Electrochemistry MCQ Quiz Exam!

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  • 1/20 Questions

    What is the charge of the cations of group 2 elements?

    • +1
    • +2
    • +3
    • +4
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About This Quiz

The 'Electrochemistry MCQ Quiz Exam' assesses understanding of key electrochemical concepts, including ion charges, electrolysis, and redox reactions. Ideal for learners aiming to master electrochemistry fundamentals and apply them in academic or professional settings.

Electrochemistry MCQ Quiz Exam! - Quiz

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  • 2. 

    What process requires electrical energy to allow a nonspontaneous chemical redox reaction to occur?

    • Oxidation

    • Reduction

    • Electrolysis

    • Combustion

    Correct Answer
    A. Electrolysis
    Explanation
    Electrolysis is the process that requires electrical energy to allow a nonspontaneous chemical redox reaction to occur. In this process, an electric current is passed through an electrolyte solution, causing the ions to move and undergo oxidation and reduction reactions. The electrical energy provides the necessary driving force to separate the ions and facilitate the nonspontaneous redox reaction.

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  • 3. 

    Given the half-reaction, Fe3+(aq)   +   e-      →  Fe2+(aq)   , identify the type of chemical reaction

    • Reduction

    • Oxidation

    • Cathodic

    • Both reduction and oxidation

    • None of the above

    Correct Answer
    A. Reduction
    Explanation
    The given half-reaction involves the gain of an electron, which indicates a reduction process. Reduction is defined as the gain of electrons by a species, resulting in a decrease in its oxidation state. Therefore, the correct answer is reduction.

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  • 4. 

    What is the unit of the electric current defined as 1coulomb per second?

    • 1 ampere

    • 1 volt

    • 1 coulomb

    • 1 ohm

    Correct Answer
    A. 1 ampere
    Explanation
    The unit of electric current defined as 1 coulomb per second is called 1 ampere. Ampere is the SI unit of electric current and represents the rate of flow of electric charge. It is commonly used to measure the intensity of electric current in a circuit.

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  • 5. 

    What is the charge of aluminum cation?

    • +1

    • +2

    • +3

    • +4

    Correct Answer
    A. +3
    Explanation
    Aluminum is a metal and tends to lose three electrons to achieve a stable electron configuration. When aluminum loses these three electrons, it forms a cation with a charge of +3. This is because the number of protons in the nucleus remains the same, but the number of electrons decreases, resulting in a net positive charge. Therefore, the correct answer is +3.

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  • 6. 

    What is the unit measure for the difference in chemical potentials between two electrodes of an electrochemical cell?

    • Volt

    • Ohm

    • Ampere

    • Coulomb

    Correct Answer
    A. Volt
    Explanation
    The unit measure for the difference in chemical potentials between two electrodes of an electrochemical cell is volt. This is because the chemical potential difference, also known as the electromotive force (EMF), is measured in volts. It represents the potential energy per unit charge that is available to drive an electric current in the cell. The volt is the standard unit for measuring electric potential difference or voltage.

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  • 7. 

    Where does oxidation half-reaction occur in a galvanic cell?

    • Anode

    • Cathode

    • Salt bridge

    • Either cathode or anode

    Correct Answer
    A. Anode
    Explanation
    In a galvanic cell, oxidation half-reaction occurs at the anode. This is because the anode is where the oxidation process takes place, where atoms or ions lose electrons and become positively charged. The electrons released during oxidation flow through the external circuit to the cathode, where reduction half-reaction occurs. Therefore, the anode is the site of oxidation, while the cathode is where reduction occurs in a galvanic cell.

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  • 8. 

    Which of the following statements is correct?

    • Reduction occurs at the anode in both electrolytic and galvanic cell.

    • Oxidation occurs at the cathode in both electrolytic and galvanic cell.

    • Reduction occurs at the cathode in both electrolytic and galvanic cell

    • Oxidation occurs at the cathode of galvanic cell only.

