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 31 of 35
- A. At the start of procedure
- B. Somewhere in the middle of balancing
- C. After Adding two half reactions
- D. Before Adding two half reactions
Explanation: For the reduction half, there are 12 positive charges on the left side of the equation and 6 positive charges on the right side of the…
Correct answer: Before Adding two half reactions- A. Just to be sure, I called three more doctors' offices.
- B. Just to be sure, I called three more doctors offices.
- C. Just to be sure, I called three more doctor's' offices.
- D. Just to be sure, I called three more doctor offices.
Explanation: This option is correct. It uses a comma correctly after the introductory phrase "Just to be sure." Additionally, "doctors' offices" is the…
Correct answer: Just to be sure, I called three more doctors' offices.603. Number of electrons added on both sides of oxidation and reduction half reactions are balanced _.
- A. At the start of procedure
- B. Somewhere in the middle of balancing
- C. After adding two half reactions
- D. Before adding two half reactions
Explanation: A redox equation can be balanced using the following stepwise procedure: (1) Divide the equation into two half-reactions.
Correct answer: Before adding two half reactions- A. Acidic Medium
- B. Basic Medium
- C. Neutral solution
- D. All of these
Explanation: The addition of OH⁻ ions can happen in all three scenarios - acidic, basic, and neutral solutions.
Correct answer: All of these- A. +7
- B. +3
- C. +6
- D. +5
Explanation: a) +7:The oxidation number of iodine in periodic acid (H5IO6) is indeed +7.
Correct answer: +7- A. +1
- B. -2
- C. -1
- D. -1/2
Explanation: The explanation for this question will be added soon.
Correct answer: -1- A. Oxidation potential
- B. Reduction potential
- C. Redox potential
- D. EMF of cell
Explanation: The stronger the oxidizing agent, the greater will be its reduction potential.
Correct answer: Reduction potential608. Reduction occurs at:
- A. Anode
- B. Cathode
- C. Salt bridge
- D. None of these options are correct
Explanation: Oxidation occurs at the anode and reduction occurs at the cathode. Since the reaction at the anode is the source of electrons for the…
Correct answer: Cathode- A. +1
- B. +2
- C. 0
- D. -1
Explanation: Free state means uncombined state, i.e., atomic state. In an uncombined state the oxidation number will always be zero.
Correct answer: 0- A. Anode
- B. Cathode
- C. SHE
- D. Salt bridge
Explanation: Reduction happens at the negative cathode because this is where positive ions gain electrons.
Correct answer: Cathode- A. Anode
- B. Cathode
- C. SHE
- D. Salt bridge
Explanation: During electrolysis, reduction occurs at cathode
Correct answer: Cathode- A. 7
- B. -7
- C. 6
- D. -6
Explanation: The oxidation number of Mn in MnO4- can be calculated by considering the overall charge of the compound and assigning oxidation numbers to…
Correct answer: 7- A. Gain
- B. Lost
- C. Accept
- D. Produced
Explanation: The number of charges present on a cation depends on the number of electrons lost by the atom.
Correct answer: Lost- A. Unity
- B. Positive
- C. Zero
- D. Negative
Explanation: Option C is correct since a free state means an 'uncombined state', i.e., atomic state.
Correct answer: Zero- A. Discharged at anode
- B. Discharge at cathode
- C. Do not discharge
- D. None of these options is correct
Explanation: At Cathode Sodium is not discharged but H2 gas is discharged. Sodium remains in the solution and more H2 gas is formed at the end of…
Correct answer: Do not discharge- A. Reduction
- B. Hydrolysis
- C. Dehydration
- D. Oxidation
Explanation: K2Cr2O7 and H2SO4 lose their oxygen while acting as a reagent hence leading to the "oxidation" of the reactant. Hence, the answer will be D.
Correct answer: Oxidation617. Pure metals_?
- A. Corrode slowly
- B. Does not corrode easily
- C. Corrode rapidly
- D. None of these
Explanation: Pure metals do not corrode. The absence of impurities and the ability to develop protective surface layers make pure metals less prone to…
Correct answer: Does not corrode easily- A. O2, H2
- B. O2, Na
- C. O2, SO2
- D. O2, S2O4-2
Explanation: Keeping in mind the electrochemical series; OH- gets discharged at the anode whereas H+ gets discharged at the cathode giving out O2 and…
Correct answer: O2, H2- A. Fe
- B. Cu
- C. Zn
- D. Sn
Explanation: d) Sn (tin):The action of nitric acid (HNO3) on tin (Sn) does indeed produce nitric oxide (NO).
Correct answer: Sn- A. Mercury
- B. Titanium
- C. Graphite
- D. Copper
Explanation: The explanation for this question will be added soon.
Correct answer: Titanium