Electrochemistry Quiz: MCQ Exam! Trivia

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  • 1/15 Questions

    The standard electrode potential is measured by:

    • Electrometer
    • Voltmeter
    • Pyrometer
    • Galvanometer
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About This Quiz

The Electrochemistry Quiz: MCQ Exam! Trivia tests your understanding of electrochemical series, conductance, electrode potentials, and the use of salt bridges. This quiz enhances your grasp of fundamental and advanced concepts in electrochemistry, suitable for academic and professional growth.

Electrochemistry Quiz: MCQ Exam! Trivia - Quiz

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  • 2. 

    Li occupies higher position in the electrochemical series of metals as compared to Cu since:

    • The standard reduction potential of Li+/Li is lower than that of Cu+/Cu

    • The standard reduction potential of Cu+/Cu is lower than that of Li+/Li

    • The standard oxidation potential of Li+/Li is lower than that of Cu+/Cu

    • Li is smaller in size as compared to Cu

    Correct Answer
    A. The standard reduction potential of Li+/Li is lower than that of Cu+/Cu
    Explanation
    The standard reduction potential is a measure of the tendency of a species to gain electrons and be reduced. A lower standard reduction potential indicates a greater tendency to be reduced. In this case, since the standard reduction potential of Li+/Li is lower than that of Cu+/Cu, it means that Li has a greater tendency to be reduced compared to Cu. Therefore, Li occupies a higher position in the electrochemical series of metals compared to Cu.

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  • 3. 
    The standard electrode potential for the half cell reactions are: The E.M.F. of the cell reaction - ProProfs

    The standard electrode potential for the half cell reactions are: The E.M.F. of the cell reaction

    • – 1.20 V

    • + 1.20 V

    • + 0.32 V

    • – 0.32 V

    Correct Answer
    A. + 0.32 V
    Explanation
    The correct answer is + 0.32 V. This means that the standard electrode potential for the half cell reactions is + 0.32 V. This indicates that the half cell reaction is favorable and has a positive potential, meaning it is more likely to occur spontaneously.

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  • 4. 

    What would happen if the electrodes used in the electrolysis process rective:

    • Electrode does not participate in the chemical reaction

    • Electrodes only act as source and sink for electrons

    • Electrodes participate in the electrode reaction

    • Electrodes are burnt up

    Correct Answer
    A. Electrodes participate in the electrode reaction
    Explanation
    When the electrodes participate in the electrode reaction, it means that they are actively involved in the chemical reactions that occur during the electrolysis process. This involvement can include either gaining or losing electrons, which allows for the transfer of charge and the breakdown of the electrolyte into its constituent ions. In other words, the electrodes play a crucial role in facilitating the electrolysis process by enabling the flow of electrons and driving the chemical reactions.

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  • 5. 

    Saturated solution of KNO3 is used to make 'salt bridge' because:

    • KNO3 is highly soluble in water

    • Velocity of (NO3)- is greater than that of K+

    • Velocity of K+ is greater than that of (NO3)-

    • Velocity of both K+ and (NO3)- are nearly the same

    Correct Answer
    A. Velocity of both K+ and (NO3)- are nearly the same
    Explanation
    The correct answer is that the velocity of both K+ and (NO3)- ions are nearly the same. A salt bridge is used to maintain electrical neutrality in a galvanic cell by allowing the flow of ions between the half-cells. If the velocities of the ions were significantly different, it could lead to an imbalance of charges and disrupt the functioning of the cell. Therefore, using a saturated solution of KNO3, where both K+ and (NO3)- ions have similar velocities, ensures that the salt bridge effectively maintains electrical neutrality in the cell.

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  • 6. 

