yellow precipitate of silver iodide

Test for Halides

Why do you add dilute nitric acid?

To remove ions which may interfere with test

Test for Halides

Write the general equation

Ag⁺(aq) + X⁻(aq) → AgX(s)
(where X⁻ = Cl⁻, Br⁻, I⁻)

Log in to view all terms

Related Flashcard Decks

Chemistry

CSEC Chemistry: States of Matter and Particulate Theory

This deck covers the fundamental concepts of states of matter and the particulate theory, including diffusion, osmosis, and the characteristics of solids, liquids, and gases.

10 cards
View Deck
Chemistry

Understanding Redox Reactions

This deck covers the fundamental concepts of redox reactions, including definitions of oxidation and reduction, electron transfer, oxidation states, and how to write ionic and half equations.

10 cards
View Deck
Chemistry

Cambridge IGCSE Chemistry Essentials

This deck covers key concepts from the Cambridge IGCSE Chemistry Coursebook, including states of matter, atomic structure, bonding, reactions, organic chemistry, and environmental chemistry.

50 cards
View Deck
Chemistry

Edexcel A Level Chemistry: Exam Questions

This flashcard set provides explanations for molecular shapes and trends in chemical behavior. It covers the geometry of phosphorus (III) chloride due to electron pair repulsion, the thermal stability of nitrates based on cation size and charge, and the increase in boiling points from chlorine to iodine due to stronger London dispersion forces.

8 cards
View Deck
Chemistry

Edexcel A Level Chemistry: 19: Modern Analytical Techniques II

This flashcard set explains the fundamental principles of mass spectrometry, including the formation of molecular ions, the meaning of m/z (mass-to-charge) values, and how high-resolution mass spectrometry provides precise molecular mass measurements.

46 cards
View Deck
Chemistry

Edexcel A Level Chemistry: 18: Organic Chemistry III

This flashcard set introduces aromatic and aliphatic compounds, highlighting the presence or absence of delocalised electrons in hydrocarbon rings. It also describes benzene as an arene and outlines its key physical properties.

89 cards
View Deck

Study Tips

  • Press F to enter focus mode for distraction-free studying
  • Review cards regularly to improve retention
  • Try to recall the answer before flipping the card
  • Share this deck with friends to study together
A-level Chemistry: 3.2.3 Group 7, The Halogens Part 2
TermDefinition

Test for Halides

What is the result for fluoride?

No precipitate

Test for Halides

What is the result for chloride?

White precipitate of silver chloride

Test for Halides

What is the result for bromide?

Cream precipitate of silver bromide

Test for Halides

What is the result for iodide?

Yellow precipitate of silver iodide

Test for Halides

Why do you add dilute nitric acid?

To remove ions which may interfere with test

Test for Halides

Write the general equation

Ag⁺(aq) + X⁻(aq) → AgX(s)
(where X⁻ = Cl⁻, Br⁻, I⁻)

Fill in the gaps (speed)

  • Chloride (Cl⁻): white precipitate → dissolves in dilute NH₃

  • Bromide (Br⁻): cream precipitate → dissolves in concentrated NH₃

  • Iodide (I⁻): yellow precipitate → insoluble in NH₃

Test for Halides

What can you do to be sure of your results & why does this work?

  • You can test your results by adding ammonia solution

  • Each silver halide has a different solubility in ammonia

Fill in the gaps

  • Chloride (Cl⁻): white precipitate → dissolves in dilute NH₃

  • Bromide (Br⁻): cream precipitate → dissolves in concentrated NH₃

  • Iodide (I⁻): yellow precipitate → insoluble in NH₃

What reactions reflect the trend in the reducing ability of halide ions?

