Free Electrochemistry MCQs with Answers
696 Electrochemistry MCQs from Chemistry, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Electrochemistry describes electron transfer through oxidation and reduction, identifies oxidizing and reducing agents, and applies oxidation numbers to balance redox equations. It also covers electrode potential and the standard hydrogen electrode, which provides the reference for comparing the reduction tendencies of other electrodes.
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Read the Electrochemistry notesFree MDCAT chapter notes with key terms696 questions · page 22 of 35
- A. 0.1 M HCl
- B. 0.1 M CH3COOH
- C. 0.1 M H3PO4
- D. 0.1 M H2SO4
Explanation: The correct answer is 0.1 M H2SO4. This solution provides a high concentration of H+ ions due to its nature as a strong diprotic acid…
Correct answer: 0.1 M H2SO4- A. − 0.30 V
- B. + 0.15 V
- C. + 0.10 V
- D. − 0.15 V
Explanation: The correct answer is -0.15 V. To find the standard potential of the Ag, AgI-/I electrode, we use the Nernst equation and consider the…
Correct answer: − 0.15 V- A. 12.7
- B. 16
- C. 31.8
- D. 63.5
Explanation: To determine the mass of copper liberated, we use Faraday's laws of electrolysis.
Correct answer: 16- A. Mercury
- B. H+
- C. Hg+2
- D. Cl-
Explanation: The correct answer is that the calomel electrode is reversible with respect to chloride ions (Cl-).
Correct answer: Cl-- A. 5.6 amp
- B. 7.2 amp
- C. 8.85 amp
- D. 11.2 amp
Explanation: To calculate the current needed, use Faraday's first law of electrolysis, which states that the mass of a substance deposited or liberated…
Correct answer: 8.85 amp- A. 1
- B. 2
- C. 3
- D. 4
Explanation: To find the valency of the metal, we use Faraday's laws of electrolysis, which state that the mass of a substance deposited at an…
Correct answer: 2- A. Increases the emf of the cell.
- B. Decreasing the emf of cell
- C. No change in emf of the cell.
- D. Unpredictable.
Explanation: Increasing the concentration of Sn2+ in an electrochemical cell decreases the EMF (cell potential).
Correct answer: Decreasing the emf of cell- A. p(H2)=1 atm and [H+]=1M
- B. p(H2)=2 atm and [H+] = 2M
- C. p(H2)=2 atm and [H+] = 1M
- D. p(H2)=1 atm and [H+]= 2M
Explanation: According to the Nernst equation, the reduction potential of a half-cell is affected by the concentration of ions and the partial pressure…
Correct answer: p(H2)=2 atm and [H+] = 1M- A. 7.77 min
- B. 9.44 min
- C. 5.24 min
- D. 11.39 min
Explanation: To solve this problem, we use Faraday's laws of electrolysis. First, calculate the amount of sodium needed to form a 10% Na-Hg amalgam on…
Correct answer: 7.77 min- A. 369000 coulombs
- B. 115800 coulombs
- C. 32100 coulombs
- D. 521900 coulombs
Explanation: To calculate the quantity of electricity, we use Faraday's law of electrolysis: Q = n × F, where Q is the quantity of electricity in…
Correct answer: 115800 coulombs- A. Hydrogen is liberated at the anode
- B. Hydrogen is liberated at the cathode
- C. No reaction takes place
- D. Hydride ions migrate towards cathode
Explanation: In the electrolysis of an ionic hydride in the molten state, hydride ions (H⁻) migrate towards the anode because they are negatively…
Correct answer: Hydrogen is liberated at the anode- A. S2
- B. SO2
- C. SO2CI2
- D. Na2S2O3
Explanation: Among these options, option C (SO2Cl2) is the one where sulfur exhibits its highest oxidation state (+6), making it the correct choice.
Correct answer: SO2CI2- A. -1/2
- B. -2
- C. -1
- D. -3
Explanation: Option A is correct. The oxidation state of oxygen in KO2 is -1/2. This can be determined by considering that potassium (K) has a +1…
Correct answer: -1/2434. Apparent charge on an atom of an element in a molecule or ion is called oxidation number. It may be:
- A. Positive
- B. Zero or fraction
- C. Negative
- D. All of these
Explanation: Option D is correct.Oxidation number is the apparent charge on an atom of an element in a molecule or ion.
Correct answer: All of these- A. Valency
- B. Coordination number
- C. Oxidation number
- D. Charge number
Explanation: The correct answer is Option C: Oxidation number. The oxidation number represents the apparent charge on an atom in a molecule, assuming…
Correct answer: Oxidation number- A. 2 e- are added on LHS
- B. 2 e- are added on RHS
- C. 4 e- are added on LHS
- D. 4 e- are added on RHS
Explanation: Option A is correctYes, the reaction you have given is a redox reaction.
Correct answer: 2 e- are added on LHS- A. MnO42-→ MnO2
- B. CrO42- → Cr+3
- C. MnO41-→ Mn+2
- D. Cr2O7-2 → 2Cr+2
Explanation: Option C is correct.The only change in the given options that involves a transfer of five electrons is MnO4^- → Mn^2+.
Correct answer: MnO41-→ Mn+2- A. Reducing agent
- B. Nitrating agent
- C. Oxidizing agent
- D. Sulphonating agent
Explanation: Option A is correct.In the reaction H2S + Cl2 → 2HCl + S, H2S acts as a reducing agent.
Correct answer: Reducing agent- A. 5e- on R.H.S
- B. 5e- on L.H.S
- C. 3e- on R.H.S
- D. 3e- on L.H.S
Explanation: Option B is correct.The balanced equation for the reaction is: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O The oxidation state of manganese changes…
Correct answer: 5e- on L.H.S- A. Cu2++ 2e- → Cu
- B. Cu + 2e- → Cu2+
- C. Hg + 1/2O2 → HgO
- D. Mg + 1/2O2 → MgO
Explanation: Option A is correct.The reaction that occurs at the cathode is the reduction reaction.
Correct answer: Cu2++ 2e- → Cu