i. Which of the following statement is not correct about the galvanic cell?
Answer: c) Electrons flow from cathode to anode
Explanation: In a galvanic cell, oxidation occurs at the anode (negative electrode) and reduction at the cathode (positive electrode). Electrons are released at the anode and flow through the external circuit to the cathode. Therefore, the flow of electrons is from anode to cathode, not cathode to anode.
ii. Which of the following is not true about the Daniel cell?
Answer: a) Half-cell of an active metal acts as a cathode
Explanation: In a Daniel cell, Zinc (Zn) is the more active metal. It undergoes oxidation and acts as the anode. The less active metal, Copper (Cu), acts as the cathode. Therefore, the active metal acts as the anode, not the cathode.
iii. Which of the following does not involve an electrolytic process?
Answer: d) Reduction of metal oxide by a reducing agent
Explanation: Electrolytic processes use electrical energy to drive non-spontaneous chemical reactions. Refining, manufacturing sodium, and electroplating all require electricity. Reduction of a metal oxide by a chemical reducing agent (like carbon) is a chemical process, not an electrolytic one.
iv. Galvanizing is:
Answer: b) Coating with Zn
Explanation: Galvanization is the process of applying a protective zinc coating to steel or iron to prevent rusting. Zinc acts as a sacrificial metal.
v. Which of the following is true for the Nelson cell?
Answer: b) Chlorine gas is produced at the anode
Explanation: In the Nelson cell (used for the electrolysis of brine), oxidation occurs at the anode. Chloride ions (Cl⁻) are oxidized to produce chlorine gas (Cl₂).
vi. Electroplating involves:
Answer: b) Coating a metal object with a thin layer of metal through electrolysis
Explanation: Electroplating is an electrolytic process where a metal object (cathode) is coated with a thin layer of another metal (anode) using electrical current.
vii. In a voltaic cell, the salt bridge serves the function of:
Answer: b) Maintaining electrical neutrality by allowing ion movement
Explanation: The salt bridge contains an inert electrolyte that allows the flow of ions between the two half-cells. This prevents the build-up of charge (maintains electrical neutrality) which would otherwise stop the reaction.
viii. If the voltage data shows metal A has a higher standard reduction potential than metal B, it can be concluded that:
Answer: b) Metal B is more reactive than metal A
Explanation: A higher standard reduction potential means a greater tendency to be reduced (gain electrons). Therefore, Metal A is less reactive (more noble) than Metal B, which has a lower reduction potential and a greater tendency to be oxidized (more reactive).
i. What is a fuel cell?
Answer: A fuel cell is an electrochemical cell that converts the chemical energy of a fuel (often hydrogen) and an oxidizing agent (often oxygen) into electricity through a pair of redox reactions. Unlike batteries, fuel cells require a continuous source of fuel and oxygen to sustain the chemical reaction. They are highly efficient and produce only water and heat as byproducts when hydrogen is used.
ii. Write chemical reactions that occur in Nelson's cell.
Answer: The Nelson cell is used for the electrolysis of brine (concentrated NaCl solution).
At Anode (Oxidation): 2Cl⁻(aq) [source symbol unavailable] Cl₂(g) + 2e⁻ At Cathode (Reduction): 2H₂O(l) + 2e⁻ [source symbol unavailable] H₂(g) + 2OH⁻(aq)
Overall Reaction: 2NaCl(aq) + 2H₂O(l) [source symbol unavailable] 2NaOH(aq) + Cl₂(g) + H₂(g)
iii. Why tin-plated steel is used to make food cans?
Answer: Tin-plated steel is used for food cans because tin is non-toxic, resistant to corrosion, and does not react with food acids. The tin coating protects the underlying steel from rusting and prevents the food from coming into contact with iron, which could cause discoloration and off-flavors.
iv. Explain one example from daily life which involves an oxidation-reduction reaction.
Answer: Rusting of iron is a common example of an oxidation-reduction reaction. Iron (Fe) is oxidized by oxygen (O₂) in the presence of moisture to form hydrated iron(III) oxide (rust). Iron loses electrons (oxidation), and oxygen gains electrons (reduction). This reaction is undesirable as it weakens iron structures.
v. Define electrochemical series.
Answer: The electrochemical series is a list of elements arranged in order of their standard electrode potentials (reduction potentials). It is used to predict the direction of redox reactions. Elements with more negative reduction potentials (e.g., Li, K) are strong reducing agents and are placed at the top, while elements with more positive potentials (e.g., F₂, Au) are strong oxidizing agents and are placed at the bottom.
vi. Why does chlorine form at the anode during the electrolysis of molten lead(II) chloride?