Reactions of solid sodium halides with concentrated sulfuric acid

All halides react with concentrated sulfuric acid to give what as a product?

hydrogen halide

Write the equations for the reaction of NaF or NaCl with H2SO4

NaF(s) + H₂SO₄(l) → HF(g) + NaHSO₄(s)
NaCl(s) + H₂SO₄(l) → HCl(g) + NaHSO₄(s)

Reaction of NaF or NaCl with H2SO4

Describe the observations

Misty (white) fumes hydrogen fluoride / hydrogen chlroide

Name 2 features of these reactions

  • HF and HCl aren't strong enough reducing agents to reduce sulfuric acid so reaction stops there

  • Acid –base reactions & not redox reactions

    • H2SO4 plays the role of an acid

    • Oxidation states of halide and sulfur stay the same (-1 and +6)

Write the equations for the reaction of NaBr with H2SO4

  1. Acid-base step

  2. Redox step
    NaBr(s) + H₂SO₄(l) → HBr(g) + NaHSO₄(s)

  3. 2HBr(g) + H₂SO₄(l) → Br₂(l) + SO₂(g) + 2H₂O(l)

Describe the observations in this reaction

Misty fumes of hydrogen bromide gas (HBr)

Write the oxidation states of S & Br for the 2nd equation

  • Sulfur (S): +6 in H₂SO₄ → +4 in SO₂

  • Bromine (Br): –1 in HBr → 0 in Br₂

Describe the reducing agent in these reactions

HBr is stronger reducing agent than HCl and reacts with H2SO4 in a redox reaction

Describe the observations in the 2nd equation

Choking fumes of SO2 and orange fumes of Br2

Write the equations for the reaction of Nal with H2SO4

  • NaI(s) + H₂SO₄(l) → HI(g) + NaHSO₄(s)

  • 2HI(g) + H₂SO₄(l) → I₂(s) + SO₂(g) + 2H₂O(l)

  • 6HI(g) + H₂SO₄(l) → 3I₂(s) + S(s) + 4H₂O(l)

  • 8HI(g) + H₂SO₄(l) → 4I₂(s) + H₂S(g) + 4H₂O(l)

Reactions of NaI with H2SO4

Write the oxidation state of S & I

  • Sulfur (S): +6 in H₂SO₄ → +4 in SO₂

  • Iodine (I): –1 in HI → 0 in I₂

Reactions of NaI with H2SO4

Write the oxidation state of S & I

  • Sulfur (S): +6 in H₂SO₄ → +4 in SO₂, 0 in S, –2 in H₂S

  • Iodine (I): –1 in HI → 0 in I₂

Describe the reducing agent in these equations

The reducing agent is iodide ion (I⁻) from NaI, because it loses electrons and is oxidised from –1 in HI to 0 in I₂.

Bromine reacts with phosphorus to form phosphorus tribromide. Write an equation for this reaction.

6Br2 + P4 → 4PBr3

Describe the role of H2SO4 in these reactions

  1. H2SO4 plays the role of acid in the first step

  2. Then acts as an oxidising agent in the second redox

State the reduction product

Sulfur dioxide

State the oxidation and reduction half equations for when bromine reacts with conc. sulfuric acid

  • Ox ½ equation: 2Br- → Br2 + 2e-

  • Re ½ equation: H2SO4 + 2H+ + 2 e- → SO2 + 2H2O

Reactions of NaI with H2SO4

Write an equation that gives sulfur

6HI + H2SO4 → 3I2 + S + 4H2O

Reactions of NaI with H2SO4

Name the products

Sulfur dioxide, sulfur and hydrogen sulfide

Reactions of NaI with H2SO4

Describe H2SO4 role in the reactions

H2SO4 plays the role of acid in the first step producing HI & then acts as an oxidising agent in the three redox steps

Reactions of NaI with H2SO4

Write the oxidation half equation and 3 reduction half equations.

Oxidation

  • 2I- → I2 + 2e-

Reduction:

  • H2SO4 + 2H+ + 2e- → SO2 + 2H2O

  • H2SO4 + 6H+ + 6e- → S + 4H2O

  • H2SO4 + 8H+ + 8e- → H2S + 4H2O

Reactions of NaI with H2SO4

Describe the observations

  • Sulfur dioxide is a choking gas

  • Sulfur is a yellow solid

  • Hydrogen sulfide has a smell of bad eggs

2NaBr + 2H2SO4 → Na2SO4 + Br2 + SO2 + 2H2O

Explain why bromide ions reacts differently from chloride ions (2)

  • Br− ions are bigger than Cl− ions

  • ∴ Br− ions are more easily oxidised/lose an electron

Write an ionic equation for the reaction between chlorine and cold dilute sodium hydroxide solution

Cl2 + 2OH- → ClO- + Cl- + H2O