Unit 7: Redox Reactions and Electrochemistry — Online MCQ Test
CHEMISTRY · JEE MAINS · JEE
Practice Unit 7: Redox Reactions and Electrochemistry with a free chapter-wise online MCQ test.
This chapter covers: Electronic concepts of oxidation and reduction redox reactions oxidation number rules for assigningoxidation number and balancing of redox reactions. Electrolytic and metallic cond....
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Unit 7: Redox Reactions and Electrochemistry — Important Questions & Answers
Which of the following electrolyte can be used to obtain H2S2O8 by the process of electrolysis?
- A. Dilute solution of sodium sulphate.
- B. Acidified dilute solution of sodium sulphate.
- C. Dilute solution of sulphuric acid
- D. Concentrated solution of sulphuric acid
Answer: D. Concentrated solution of sulphuric acid
Peroxydisulphuric acid (H2S2O8) is obtained by the electrolysis of concentrated sulphuric acid (Marshall's acid preparation).
Peroxydisulphuric acid (H2S2O8) is obtained by the electrolysis of concentrated sulphuric acid (Marshall's acid preparation).
A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A. The amount of the aluminium deposited at the cathode is [Given: molar mass of aluminium is 27 g mol^-1, Faraday constant = 96500 C mol^-1]
- A. 1.660 g
- B. 0.336 g
- C. 0.441 g
- D. 1.007 g
Answer: B. 0.336 g
Mass = (I * t * M) / (n * F) = (2 * 1800 * 27) / (3 * 96500) = 97200 / 289500 ≈ 0.3357 g.
Mass = (I * t * M) / (n * F) = (2 * 1800 * 27) / (3 * 96500) = 97200 / 289500 ≈ 0.3357 g.
A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A. The amount of the aluminium deposited at the cathode is (Given: molar mass of aluminium is 27 g/mol, Faraday constant = 96500 C/mol):
- A. 1.660 g
- B. 0.336 g
- C. 0.441 g
- D. 1.007 g
Answer: B. 0.336 g
Using Faraday's Law, mass m = (I * t * M) / (n * F). Here I = 2 A, t = 1800 s, M = 27 g/mol, n = 3 (for Al3+), F = 96500. m = (2 * 1800 * 27) / (3 * 96500) ≈ 0.336 g.
Using Faraday's Law, mass m = (I * t * M) / (n * F). Here I = 2 A, t = 1800 s, M = 27 g/mol, n = 3 (for Al3+), F = 96500. m = (2 * 1800 * 27) / (3 * 96500) ≈ 0.336 g.
Aqueous HCl reacts with MnO2(s) to form MnCl2(aq), Cl2(g) and H2O(l). What is the weight (in g) of Cl2 liberated when 8.7 g of MnO2(s) is reacted with excess aqueous HCl solution? (Given Molar mass in g mol–1 Mn = 55, Cl = 35.5, O = 16, H = 1)
- A. (1) 7.1
- B. (2) 71
- C. (3) 21.3
- D. (4) 14.2
Answer: A. (1) 7.1
MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O. Moles of MnO2 = 8.7 / 87 = 0.1 mol. Since the stoichiometry is 1:1, moles of Cl2 produced = 0.1 mol. Weight of Cl2 = 0.1 * 71 = 7.1 g.
MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O. Moles of MnO2 = 8.7 / 87 = 0.1 mol. Since the stoichiometry is 1:1, moles of Cl2 produced = 0.1 mol. Weight of Cl2 = 0.1 * 71 = 7.1 g.
For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?
- A. Plot 1
- B. Plot 2
- C. Plot 3
- D. Plot 4
Answer: B. Plot 2
In an ideal closed-circuit Daniell cell operation (assuming constant conditions and standard states), the potential E_cell remains constant with time during the initial discharge phase before polarization or concentration changes significantly alter it.
In an ideal closed-circuit Daniell cell operation (assuming constant conditions and standard states), the potential E_cell remains constant with time during the initial discharge phase before polarization or concentration changes significantly alter it.