Question 374 marksSection Section E

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) Name the process of reduction used for a metal that gives vigorous reaction with air and water both.

(b) Carbon cannot be used as a reducing agent to obtain aluminium from its oxide? Why?

(c) Describe briefly the method to obtain mercury from cinnabar. Write the chemical equation for the reactions involved in the process.

OR

(c) Differentiate between roasting and calcination giving chemical equation for each.

Metals and Non-metals
Metals and Non-metals

Official answer

The process of reduction used for a metal that gives vigorous reaction with air and water both is electrolysis. This is because metals that react vigorously with air and water are highly reactive and cannot be reduced by simple heating or using carbon as a reducing agent. Electrolysis is the process of using an electric current to drive a chemical reaction, and it is often used to extract highly reactive metals from their ores.


Carbon cannot be used as a reducing agent to obtain aluminium from its oxide because aluminium is more reactive than carbon. In order to reduce aluminium oxide to aluminium, a more reactive metal such as sodium or potassium is needed, or electrolysis must be used.


Mercury is obtained from cinnabar (mercury sulphide) through the process of roasting, followed by reduction. The chemical equation for the reaction involved in the process is:


2HgS + 3O2 →\rightarrow 2HgO + 2SO2


2HgO →\rightarrow 2Hg + O2


In the first equation, cinnabar is roasted in the presence of oxygen to produce mercury oxide and sulphur dioxide. In the second equation, the mercury oxide is reduced to mercury through simple heating.


OR


Roasting and calcination are two different processes used to convert ores into their oxides. Roasting is the process of heating an ore in the presence of air to convert it into its oxide, while calcination is the process of heating an ore in the absence of air to convert it into its oxide. The chemical equation for roasting is:


2ZnS + 3O2 →\rightarrow 2ZnO + 2SO2


The chemical equation for calcination is:


CaCO3 →\rightarrow CaO + CO2

Answer keywords:electrolysisreductionaluminiumcarbonmercurycinnabarroastingcalcination

Marking scheme

  1. 1.(a) 1 mark
  2. 2.(b) 1 mark
  3. 3.(c) 2 marks

Hint

The student should refer to the chapter on Metals and Non-metals to answer this question. They should review the processes of extraction of metals, the chemical properties of metals, and the differences between roasting and calcination.

Quick oral answer

The student should be able to answer the questions clearly and concisely, using the correct terminology and chemical equations. They should be able to explain the processes of extraction of metals and the chemical properties of metals, and differentiate between roasting and calcination.

Common mistakes

  1. 1.The student may confuse the processes of roasting and calcination.
  2. 2.The student may not be able to write the correct chemical equation for the reactions involved in the process.

Analysis and explanation

This question requires the student to apply their knowledge of the extraction of metals and the chemical properties of metals to answer the questions. The student must be able to identify the process of reduction used for a metal that gives vigorous reaction with air and water both, explain why carbon cannot be used as a reducing agent to obtain aluminium from its oxide, describe briefly the method to obtain mercury from cinnabar and write the chemical equation for the reactions involved in the process, and differentiate between roasting and calcination giving chemical equation for each.

Interesting facts

The student can be encouraged to research more about the extraction of metals and the chemical properties of metals.

The student can be encouraged to think about the importance of the extraction of metals in our daily lives.

Question details

Q37 of 39 questions

Board
CBSE
Class
Class 10
Subject
Science
Year
2023
Section
Section Section E
Marks
4
Chapter
Metals and Non-metals
Topic
Metals and Non-metals
Bloom's level
Analyzing
Syllabus topic
Metals and Non-metals

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Frequently asked questions

How many marks does this question carry in CBSE Class 10 Science 2023?

This question carries 4 marks in the CBSE Class 10 Science 2023 examination.

Which chapter does this question come from in Science?

This question is from the chapter "Metals and Non-metals" in the CBSE Class 10 Science syllabus.

What topic does this question cover in Science?

This question covers the topic "Metals and Non-metals" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2023 paper?

This question appears in Section Section E in the CBSE Class 10 Science 2023 paper.