    When KMnO4 acts as an oxidizing agent and ultimately forms MnO42–, MnO2, Mn2O3 and Mn2+, then the number of electrons transferred in each case is:

    • 1, 3, 4, 5

    • 3, 7, 1, 5

    • 4, 3, 1, 5

    • 1, 5, 7, 3

    Correct Answer
    A. 1, 3, 4, 5
    Explanation
    The correct answer is 1, 3, 4, 5. When KMnO4 acts as an oxidizing agent, it undergoes reduction itself. In each case, the Mn in KMnO4 is reduced to a different oxidation state. MnO42- has Mn in the +7 oxidation state, MnO2 has Mn in the +4 oxidation state, Mn2O3 has Mn in the +3 oxidation state, and Mn2+ has Mn in the +2 oxidation state. The number of electrons transferred in each case corresponds to the change in oxidation state of Mn, which is 1, 3, 4, and 5 respectively.

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  • 7. 

    Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminum is produced? (Assume 100% current efficiency, atomic mass of Al = 27 g/mol)

    • 9.0 x 103 g

    • 8.1 x 104 g

    • 2.4 x 105 g

    • 9.0 x 103 g

    Correct Answer
    A. 8.1 x 104 g
    Explanation
    Aluminum oxide (Al2O3) can be reduced to aluminum (Al) through the process of electrolysis. In this process, a high current is passed through molten Al2O3 at low potentials. The amount of aluminum produced can be calculated using Faraday's law of electrolysis, which states that the amount of substance produced is directly proportional to the amount of charge passed through the electrolyte.

    Given that 4.0 x 10^4 amperes of current is passed through molten Al2O3 for 6 hours, we can calculate the total charge passed using the formula Q = I x t, where Q is the charge, I is the current, and t is the time.

    Next, we need to convert the charge to moles of electrons using Faraday's constant (F = 96,485 C/mol). The number of moles of electrons is equal to Q/F.

    Since the reaction is assumed to have 100% current efficiency, each mole of electrons is used to produce one mole of aluminum. Therefore, the number of moles of aluminum produced is equal to the number of moles of electrons.

    Finally, we can calculate the mass of aluminum produced using the molar mass of aluminum (27 g/mol) and the number of moles of aluminum.

    Therefore, the correct answer is 8.1 x 10^4 g of aluminum.

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  • 8. 

    Which of the following decreases with dilution:

    • Molar Conductance

    • Conductance

    • Specific Conductance

    • Equivalent Conductance

    Correct Answer
    A. Specific Conductance
    Explanation
    Specific conductance is a measure of the conductivity of a solution per unit volume. When a solution is diluted, the total amount of ions or charged species in the solution decreases, resulting in a decrease in the number of charged particles available to carry an electric current. As a result, the specific conductance of the solution decreases with dilution.

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  • 9. 

    The hydrogen electrode is dipped in a solution of pH 3 at 25°C. The potential would be (the value of 2.303 RT/F is 0.059 V)

    • 0.177 V

    • - 0.177 V

    • 0.059 V

    • - 0.059 V

    Correct Answer
    A. - 0.177 V
    Explanation
    The potential of the hydrogen electrode is determined by the difference in hydrogen ion concentration between the electrode and the solution it is immersed in. Since the solution has a pH of 3, it indicates a high concentration of hydrogen ions. The potential of the hydrogen electrode is negative when the concentration of hydrogen ions is high, so the correct answer is -0.177 V.

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  • 10. 

    An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:

    • Increase in number of ions

    • Increase in ionic mobility of ions

    • 100% ionisation of electrolyte at normal dilution

    • Increase in both i.e., number of ions and ionic mobility of ions

    Correct Answer
    A. Increase in ionic mobility of ions
    Explanation
    An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to an increase in ionic mobility of ions. As the concentration of the electrolyte decreases, the ions become more dispersed and have more space to move freely. This leads to an increase in their mobility, allowing them to carry more charge and contribute to a higher conductance. The increase in the number of ions alone does not necessarily result in a higher conductance, as their mobility also plays a crucial role in the overall conductivity of the solution.

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  • 11. 

