Mr Cole Chemistry

Reducing Power of Halides Questions

Reducing Power of Halides

Group 2 and Group 7 Worksheet

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1. Solid sodium chloride will react with concentrated sulfuric acid to produce an acidic gas.

a) Write an equation for the reaction. Include state symbols.

NaCl_(s) + H₂SO_4(l)

→

NaHSO_4(s) + HCl_(g)

b) How can you test for the presence of the acidic gas produced?

Damp blue litmus paper held at the mouth of the test tube will turn red.

2. Sodium bromide will also react with concentrated sulfuric acid. However, it undergoes several different possible reactions.

a) Write the equation for the reaction between sodium bromide and concentrated sulfuric acid that is not a redox reaction.

NaBr_(s) + H₂SO_4(l)

→

NaHSO_4(s) + HBr_(g)

b) Sodium bromide will also undergo a redox reaction. Write two half equations for the reaction that will occur.

Reduction:

SO₄^2- + 4H⁺ + 2e^–

→

SO₂ + 2H₂O

Oxidation:

2Br^–

→

Br₂ + 2e^–

c) Combine the two half equations to form an equation for the redox reaction that will occur.

SO₄^2- + 4H⁺ + 2Br^–

→

SO₂ + 2H₂O + Br₂

d) Show how it is a redox reaction using oxidation numbers.

Sulfur is reduced: Oxidation number decreases from +6 (in SO₄^2-) to +4 (in SO₂).

Bromine is oxidised: Oxidation number increases from -1 (in Br^–) to 0 (in Br₂).

e) State how you would test for the presence of the sulfur-based product.

A piece of filter paper dipped in acidified potassium dichromate(VI) solution will turn from orange to green in the presence of Sulfur Dioxide gas.

3. Sodium iodide will also react with concentrated sulfuric acid. However, it can undergo multiple different redox reactions.

a) Write the half equation for the reduction of sulfur from sulfuric acid to: (i) Sulfur Dioxide, (ii) Sulfur, (iii) Hydrogen Sulfide.

i) To Sulfur Dioxide:

SO₄^2- + 4H⁺ + 2e^–

→

SO₂ + 2H₂O

ii) To Sulfur:

SO₄^2- + 8H⁺ + 6e^–

→

S + 4H₂O

iii) To Hydrogen Sulfide:

SO₄^2- + 10H⁺ + 8e^–

→

H₂S + 4H₂O

b) Write the half equation for the oxidation of iodide to iodine.

2I^–

→

I₂ + 2e^–

c) Write the full equations for the production of sulfur dioxide, sulfur, and hydrogen sulfide.

i) Production of Sulfur Dioxide:

2I^– + SO₄^2- + 4H⁺

→

I₂ + SO₂ + 2H₂O

ii) Production of Sulfur:

6I^– + SO₄^2- + 8H⁺

→

3I₂ + S + 4H₂O

iii) Production of Hydrogen Sulfide:

8I^– + SO₄^2- + 10H⁺

→

4I₂ + H₂S + 4H₂O

4. Explain why sodium iodide can reduce sulfur to hydrogen sulfide whereas bromide can only reduce it to sulfur dioxide.

Iodide is a stronger reducing agent than bromide.

The iodide ion (I^–) has a larger ionic radius and more shielding than the bromide ion (Br^–). This means the electrostatic attraction between the nucleus and the outer electron is weaker in iodide.

Therefore, it is easier for iodide to lose an electron (be oxidised) and reduce the sulfur in sulfuric acid further (from +6 all the way to -2 in H₂S), whereas bromide can only reduce it to +4 (in SO₂).