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9701 · 11.3

Some reactions of the halide ions flashcards

Revision flashcards for Cambridge 9701 Some reactions of the halide ions (syllabus 11.3). Flip, recall, then mark a real past-paper question.

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    What is the colour of the precipitate formed when aqueous silver nitrate is added to a solution containing chloride ions, Cl⁻?

    A white precipitate of silver chloride (AgCl).

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    What is the colour of the precipitate formed when aqueous silver nitrate is added to a solution containing bromide ions, Br⁻?

    A cream precipitate of silver bromide (AgBr).

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    What is the colour of the precipitate formed when aqueous silver nitrate is added to a solution containing iodide ions, I⁻?

    A yellow precipitate of silver iodide (AgI).

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    How does the solubility of silver chloride (AgCl) in ammonia compare to silver iodide (AgI)?

    AgCl is soluble in dilute aqueous ammonia, forming the complex ion [Ag(NH₃)₂]⁺. AgI is insoluble in even concentrated aqueous ammonia.

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    Why is dilute nitric acid added before silver nitrate in the test for halide ions?

    To acidify the solution and react with any carbonate ions (CO₃²⁻) present, which would otherwise form a white precipitate of silver carbonate (Ag₂CO₃) and give a false positive result.

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    What are the products when solid sodium chloride reacts with concentrated sulfuric acid?

    Sodium hydrogensulfate (NaHSO₄) and hydrogen chloride gas (HCl). This is an acid-base reaction, not a redox reaction.

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    What are the main products when solid sodium bromide reacts with concentrated sulfuric acid?

    HBr gas (misty fumes), Br₂ (brown fumes), and SO₂ gas (choking smell). The Br⁻ ion is a strong enough reducing agent to reduce H₂SO₄.

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    Name three reduction products of sulfuric acid formed when it reacts with solid sodium iodide.

    Sulfur dioxide (SO₂), elemental sulfur (S, a yellow solid), and hydrogen sulfide (H₂S, a gas with a 'bad egg' smell).

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    Explain the trend in reducing power of the halide ions down Group 17.

    Reducing power increases down the group (I⁻ > Br⁻ > Cl⁻). The ionic radius increases and there is more electron shielding, so the outermost electron is held less strongly by the nucleus and is more easily lost (oxidised).

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    Complete the ionic equation: $Cl_2(aq) + 2I^-(aq) \rightarrow$ ?

    $Cl_2(aq) + 2I^-(aq) \rightarrow 2Cl^-(aq) + I_2(aq)$. Chlorine is a stronger oxidising agent than iodine, so it displaces iodide ions from solution.

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    What is observed when aqueous bromine is added to aqueous potassium iodide?

    The solution turns from colourless (or pale orange if Br₂ is concentrated) to brown/black due to the formation of iodine (I₂).