    Resistance of a conductive cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω. The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 Ω. The molar conductivity of 0.02 M solution of the electrolyte will be:

    • 1240 x 10-4 S m2 mol-1

    • 124 x 10-4 S m2 mol-1

    • 12.4 x 10-4 S m2 mol-1

    • 1.24 x 10-4 S m2 mol-1

    Correct Answer
    A. 12.4 x 10-4 S m2 mol-1
    Explanation
    The resistance of a conductive cell is directly proportional to the length of the cell and inversely proportional to the cross-sectional area of the cell. Therefore, when the concentration of the electrolyte solution is doubled, the resistance of the cell also doubles. This means that the conductivity of the solution is halved. The molar conductivity is defined as the conductivity of a solution divided by the concentration of the electrolyte. Since the conductivity is halved when the concentration is doubled, the molar conductivity of the 0.02 M solution will be 12.4 x 10-4 S m2 mol-1.

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  • 12. 

    9.65 C of electric current is passed through fused anhydrous magnesium chloride. The magnesium metal thus, obtained is completely converted into a Grignard reagent. The number of moles of the Grignard reagent obtained is:

    • 5 x 10-4

    • 1 x 10-4

    • 5 x 10-5

    • 1 x 10-5

    Correct Answer
    A. 5 x 10-5
    Explanation
    When 9.65 C of electric current is passed through fused anhydrous magnesium chloride, the magnesium metal is completely converted into a Grignard reagent. The number of moles of the Grignard reagent obtained can be calculated using Faraday's law of electrolysis. According to this law, the amount of substance produced at an electrode is directly proportional to the quantity of electricity passed through it. The formula to calculate the number of moles is moles = charge / Faraday's constant. In this case, the charge is 9.65 C and the Faraday's constant is 96485 C/mol. By substituting these values into the formula, we get 9.65 / 96485 = 5 x 10-5 mol.

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  • 13. 

    Which of the following reaction is preferred during electrolysis of dilute sulphuric acid:

    • 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-

    • 2SO42- (g) → S2O82- (g) + 2e-

    • 2H2 (g) + 4OH- (aq) → 4H2O (l) + 4e-

    • 2SO42- (aq) → S2O82- (aq) + 2e-

    Correct Answer
    A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-
    Explanation
    During the electrolysis of dilute sulphuric acid, the preferred reaction is 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-. This is because water molecules are broken down into oxygen gas, hydrogen ions, and electrons. This reaction occurs at the anode, where oxidation takes place. The production of oxygen gas is preferred as it can be collected and used for various applications. The other reactions listed do not involve the production of oxygen gas and are therefore not preferred during the electrolysis of dilute sulphuric acid.

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  • 14. 

    For a cell reaction involving two electrons change, the standard E.M.F. of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C is:

    • 2.95 x 10-2

    • 2.95 x 102

    • 1 x 1010

    • 1 x 10-10

    Correct Answer
    A. 1 x 1010
    Explanation
    The equilibrium constant of a cell reaction is related to the standard E.M.F. of the cell through the Nernst equation. The Nernst equation states that the standard E.M.F. of the cell is equal to the equilibrium constant multiplied by the natural logarithm of the reaction quotient. In this case, the standard E.M.F. is given as 0.295 V, and since the reaction involves two electrons, the equilibrium constant can be calculated as 10^(0.295/0.059) = 1 x 10^10.

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  • 15. 

    The reduction potential of hydrogen half-cell will be negative if:

    • P(H2) = 1 atm and [H+] =  M

    • P(H2) = 1 atm and [H+] = 2 M

    • P(H2) = 2 atm and [H+] = 1 M

    • P(H2) = 2 atm and [H+] = 2 M

    Correct Answer
    A. P(H2) = 2 atm and [H+] = 2 M
    Explanation
    The reduction potential of the hydrogen half-cell will be negative if the partial pressure of hydrogen (p(H2)) is increased to 2 atm and the concentration of hydrogen ions ([H+]) is increased to 2 M. This is because an increase in p(H2) and [H+] will shift the equilibrium towards the reactants, causing a decrease in the reduction potential.

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  • Current Version
  • May 26, 2024
    Quiz Edited by
    ProProfs Editorial Team
  • Nov 30, 2013
    Quiz Created by
    Tanmay Shankar
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