    Correct Answer
    A. Reduction occurs at the cathode in both electrolytic and galvanic cell
    Explanation
    In both electrolytic and galvanic cells, reduction occurs at the cathode. In an electrolytic cell, electrical energy is used to drive a non-spontaneous redox reaction, causing reduction to occur at the cathode. In a galvanic cell, the spontaneous redox reaction produces electrical energy, and reduction also occurs at the cathode. Therefore, reduction always takes place at the cathode in both types of cells.

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  • 9. 

    What is the type of electrochemical cell that produces electricity from a spontaneous chemical redox reaction?

    • Electrolytic cell

    • Downs cell

    • Electroplating cell

    • Galvanic cell

    Correct Answer
    A. Galvanic cell
    Explanation
    A galvanic cell is a type of electrochemical cell that produces electricity from a spontaneous chemical redox reaction. In a galvanic cell, chemical energy is converted into electrical energy through the transfer of electrons from the anode (where oxidation occurs) to the cathode (where reduction occurs). This flow of electrons creates an electric current that can be used to power devices. Unlike an electrolytic cell, which requires an external power source to drive a non-spontaneous redox reaction, a galvanic cell operates spontaneously without the need for an external power supply.

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  • 10. 

    Which is is the anode of the galvanic cell, given the electrochemical notation. Fe (s) │Fe2+(aq)   ││ Cu2+(aq)  │Cu (s)

    • Fe (s)

    • Fe2+(aq)

    • Cu2+(aq)

    • Cu (s)

    Correct Answer
    A. Fe (s)
    Explanation
    The anode of a galvanic cell is the electrode where oxidation occurs. In the given electrochemical notation, Fe (s) is placed before Fe2+(aq), indicating that Fe (s) is being oxidized to Fe2+(aq). Therefore, Fe (s) is the anode of the galvanic cell.

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  • 11. 

    Which is the anode , given the electrochemical notation of a galvanic cell,  Ni(s) │ Ni2+(aq)    ││  Ag+(aq)    │ Ag (s

    • Ni(s)

    • Ni2+(aq)

    • Ag+(aq)

    • Ag (s

    Correct Answer
    A. Ni(s)
    Explanation
    In the given electrochemical notation, the anode is the electrode where oxidation occurs. In this case, Ni(s) is the anode because it is being oxidized to Ni2+(aq). The Ag(s) electrode is the cathode because it is being reduced to Ag+(aq).

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  • 12. 

    What is the charge that passes through a point when one ampere of electric current flows for one second?

    •  1 coulomb

    • 1 faraday

    • 1 volt

    • 1 ohm

    Correct Answer
    A.  1 coulomb
    Explanation
    When one ampere of electric current flows for one second, the charge that passes through a point is defined as one coulomb. This is because the coulomb is the unit of electric charge in the International System of Units (SI). It is equivalent to the charge transported by a constant current of one ampere in one second. Therefore, the correct answer is 1 coulomb.

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  • 13. 

     What is the product of electrolysis of  molten sodium chloride at the cathode?

    • Chlorine gas

    • Sodium metal

    • Water vapor

    • Hydrogen as

    Correct Answer
    A. Sodium metal
    Explanation
    During the electrolysis of molten sodium chloride, the positive sodium ions migrate towards the cathode (negative electrode) where they gain electrons and are reduced to form sodium metal. Therefore, the product of electrolysis of molten sodium chloride at the cathode is sodium metal.

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  • 14. 

    In the electroplating process at which metal electrode surface can aluminum deposit coming from a molten aluminum oxide?

    • Anode

    • Cathode

    • Either anode or cathode

    • Neither anode nor cathode

    Correct Answer
    A. Cathode
    Explanation
    In the electroplating process, the metal electrode surface can deposit aluminum when it is the cathode. The cathode is the negatively charged electrode where reduction reactions occur. During electroplating, the metal ions in the molten aluminum oxide are reduced and deposited onto the cathode, resulting in the formation of a layer of aluminum on the surface of the electrode. Therefore, the correct answer is cathode.

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  • 15. 

    Given the galvanic cell reaction,  Cu (s) + 2Ag+ (aq) → Cu2+(aq) + 2 Ag(s)  , which is the reduced product?

