Unit 7: Redox Reactions and Electrochemistry — Online MCQ Test
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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 assigning oxidation number balancing of redox reactions. Electrolytic and metallic conduct....
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Unit 7: Redox Reactions and Electrochemistry — Important Questions & Answers
What is the change in oxidation number of carbon in the following reaction? CH4(g) + 4Cl2(g) -> CCl4(l) + 4HCl(g)
- A. 0 to +4
- B. -4 to +4
- C. 0 to -4
- D. +4 to +4
Answer: B. -4 to +4
In CH4, C is in -4 oxidation state. In CCl4, C is in +4 oxidation state.
In CH4, C is in -4 oxidation state. In CCl4, C is in +4 oxidation state.
On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be:
- A. Oxygen gas
- B. H2S gas
- C. SO2 gas
- D. Hydrogen gas
Answer: A. Oxygen gas
During electrolysis of dilute sulfuric acid, hydroxyl ions (OH-) or water molecules are discharged at the anode to produce oxygen gas.
During electrolysis of dilute sulfuric acid, hydroxyl ions (OH-) or water molecules are discharged at the anode to produce oxygen gas.
If the molar conductivity (Λm) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be [Assume Λ°+ = 349.6 S cm² mol⁻¹ and Λ°– = 50.4 S cm² mol⁻¹]
- A. 0.115
- B. 0.125
- C. 0.225
- D. 0.215
Answer: C. 0.225
Degree of dissociation α = Λm / Λ°m. Λ°m = 349.6 + 50.4 = 400. α = 90 / 400 = 0.225.
Degree of dissociation α = Λm / Λ°m. Λ°m = 349.6 + 50.4 = 400. α = 90 / 400 = 0.225.
If the molar conductivity (Λm) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be [Assume Λ⁰₊ = 349.6 S cm² mol⁻¹ and Λ⁰₋ = 50.4 S cm² mol⁻¹]
- A. 0.115
- B. 0.125
- C. 0.225
- D. 0.215
Answer: C. 0.225
Degree of dissociation α = Λm / Λ⁰m. Λ⁰m = Λ⁰₊ + Λ⁰₋ = 349.6 + 50.4 = 400. α = 90 / 400 = 0.225.
Degree of dissociation α = Λm / Λ⁰m. Λ⁰m = Λ⁰₊ + Λ⁰₋ = 349.6 + 50.4 = 400. α = 90 / 400 = 0.225.
Among the following the correct order of acidity is :
- A. HClO < HClO2 < HClO3 < HClO4
- B. HClO4 < HClO < HClO3 < HClO4
- C. HClO4 < HClO2 < HClO < HClO3
- D. HClO3 < HClO4 < HClO2 < HClO
Answer: A. HClO < HClO2 < HClO3 < HClO4
Acidity of oxyacids increases with the oxidation state of the central atom. The O.S. of Cl increases from +1 in HClO to +7 in HClO4.
Acidity of oxyacids increases with the oxidation state of the central atom. The O.S. of Cl increases from +1 in HClO to +7 in HClO4.
The pressure of H2 required to make the potential of H2-electrode zero in pure water at 298 K is :
- A. 10^-12 atm
- B. 10^-10 atm
- C. 10^-4 atm
- D. 10^-14 atm
Answer: D. 10^-14 atm
Using the Nernst equation for H2/H+ electrode, for E=0, pH=7 (H+=10^-7), P(H2) = (H+)^2 = (10^-7)^2 = 10^-14 atm.
Using the Nernst equation for H2/H+ electrode, for E=0, pH=7 (H+=10^-7), P(H2) = (H+)^2 = (10^-7)^2 = 10^-14 atm.
Match List-I with List-II. List-I (Conversion) A. 1 mol of H2O to O2 B. 1 mol of MnO4^- to Mn2+ C. 1.5 mol of Ca from molten CaCl2 D. 1 mol of FeO to Fe2O3 List-II (Number of Faraday required) I. 3F II. 2F III. 1F IV. 5F Choose the correct answer from the options given below:
- A. A-II, B-III, C-I, D-IV
- B. A-III, B-IV, C-II, D-I
- C. A-II, B-IV, C-I, D-III
- D. A-III, B-IV, C-I, D-II
Answer: C. A-II, B-IV, C-I, D-III
H2O → O2 needs 2 electrons per mole of O2; for 1 mol H2O, 1/2 mol O2 is formed, so 1F. MnO4^- to Mn2+ needs 5F per mole. Ca2+ to Ca requires 2F per mole, so 1.5 mol requires 3F. FeO to Fe2O3 corresponds to Fe2+ to Fe3+, so 1F per Fe; for 1 mol FeO, 1F.
H2O → O2 needs 2 electrons per mole of O2; for 1 mol H2O, 1/2 mol O2 is formed, so 1F. MnO4^- to Mn2+ needs 5F per mole. Ca2+ to Ca requires 2F per mole, so 1.5 mol requires 3F. FeO to Fe2O3 corresponds to Fe2+ to Fe3+, so 1F per Fe; for 1 mol FeO, 1F.
Which reaction is NOT a redox reaction?
- A. H2 + Cl2 → 2 HCl
- B. BaCl2 + Na2SO4 → BaSO4 + 2 NaCl
- C. Zn + CuSO4 → ZnSO4 + Cu
- D. 2 KClO3 + I2 → 2 KIO3 + Cl2
Answer: B. BaCl2 + Na2SO4 → BaSO4 + 2 NaCl
In BaCl2 + Na2SO4 → BaSO4 + 2NaCl, oxidation states do not change; it is a double displacement precipitation reaction, not a redox reaction.
In BaCl2 + Na2SO4 → BaSO4 + 2NaCl, oxidation states do not change; it is a double displacement precipitation reaction, not a redox reaction.
Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is: (Given: Molar mass of Cu : 63 g mol−1, 1F = 96487 C)
- A. 31.5 g
- B. 0.0315 g
- C. 3.15 g
- D. 0.315 g
Answer: D. 0.315 g
Charge passed = It = 9.6487 × 100 = 964.87 C = 0.01 F. For Cu2+ + 2e− → Cu, equivalent mass deposited = 63/2 = 31.5 g per Faraday. So mass = 31.5 × 0.01 = 0.315 g.
Charge passed = It = 9.6487 × 100 = 964.87 C = 0.01 F. For Cu2+ + 2e− → Cu, equivalent mass deposited = 63/2 = 31.5 g per Faraday. So mass = 31.5 × 0.01 = 0.315 g.
The number of Faradays(F) required to produce 20 g of calcium from molten CaCl2 (Atomic mass of Ca = 40 g mol–1) is
- A. 3
- B. 4
- C. 1
- D. 2
Answer: C. 1
Ca2+ + 2e- -> Ca. 40g requires 2 Faradays. 20g requires 1 Faraday.
Ca2+ + 2e- -> Ca. 40g requires 2 Faradays. 20g requires 1 Faraday.