Answer: During the electrolysis of molten lead(II) chloride, the ions present are Pb²⁺ and Cl⁻. At the anode (positive electrode),
oxidation occurs. The chloride ions (Cl⁻) are the only negatively charged species available to lose electrons. Therefore, they are oxidized to form chlorine gas (Cl₂).
vii. How do hydrogen-oxygen fuel cells benefit the environment compared to gasoline engines?
Answer: Hydrogen-oxygen fuel cells benefit the environment because their only byproduct is water (H₂O) and heat, with no emission of carbon dioxide (CO₂), carbon monoxide (CO), nitrogen oxides (NOx), or particulate matter. Gasoline engines, on the other hand, emit greenhouse gases and toxic air pollutants that contribute to climate change and respiratory diseases.
viii. Sketch a Daniel cell with labeled components and indicate the direction of electron flow.
Answer: (Diagram Description)
Left Half-Cell: Zinc electrode immersed in ZnSO₄ solution. Labeled as Anode (-). Right Half-Cell: Copper electrode immersed in CuSO₄ solution. Labeled as Cathode (+). Salt Bridge: Connects the two solutions. External Circuit: A wire connecting the two electrodes with a voltmeter. Electron Flow: An arrow on the wire pointing from the Zinc electrode (Anode) to the Copper electrode (Cathode).
1. Define oxidation and reduction in terms of loss or gain of oxygen or hydrogen.
Answer:
Oxidation and Reduction in terms of Oxygen:
Oxidation: Oxidation is the gain of oxygen by a substance. For example, when magnesium burns in air, it gains oxygen to form magnesium oxide: 2Mg + O₂ [source symbol unavailable] 2MgO. Here, magnesium is oxidized. Reduction: Reduction is the loss of oxygen from a substance. For example, when copper(II) oxide is heated with hydrogen, it loses oxygen to form copper: CuO + H₂ [source symbol unavailable] Cu + H₂O. Here, copper(II) oxide is reduced.
Oxidation and Reduction in terms of Hydrogen:
Oxidation: Oxidation is the loss of hydrogen from a substance. For example, when ethanol is oxidized to ethanal, it loses hydrogen: CH₃CH₂OH [source symbol unavailable] CH₃CHO + 2H. Here, ethanol is oxidized. Reduction: Reduction is the gain of hydrogen by a substance. For example, when ethene reacts with hydrogen to form
ethane, it gains hydrogen: CH₂=CH₂ + H₂ [source symbol unavailable] CH₃CH₃. Here, ethene is reduced.
2. Sketch a Daniel cell, labelling the cathode, the anode, and the direction of the flow of electrons.
Answer: (Diagram Description)
Anode: The Zinc (Zn) electrode. It is the negative electrode where oxidation occurs: Zn(s) [source symbol unavailable] Zn²⁺(aq) + 2e⁻. Cathode: The Copper (Cu) electrode. It is the positive electrode where reduction occurs: Cu²⁺(aq) + 2e⁻ [source symbol unavailable] Cu(s). Electron Flow: Electrons flow from the Zinc anode to the Copper cathode through the external wire. Salt Bridge: A U-shaped tube containing an inert electrolyte (e.g., KCl) that connects the two half-cells and maintains electrical neutrality. Ion Movement: Zn²⁺ ions are produced at the anode, and Cu²⁺ ions are consumed at the cathode. The salt bridge allows ions to migrate to balance the charge.
3. Describe the method of recovering metal from lead(II) chloride.
Answer: The recovery of lead metal from lead(II) chloride (PbCl₂) is carried out by electrolysis of its molten state.
Preparation: Lead(II) chloride is heated until it melts. In the molten state, it dissociates into Pb²⁺ and Cl⁻ ions. Electrolysis: The molten PbCl₂ is placed in an electrolytic cell. A direct current is passed through it using inert electrodes (e.g., graphite). At Cathode (Reduction): Pb²⁺ ions migrate to the cathode and gain electrons to form molten lead metal: Pb²⁺(l) + 2e⁻ [source symbol unavailable] Pb(l). The molten lead is collected at the bottom. At Anode (Oxidation): Cl⁻ ions migrate to the anode and lose electrons to form chlorine gas: 2Cl⁻(l) [source symbol unavailable] Cl₂(g) + 2e⁻. Overall Reaction: PbCl₂(l) [source symbol unavailable] Pb(l) + Cl₂(g).
4. Construct ionic half-equations for reactions at electrodes in the electrolysis of copper(II) sulphate solution using copper electrodes.
Answer: When copper electrodes are used in the electrolysis of copper(II) sulphate solution:
At Cathode (Reduction): Cu²⁺ ions from the solution are reduced and deposited as copper metal on the cathode.
