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Secondary 4 Pure Chemistry Redox Electrochemistry Quiz

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Questions

Secondary 4 Pure Chemistry Quiz - Redox Electrochemistry

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a calculator.

Section A: Multiple Choice & Short Concepts (Questions 1-5)

1. Which statement correctly describes oxidation and reduction in terms of electron transfer?
A. Oxidation is the gain of electrons; reduction is the loss of electrons.
B. Oxidation is the loss of electrons; reduction is the gain of electrons.
C. Oxidation is the gain of hydrogen; reduction is the loss of oxygen.
D. Oxidation is the decrease in oxidation state; reduction is the increase in oxidation state.

[1]

2. In the reaction below, which species acts as the oxidising agent?
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$
A. $Zn$
B. $Cu^{2+}$
C. $Zn^{2+}$
D. $Cu$

[1]

3. What is the oxidation state of manganese in potassium manganate(VII), $KMnO_4$ ?
A. +2
B. +5
C. +6
D. +7

[1]

4. During the electrolysis of molten lead(II) bromide, $PbBr_2$ , which product is formed at the anode?
A. Lead metal
B. Bromine gas
C. Hydrogen gas
D. Oxygen gas

[1]

5. Which of the following ions will be preferentially discharged at the cathode during the electrolysis of dilute aqueous sodium chloride?
A. $Na^+$
B. $H^+$
C. $Cl^-$
D. $OH^-$

[1]

Section B: Electrolysis Concepts & Cells (Questions 6-10)

6. In a simple chemical cell consisting of magnesium and copper strips dipped in dilute sulfuric acid, electrons flow from:
A. Copper to Magnesium through the wire.
B. Magnesium to Copper through the wire.
C. Copper to Magnesium through the solution.
D. Magnesium to Copper through the solution.

[1]

7. Identify the change that represents reduction.
A. $Fe^{2+} \rightarrow Fe^{3+}$
B. $Cl_2 \rightarrow 2Cl^-$
C. $Na \rightarrow Na^+$
D. $SO_2 \rightarrow SO_3$

[1]

8. Why is graphite often used as an electrode in electrolysis?
A. It is a metal that conducts electricity well.
B. It is unreactive and conducts electricity due to delocalised electrons.
C. It has a high melting point and forms ionic bonds.
D. It is cheap and dissolves in the electrolyte.

[1]

9. What is the observation at the anode during the electrolysis of concentrated aqueous copper(II) chloride using inert electrodes?
A. A brown solid deposits.
B. A colourless gas bubbles off.
C. A greenish-yellow gas bubbles off.
D. The blue solution fades.

[1]

10. Which metal would provide the highest voltage when paired with copper in a simple cell, given the reactivity series: $Mg > Zn > Fe > Pb > Cu$ ?
A. Zinc
B. Iron
C. Magnesium
D. Lead

[1]

Section C: Structured Questions - Electrolysis of Brine (Questions 11-15)

11. The diagram below shows the setup for the electrolysis of concentrated aqueous sodium chloride (brine) using inert graphite electrodes.

Image pending generation for this question.

Name the cations present in the solution.
[1]

12. Referring to the setup in Question 11, name the anions present in the solution.
[1]

13. Write the half-equation for the reaction occurring at the anode in the electrolysis of brine.
[1]

14. Explain why chlorine is formed at the anode instead of oxygen, despite the position in the electrochemical series.
[2]

15. State the observation at the cathode and write the half-equation for the reaction occurring there.
[2]
Observation: _______________________________________________________________
Half-equation: _____________________________________________________________

Section D: Structured Questions - Simple Cells & Redox Applications (Questions 16-20)

16. A student sets up a simple chemical cell using Zinc and Copper electrodes in dilute sulfuric acid. Identify the negative terminal and explain your answer in terms of reactivity and electron loss.
[2]
Negative Terminal: _________________________
Explanation: __________________________________________________________________

17. In the Zinc-Copper cell described in Question 16, gas bubbles are observed at one electrode. Identify the gas and the electrode where it is observed.
[2]
Gas: _________________________
Electrode: _________________________

