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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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Redox ReactionsOxidation and ReductionBalancing Redox EquationsStandard Hydrogen Electrode
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696 questions · page 20 of 35

381. Which one of the following is a strong electrolyte in solution?

  • A. Acetic acid
  • B. Ammonium hydroxide
  • C. Carbonic acid
  • D. Potassium iodide

Explanation: Potassium iodide (KI) is considered a strong electrolyte because it dissociates completely into its constituent ions when dissolved in…

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Correct answer: Potassium iodide

382. The electrode potential of the standard hydrogen electrode is chosen as

  • A. -1V
  • B. 0V
  • C. 1V
  • D. 2V

Explanation: The potential of S.H.E is chosen as 0 volts.

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Correct answer: 0V

383. What change in oxidation number of Cu takes place in the following reaction: Cu+4HNO3 →Cu(NO3 )2 +2NO2 +2H2 O

  • A. 0 to -1
  • B. 0 to +1
  • C. 0 to -2
  • D. 0 to +2

Explanation: Elemental copper (Cu) has an oxidation number of 0. In Cu(NO3)2, copper has an oxidation number of +2.

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Correct answer: 0 to +2

384. Electric current in solutions is carried by:

  • A. Ions
  • B. Canal rays
  • C. Free protons
  • D. Free electrons

Explanation: Correct. In solutions, ions (positive and negative) carry the electrical charge.

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Correct answer: Ions

385. In an experiment, 50 ml of 0.1M solution of a salt reacted with 25 ml of 0.1M solution of Sodium sulfite. The half equation for the oxidation of sulfite ion is: SO32-(Aq) H2O→ SO42- + 2H+ + 2e- If the oxidation number of the metal in the salt was 3, what would be the new oxidation number of the metal?

  • A. 0
  • B. 1
  • C. 2
  • D. 4

Explanation: In this redox reaction, the sulfite ion (SO32-) is oxidized to sulfate (SO42-), releasing 2 electrons.

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Correct answer: 2

386. In the balanced chemical reaction, IO3- + aI- + bH+→ cH2O + dI2 a, b, c and d respectively correspond to:

  • A. 5, 6, 3, 3
  • B. 5, 3, 6, 3
  • C. 3, 5, 3, 6
  • D. 5, 6, 5, 5

Explanation: Following is the solution to this question:

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Correct answer: 5, 6, 3, 3

387. Oxidation state of 'V' in Rb4Na[HV10O28] is:

  • A. +5
  • B. +6
  • C. +7/5
  • D. +4

Explanation: The oxidation state of vanadium in Rb4Na[HV10O28] can be determined by considering the overall charge of the compound.

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Correct answer: +5

388. In which of the following reaction(s) does H2SO4 act as an oxidizing agent and as an acid?

  • A. Option A
  • B. Option B
  • C. Option C
  • D. Option D

Explanation: In Option C, H₂SO₄ functions as both an oxidizing agent and an acid. It oxidizes copper (Cu) by taking electrons from it, which is…

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Correct answer: Option C

389. In a reaction, 3 moles of electrons are gained by 1 mole of HNO3. Assuming no change in the oxidation number of hydrogen and oxygen, the possible product obtained due to reduction will be:

  • A. 1 mole of NO2
  • B. 0.5 mole of N2O
  • C. 0.5 mole of NO
  • D. 0.5 mole of N2O3

Explanation: The key to solving this question lies in determining the change in oxidation state of nitrogen in HNO3 during reduction.

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Correct answer: 0.5 mole of NO

390. 10-2 mole of Fe3O4 is treated with excess KI solution in presence of dilute H2SO4, the products are Fe2+ and I2(g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2(g)?

  • A. 50 ml
  • B. 100 ml
  • C. 200 ml
  • D. 400 ml

Explanation: When Fe3O4 reacts with excess KI and dilute H2SO4, it produces Fe2+ and I2. For every mole of Fe3O4, 4 moles of I2 are produced.

