CBSE Class 10 Science 2023: question 37
Set in the CBSE Class 10 board exam of 2023, paper 31/6/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.
Q37 · 4 marks · Case-basedCompetency-basedMetals and Non-metals
On the basis of reactivity metals are grouped into three categories:
(i) Metals of low reactivity
(ii) Metals of medium reactivity
(iii) Metals of high reactivity
Therefore metals are extracted in pure form from their ores on the basis of their chemical properties.
Metals of high reactivity are extracted from their ores by electrolysis of the molten ore.
Metals of low reactivity are extracted from their sulphide ores, which are converted into their oxides. The oxides of these metals are reduced to metals by simple heating.
- (a)0 marks
Name the process of reduction used for a metal that gives vigorous reaction with air and water both.
- (b)0 marks
Carbon cannot be used as a reducing agent to obtain aluminium from its oxide? Why?
- (c)0 marks
Describe briefly the method to obtain mercury from cinnabar. Write the chemical equation for the reactions involved in the process.
- (c)0 marks
Differentiate between roasting and calcination giving chemical equation for each.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 31/6/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
Place the metal in the reactivity series: very reactive metals cannot be reduced chemically.
Hint 2
For (b), compare how strongly aluminium and carbon hold on to oxygen.
Hint 3
For (c), mercury is low in the series, so heating alone works: first convert the sulphide into an oxide, then heat the oxide again.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
(a) A metal that reacts vigorously with both air and water is highly reactive (e.g. Na, K). It is obtained by electrolytic reduction, i.e. electrolysis of its molten chloride or oxide; the metal is deposited at the cathode.
- 1 mark
(b) Aluminium is a highly reactive metal and has a greater affinity for oxygen than carbon has. Carbon therefore cannot remove oxygen from , so it cannot reduce aluminium oxide to aluminium.
- 1 mark
(c) [Option A] Mercury is a metal of low reactivity. Cinnabar (HgS) is heated strongly in air, which converts it into mercuric oxide:
- 1 mark
On further heating, mercuric oxide is reduced to mercury:
(c) [Option B] Roasting: The sulphide ore is heated strongly in the presence of excess air, which converts it into the metal oxide.
Calcination: The carbonate ore is heated strongly in a limited supply of air, which converts it into the metal oxide and releases CO₂.
(Option B carries the same 2 marks as Option A.)
Answer: (a) Electrolytic reduction. (b) Al has more affinity for oxygen than carbon, so carbon cannot reduce Al₂O₃. (c) A: 2HgS + 3O₂ → 2HgO + 2SO₂, then 2HgO → 2Hg + O₂ on heating. (c) B: roasting = sulphide ore heated in excess air (2ZnS + 3O₂ → 2ZnO + 2SO₂); calcination = carbonate ore heated in limited air (ZnCO₃ → ZnO + CO₂).
Where the marks are
- (a) Electrolytic reduction (1 mark)
- (b) Aluminium has a greater affinity for oxygen than carbon / is highly reactive (1 mark)
- (c) Describes heating cinnabar in air to form HgO, then heating HgO to give Hg (1 mark)
- (c) Both balanced equations (½ + ½ marks)
- OR (c) Correct difference between roasting and calcination with an equation for each (2 marks)
Revise the chapter
- Chapter: Metals and Non-metals (notes, formulas, where marks are lost and every past question)
- Topic: Metallurgy: Extraction of Metals from Ores
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