18. Iron(III) oxide is reduced to iron using carbon monoxide: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$ .
Determine the oxidation state of iron in $Fe_2O_3$ and in the product $Fe$ .
[2]
Oxidation state in $Fe_2O_3$ : _______________
Oxidation state in $Fe$ : _______________

19. Referring to the reaction in Question 18, identify the reducing agent and explain why it is named as such in terms of electron transfer or oxygen gain/loss.
[2]
Reducing Agent: _________________________
Explanation: __________________________________________________________________

20. Electrolysis is used to purify copper using impure copper as the anode and pure copper as the cathode. Explain why the concentration of copper(II) ions in the electrolyte remains constant during this process.
[2]

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Answers

Secondary 4 Pure Chemistry Quiz - Redox Electrochemistry (Answer Key)

Total Marks: 40

Section A: Multiple Choice & Short Concepts (Questions 1-5)

1. B
[1] Oxidation is loss of electrons (OIL), Reduction is gain of electrons (RIG).

2. B
[1] $Cu^{2+}$ gains electrons to form $Cu$ . The species gaining electrons is the oxidising agent.

3. D
[1] $K (+1) + Mn (x) + 4 \times O (-2) = 0 \Rightarrow 1 + x - 8 = 0 \Rightarrow x = +7$ .

4. B
[1] At the anode (positive), anions are attracted. $Br^-$ is discharged to form $Br_2$ (bromine gas/liquid vapour).

5. B
[1] In dilute aqueous solutions, $H^+$ is preferentially discharged over $Na^+$ because hydrogen is lower in the electrochemical series (easier to reduce).

Section B: Electrolysis Concepts & Cells (Questions 6-10)

6. B
[1] Magnesium is more reactive than copper. It loses electrons more readily, becoming the negative terminal. Electrons flow from negative (Mg) to positive (Cu) via the wire.

7. B
[1] $Cl_2$ (oxidation state 0) gains electrons to become $Cl^-$ (oxidation state -1). Gain of electrons is reduction.

8. B
[1] Graphite is a non-metal but conducts electricity due to delocalised electrons between layers. It is also inert/unreactive.

9. C
[1] In concentrated $CuCl_2$ , $Cl^-$ is discharged at the anode in preference to $OH^-$ . Chlorine gas is greenish-yellow.

10. C
[1] The greater the difference in reactivity between the two metals, the higher the voltage. Magnesium is furthest from Copper in the given series.

Section C: Structured Questions - Electrolysis of Brine (Questions 11-15)

11. Cations: $Na^+$ , $H^+$
[1] (Both required)

12. Anions: $Cl^-$ , $OH^-$
[1] (Both required)

13. Anode Half-equation: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$
[1]

14. Explanation: Although $OH^-$ is lower in the electrochemical series, the concentration of $Cl^-$ is much higher in concentrated brine. Therefore, $Cl^-$ is preferentially discharged.
[2] (1 mark for concentration factor, 1 mark for linking to discharge)

15. Observation: Effervescence / Colourless gas bubbles.
Half-equation: $2H^+(aq) + 2e^- \rightarrow H_2(g)$
[2] (1 mark for observation, 1 mark for equation)

Section D: Structured Questions - Simple Cells & Redox Applications (Questions 16-20)

16. Negative Terminal: Zinc.
Explanation: Zinc is more reactive than copper. It has a greater tendency to lose electrons (oxidise) to form ions, leaving electrons on the electrode.
[2] (1 mark for identification, 1 mark for explanation)

17. Gas: Hydrogen.
Electrode: Copper (Positive terminal/Cathode).
[2] (1 mark for gas, 1 mark for electrode)

18. Oxidation state in $Fe_2O_3$ : +3
Oxidation state in $Fe$ : 0
[2] (1 mark for each)

19. Reducing Agent: Carbon monoxide ( $CO$ ).
Explanation: It causes reduction in another substance by donating electrons (or gaining oxygen to form $CO_2$ ).
[2] (1 mark for identification, 1 mark for explanation)

20. Explanation: For every copper atom that oxidises at the anode to form a $Cu^{2+}$ ion, one $Cu^{2+}$ ion is reduced at the cathode to form a copper atom. The rate of formation equals the rate of removal, so the net concentration remains constant.
[2] (1 mark for 1:1 ratio/process description, 1 mark for conclusion on concentration)