Electrochemistry Chapter 10

This 53 MCQ test checks your understanding of redox reactions, oxidation numbers, electrolysis, Faraday’s laws and electrode potentials. It also covers galvanic cells, the Nernst equation, the activity series, photovoltaic cells and the Winkler method for BOD and DO. Each question is worth 1 mark, for a total of 55 marks. All questions are based only on the text you provided.

MCQs

53 questions · 35 min · 53 marks

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Sample questions

The first 5 of 53. Try each one, then tap to see the answer.

  1. Question 1

    The activity series of metals arranges metals in order of their:

    • AAtomic mass
    • BDensity
    • CEase of oxidation
    • DEase of reduction
    Show answer

    Answer: C

    The activity series ranks metals by their tendency to lose electrons and undergo oxidation.

  2. Question 2

    According to the activity series which of the following metals would most readily displace hydrogen gas from dilute acids?

    • ACopper (Cu)
    • BSilver (Ag)
    • CMagnesium (Mg)
    • DPlatinum (Pt)
    Show answer

    Answer: C

    Mg has a very negative E° value and is high in the series so it reacts with acids replacing hydrogen. Cu and Ag and Pt are unreactive with acids.

  3. Question 3

    The salt bridge allows transfer of ------------ in Zn-Cu voltaic cell.

    • AZn2+ ions
    • BSO4 2- ions
    • CBoth
    • DNone of these
    Show answer

    Answer: C

    A salt bridge allows the movement of ions between the two half-cells so that ionic balance is maintained but it does not allow movement of electrons.

  4. Question 4

    If a Zn-Cu galvanic cell works ideally after complete discharge both compartments will have:

    • ACuSO4 solution
    • BZnSO4 solution
    • CCu2+ ions
    • DZn solid
    Show answer

    Answer: B

    The cell reaction Zn + Cu2+ -> Zn2+ + Cu continues until all Cu2+ is used so both compartments end with ZnSO4 solution.

  5. Question 5

    Which of the following half-reactions represents a reduction process?

    • AZn(s) -> Zn2+(aq) + 2e-
    • BFe2+(aq) -> Fe3+(aq) + e-
    • CCl2(g) + 2e- -> 2Cl-(aq)
    • D2H+(aq) + 2e- <- H2(g) written as H2(g) -> 2H+(aq) + 2e-
    Show answer

    Answer: C

    Gain of electrons is reduction. Cl2 gains 2 electrons to form Cl-.

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