Oxidation and reduction III
Francine Taylor-Campbell, Contributor

We will now examine and attempt questions on the topic.

Question 1

1. Sulphuric acid has a variety of uses.

Sulphur reacts with hot concentrated sulphuric acid to form sulphur dioxide and water.

Write a balanced equation for this reaction. (2 marks)

Identify the oxidising agent showing the change in oxidation number that occurs. (3 marks)

Identify one source of sulphur dioxide in the atmosphere and state ONE adverse effect of this gas on the environment. (2 marks)

A sample of sulphur dioxide is bubbled into acidified potassium manganate (VII). What would you expect to observe? Give a reason for your answer. (2 marks)

(b) Calculate the oxidation number of sulphur in:

i) sodium sulphate (Na2SO4)

ii) sodium sulphide (Na2S)

(2 marks)

ANSWERS

1a. (i) S + 2H₂SO₄ == 3SO₂ + 2H₂O

(ii) S changes from 0 to +4

In H₂SO₄, S changes from +6 to +4

H₂SO₄ is reduced, hence it is the oxidising agent.

Sulphur dioxide is produced from the burning of fossil fuels and is an important contributor to acid rain, which damages and corrodes buildings, animals and plants.

Potassium manganate (VII) would change colour from purple to colourless. This would suggest that sulphur dioxide is acting as a reducing agent since potassium manganate changes colour.

(i) In Na₂SO₄, S is in oxidation state +6

(ii) In Na₂S, S is in oxidation state -2.

Question 2

2. When orange crystals of ammonium dichromate (VI), (NH₄)₂Cr₂O7 are heated, the products are green chromium(III) oxide, nitrogen and water.

Construct the equation for this reaction.

By naming the reagents and giving the observations, explain how you would confirm that i) the orange crystals contain the ammonium ion ii) the reaction products contain water.

The action of heat on the crystals starts a redox reaction. Has the chromium been oxidised or reduced? Explain your answer.

(5 marks)

(b) The following reaction occurs when chlorine is bubbled into aqueous iron(II) chloride. 2Fe2+ + Cl₂ === 2Fe³+ + 2Cl-

Explain, in terms of electrons, why chlorine has been reduced.

Name the salt formed in the reaction

(2 marks)

Determine the oxidation state of (i) nitrogen in manganese (ii) nitrate nitrogen in sodium nitrite.

ANSWERS

2. (i) (NH₄)₂Cr₂O7 (s) === Cr₂O3 (s) + N₂ (g) + 4H₂O (l)

(ii) To confirm the presence of ammonium ions, warm the crystals with aqueous sodium hydroxide and test the gas produced with damp, red litmus paper and a stopper with hydrogen chloride gas. Ammonia should turn the litmus paper blue and produce dense, white fumes of ammonium chloride when it comes in contact with hydrogen chloride gas. NH₄+ (s) + OH- (aq) == NH₃ (g) + H₂O (l)

To confirm the presence of water in the product, add anhydrous copper sulphate to the product. The copper sulphate should change from white to blue. CuSO4 (s) white + 5H₂O (l) === CuSO4.5H₂O (s) blue

(iii) Chromium changes from +6 in ammonium dichromate(VI) to +3 in chromium(III) oxide. This means that the chromium has been reduced.

(b) Chlorine gains two electrons to form 2Cl-, thus it has been reduced.

(ii) The salt formed in the reaction is Iron(III) chloride.

(c) In Mn(NO₃)₂, nitrogen is in +5 state.

(ii) In NaNO₂, nitrogen is in +3 state.

James Kerr (right), manager, metrology and testing-analytical services at the Bureau of Standards Jamaica, demonstrates equipment used in the chemistry department to Michael Stern (centre), the minister of state in the Ministry of Industry and Commerce. - Junior Dowie/Staff Photographer

Francine Taylor-Campbell is an independent contributor.