Group 7: Tests Flashcards
What is the trend in “reducing power” of halide ions as you go down group 7 and why? Why is I- a better reducing agent than F-?
As we go down group, ionic radius increases.
Distance between nucleus and outer electron becomes larger and there is more electronic shielding, force of attraction gets weaker.
Outer electron is lost more readily, this is why I- is a more powerful reducing agent than F-.
Why are halide ions reducing agents?
Halide ions are reducing agents because they lose an electron.
What are the two tests that prove the trend that reducing power increases as you go down group 7? Ie. to test whether halide solution present is I-, Br- OR Cl-.
Reaction with sulphuric acid.
Reaction with silver nitrate solution.
What are the different reduction products you get when halide ions reduce concentrated sulphuric acid? What is the oxidation state for S in all of these different products?
NaHSO₄ (+6)
SO₂ (+4)
S (0)
H₂S (-2)
True or False
All halide ions can reduce dilute sulphuric acid.
False.
Only SOME halide ions can reduce CONCENTRATED sulphuric acid.
What is point A (ie. first stage of reduction) of reduction of sulphuric acid for Cl-, Br- and I- ions? Give the three different equations.
Cl⁻ ions: H₂SO₄ + NaCl –> NaHSO₄ + HCl
Br⁻ ions: H₂SO₄ + NaBr –> NaHSO₄ + HBr
I⁻ ions: H₂SO₄ + NaI –> NaHSO₄ + HI
Point A (ie. first stage of reduction of sulphuric acid) with Cl- ions, what is produced and why?
- White misty fumes produced.
- From the HCl.
What are the half equations and overall ionic equation for point B of reduction of sulphuric acid (ie. second stage of reduction) - using NaBr? What will be formed in this reaction? How about using NaI?
- 2Br⁻ –> Br₂ + 2e⁻ (Br⁻ ions are oxidised.)
- H₂SO₄ + 2H⁺ + 2e⁻ –> SO₂ + 2H₂O (S being reduced.)
- Overall ionic: H₂SO₄ + 2H⁺+ 2Br⁻–> Br₂ + SO₂ + 2H₂O
- What will be formed: Orange vapour of Br2 produced.
- Exact same reaction with NaI, just replace Br’s in equation to I*
What are the half equations and overall ionic equation for point C of reduction of sulphuric acid (ie. third stage of reduction) - using NaI? What will be formed in this reaction?
- 6I⁻ –> 3I₂ + + 6e⁻ (I⁻ ions are oxidised.)
- H₂SO₄ + 6H⁺ +6e⁻—> S +4H₂O (S being reduced.)
- H₂SO₄ + 6H⁺ + 6I⁻ –> 3I₂ + S+ 4H₂O
Formed: Yellow solid of S formed.
Diagrams of the points in folder.
What are the half equations and overall ionic equation for point D of reduction of sulphuric acid (ie. fourth stage of reduction) - using NaI? What will be formed in this reaction?
- 8I⁻ –> 4I₂ + 8e⁻ (I⁻ ions are oxidised.)
- H₂SO₄ +8H⁺ + 8e⁻ –> H₂S + 4H₂O (S being reduced.)
- H₂SO₄ +8H⁺ + 8I⁻ –> 4I₂ + H₂S + 4H₂O
- Rotten egg smell of H₂S produced.
Diagrams of the points in folder!
For oxidation of sulphuric acid using Cl-, Br- and I- ions, what compound is used?
NaCl
NaBr
NaI
True or False
Reduction of sulphuric acid using NaBr to produce SO2 will be the only reduction that the Br- ions do?
- False.
- NaBr will also reduce sulphuric acid to form NaHSO4 at point A.
Diagram of the points in folder!
True or False
Reduction of sulphuric acid using NaI to produce S will be the only reduction that the I- ions do?
- False
- NaI will also reduce sulphuric acid to form NaHSO4 (at point A), SO2 (at point B), H2S (at point C.)
Diagram of the points in electronic folder!!
True or False
Reduction of sulphuric acid using NaCl to produce NaHSO4 will be the only reduction that NaCl does
- True.
- NaCl will not further reduce the H2SO4.
What do HI, HBr and HCl produced in first reduction stage of sulphuric acid produce?
- White misty fumes.
After sulphuric acid is reduced by NaI, what will reduce the sulphuric acid further in the later stages of reduction.
I- ions, that come from HI produced in first stage, will reduce any excess sulphuric acid left over.
What are the different results when testing for three halide ions I-, Br- and Cl- using silver nitrate? Give ionic equations.
- Chloride ions: white precipitate, silver chloride, forms.
Ag⁺ (aq) + Cl⁻(aq) –> AgCl (s) - Iodide ions: yellow precipitate, silver iodide, forms.
Ag⁺ (aq) + l⁻(aq) –> Agl (s) - Bromide ions: cream precipitate, silver bromide, forms.
Ag⁺ (aq) + Br⁻(aq) –> AgBr (s)
True or False
When we test for halide ions using silver nitrate solution, we ONLY add silver nitrate solution to the sample we are testing. If not what else do we add and WHY?
- False!
- Add dilute nitric acid (HNO₃) and THEN add silver nitrate solution (AgNO₃.)
- We add nitric acid to react with any anions, other than halides - such as carbonates. If not, this could give a false result otherwise
We have added dilute nitric acid and silver nitrate solution to test a sample to see if it contains Br-, I- or Cl- ions, a precipitate has been formed but the colour isn’t that clear, what further test can be done to determine what ion is in the sample?
- Add ammonia (NH3) solution to the precipitates.
Cl-: the white precipitate will dissolve in dilute NH3.
Br-: cream precipitate dissolves in concentrated NH3.
I-: yellow precipitate insoluble in concentrated NH3.