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Correct answer: 200 ml

391. Find the weight of KOH in its 50 milli equivalents:

  • A. 1.6
  • B. 2.2
  • C. 2.8
  • D. 4.8

Explanation: To find the weight of KOH in its 50 milliequivalents, we use the formula:Weight (in grams) = Milliequivalents × Equivalent weightThe…

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Correct answer: 2.8

392. A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate the calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaCl2 in the original mixture is:

  • A. 15.2%
  • B. 32.1%
  • C. 21.8%
  • D. 11.07%

Explanation: The problem involves determining the percentage by mass of CaCl2 in a mixture.

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Correct answer: 32.1%

393. Equal volumes of 1 M each of KMnO4 and K2Cr2O7 are used to oxidise Fe(II) solution in acidic medium. The amount of Fe oxidized will be :

  • A. More with KMnO4
  • B. Equal with both oxidizing agents
  • C. More with K2Cr2O7
  • D. Cannot be determined

Explanation: Both KMnO4 and K2Cr2O7 are strong oxidizing agents that can oxidize Fe(II) to Fe(III) in acidic medium.

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Correct answer: More with K2Cr2O7

394. How many millilitre of 0.5 M H2SO4 are needed to dissolve 0.5 g of Cu(II) carbonate?

  • A. 6.01
  • B. 4.5
  • C. 8.1
  • D. 11.1

Explanation: To determine the volume of 0.5 M H2SO4 needed, we first calculate the moles of Cu(II) carbonate (CuCO3) using its molar mass (123.55…

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Correct answer: 8.1

395. O.N. of V in Rb4 Na [HV10O28] is:

  • A. +3
  • B. +5
  • C. +4
  • D. Zero

Explanation: To determine the oxidation number of vanadium (V) in the compound Rb4Na[HV10O28], consider the following: - Rubidium (Rb) has an oxidation…

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Correct answer: +5

396. When an alkaline solution of K2CrO4 is treated with 3% H2O2 solution, red brown paramagnetic peroxochromate is obtained as per following equation: 2K2CrO4 + 7H2O2 + 2KOH → 2K3CrO8 + 8H2O The equivalent weight of K2CrO4 for above transformation must be (assuming M is the molar mass of K2CrO4):

  • A. M/16
  • B. M
  • C. M/12
  • D. M/2

Explanation: The equivalent weight of a substance in a reaction is given by its molar mass divided by the n-factor.

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Correct answer: M

397. In a reaction, 3 moles of electrons are gained by 1 mole of HNO3. Assuming no change in O.N. of hydrogen and 23 oxygen, the possible product obtained due to reduction will be :

  • A. 1 mole of NO2
  • B. 0.5 mole of N2O
  • C. 1 mole of NO
  • D. 0.5 mole of N2O3

Explanation: In this reaction, the reduction of HNO3 involves a change in the oxidation state of nitrogen from +5 to +2, which corresponds to the…

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Correct answer: 1 mole of NO

398. Which reaction does not involve either in oxidation or reduction?

  • A. VO2+ → V2O3
  • B. Na → Na+
  • C. CrO42- → Cr2O72-
  • D. Zn2+ → Zn

Explanation: The correct answer is option C: CrO42- → Cr2O72-. In this reaction, there is no change in the oxidation state of chromium, indicating…

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Correct answer: CrO42- → Cr2O72-

399. How many kJ of energy is spent when a current of 4 amp passes for 200 second under a potential of 115 V?

  • A. 52 kJ
  • B. 72 kJ
  • C. 82 kJ
  • D. 92 kJ

Explanation: To calculate the energy spent, use the formula for electrical energy: Energy (in joules) = Voltage (V) × Current (I) × Time (t).

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Correct answer: 92 kJ

400. The minimum equivalent conductance in fused state is shown by:

  • A. MgCl2
  • B. BeCl2
  • C. CrCl2
  • D. SrCl2

Explanation: Beryllium chloride (BeCl2) has a significant covalent character, meaning it does not dissociate completely into ions in the fused state.

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Correct answer: BeCl2
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