    • Cu (s)

    • 2Ag+ (aq)

    • Cu2+(aq)

    • 2 Ag(s)

    Correct Answer
    A. 2 Ag(s)
    Explanation
    In the given galvanic cell reaction, Cu (s) is being oxidized to Cu2+(aq), while 2 Ag+ (aq) is being reduced to 2 Ag(s). The reduced product refers to the substance that gains electrons and is therefore being reduced. In this case, the reduced product is 2 Ag(s) because Ag+ ions gain electrons to form Ag atoms.

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  • 16. 

    If a galvanic cell is set up with Nickel as an anode, which of these can be the metal cathode? 

    • Magnesium

    • Zinc

    • Copper

    • Aluminum

    Correct Answer
    A. Copper
    Explanation
    Copper can be the metal cathode in a galvanic cell with Nickel as an anode because copper has a lower reduction potential than nickel. In a galvanic cell, the metal with the higher reduction potential will act as the cathode, while the metal with the lower reduction potential will act as the anode. Since copper has a lower reduction potential than nickel, it can accept electrons and be reduced at the cathode while nickel is oxidized at the anode.

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  • 17. 

    Which of these metals has the highest oxidation potential in the electrochemical series?

    • Magnesium

    • Aluminum

    • Lithium

    • Copper

    Correct Answer
    A. Lithium
    Explanation
    Lithium has the highest oxidation potential in the electrochemical series. This means that lithium is the most easily oxidized metal, or in other words, it readily loses electrons in a chemical reaction. This high oxidation potential indicates that lithium has a strong tendency to form positive ions, making it a highly reactive metal.

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  • 18. 

    What is the number of mole aluminum metal produced in electrolysis if 96485 C/s  of electricity is passed in 1 second?

    • 1/3 mole

    •  1/2 mole

    • 1 mole

    •  2.5 moles

    Correct Answer
    A. 1 mole
    Explanation
    To determine the number of moles of aluminum metal produced in electrolysis, we'll use Faraday's constant, which is 96485 C/mol.
    Given that 96485 C/s of electricity is passed in 1 second, we can calculate the number of moles of aluminum produced as follows:
    Number of moles = (Amount of charge passed) / (Faraday's constant) = (96485 C/s) / (96485 C/mol) = 1 mol
    Therefore, 1 mole of aluminum metal is produced in electrolysis if 96485 C/s of electricity is passed in 1 second. So, the correct answer is 1 mole.

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  • 19. 

    What is the amount of copper metal produced when one faraday of electricity is passed through a copper sulfate solution in an electrolytic cell?

    • 2 moles

    • 1 mole

    • 1/2 mole

    • 1/4 mole

    Correct Answer
    A. 1/2 mole
    Explanation
    When one faraday of electricity is passed through a copper sulfate solution in an electrolytic cell, it results in the production of 1/2 mole of copper metal. This is because one faraday of electricity is equivalent to the charge required to convert one mole of a substance during an electrochemical reaction. In the case of copper, the reaction involves the reduction of Cu2+ ions to Cu atoms, and since the molar ratio between Cu2+ and Cu is 2:1, only 1/2 mole of copper metal is produced.

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  • 20. 

    Given the electrochemical notation of a galvanic cell, Cd(s) │ Cd2+(aq)        ││  Pb2+(aq)    │ Pb(s) which is the oxidized product at one of the electrodes?

    • Cd(s)

    • Cd2+(aq)

    • Pb2+(aq)

    • Pb(s)

    Correct Answer
    A. Cd(s)
    Explanation
    In the electrochemical notation of a galvanic cell, the electrode on the left side (anode) undergoes oxidation, while the electrode on the right side (cathode) undergoes reduction.
    So, looking at the given notation:
    The left electrode (anode) is: Cd(s) │ Cd2+(aq)
    The right electrode (cathode) is: Pb2+(aq) │ Pb(s)
    Therefore, in this galvanic cell, the oxidized product at one of the electrodes is Cd(s).

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  • Current Version
  • Feb 19, 2024
    Quiz Edited by
    ProProfs Editorial Team
  • Sep 27, 2020
    Quiz Created by
    Leonida Norma
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