Half-equation: Cu²⁺(aq) + 2e⁻ [source symbol unavailable] Cu(s)
At Anode (Oxidation): The copper anode dissolves. Copper atoms lose electrons to form Cu²⁺ ions, which go into the solution.
Half-equation: Cu(s) [source symbol unavailable] Cu²⁺(aq) + 2e⁻
Overall Reaction: Cu(s) (anode) [source symbol unavailable] Cu(s) (cathode). This process is used in the refining of copper.
5. For each of the following cells:
(a) Sketch an electrolytic cell, label the anode and cathode and indicate the direction of electron transfer.
Answer: (Diagram Description)
Anode: The positive electrode, connected to the positive terminal of the battery. Oxidation occurs here. Cathode: The negative electrode, connected to the negative terminal of the battery. Reduction occurs here. Electron Transfer: Electrons are drawn away from the anode by the positive terminal of the battery and are pushed towards the cathode by the negative terminal. Therefore, the flow of electrons in the external circuit is from the anode to the cathode.
(b) Describe the nature of the electrochemical process.
Answer: An electrolytic process is a non-spontaneous chemical reaction driven by an external source of electrical energy. Electrical energy is converted into chemical energy. It is used to decompose compounds into their elements or to coat objects with a metal.
(c) Distinguish between electrolytic and voltaic cells.
Answer:
| Feature | Electrolytic Cell | Voltaic (Galvanic) Cell |
| Energy Conversion | Electrical energy [source symbol unavailable] Chemical energy | Chemical energy [source symbol unavailable] Electrical energy |
| Spontaneity | Non-spontaneous reaction | Spontaneous reaction |
| Electrodes | Anode is positive (+), Cathode is negative (-) | Anode is negative (-), Cathode is positive (+) |
| External Power | Requires an external power source | No external power source required |
| Example | Electrolysis of water | Daniel cell |
6. For each of the following electrolytes, write electrode half-reactions in each case that would occur during their electrolysis.
(a) Copper(II) chloride using copper electrodes
Answer:
Cathode (Reduction): Cu²⁺(aq) + 2e⁻ [source symbol unavailable] Cu(s) Anode (Oxidation): Cu(s) [source symbol unavailable] Cu²⁺(aq) + 2e⁻ (Copper anode dissolves)
(b) Copper(II) chloride using carbon electrodes
Answer:
Cathode (Reduction): Cu²⁺(aq) + 2e⁻ [source symbol unavailable] Cu(s) Anode (Oxidation): 2Cl⁻(aq) [source symbol unavailable] Cl₂(g) + 2e⁻
7. Predict the products of electrolysis for a dilute solution of potassium iodide.
Answer: In a dilute solution of potassium iodide (KI), the ions present are K⁺, I⁻, H⁺, and OH⁻.
At Cathode: H⁺ ions are reduced in preference to K⁺ ions because hydrogen is less reactive than potassium. Hydrogen gas is produced. 2H₂O(l) + 2e⁻ [source symbol unavailable] H₂(g) + 2OH⁻(aq) At Anode: I⁻ ions are oxidized in preference to OH⁻ ions because iodide ions are more easily oxidized. Iodine is produced. 2I⁻(aq) [source symbol unavailable] I₂(aq) + 2e⁻ Overall Products: Hydrogen gas at the cathode and Iodine at the anode.
8. Why does chlorine form at the anode during the electrolysis of molten lead(II) chloride?
Answer: During the electrolysis of molten lead(II) chloride (PbCl₂), the compound dissociates into Pb²⁺ and Cl⁻ ions. When an electric current is passed, the positive Pb²⁺ ions migrate to the cathode (reduction) and the negative Cl⁻ ions migrate to the anode. At the anode, oxidation occurs. Since Cl⁻ is the only negative ion present, it loses electrons to form chlorine gas (Cl₂). The half-equation is: 2Cl⁻(l) [source symbol unavailable] Cl₂(g) + 2e⁻.
9. What changes in observations might you expect when comparing electrolysis of (i) dilute aqueous solution and concentrated aqueous solution?
Answer:
Dilute Solution: In a dilute solution, water is the major source of ions (H⁺ and OH⁻). At the anode, OH⁻ ions are discharged in preference to the halide ions, producing oxygen gas. At the cathode, H⁺ ions are discharged, producing hydrogen gas. Concentrated Solution: In a concentrated solution, the concentration of halide ions is high. At the anode, the halide ions (e.g., Cl⁻, Br⁻, I⁻) are discharged in preference to OH⁻ ions, producing the corresponding halogen gas. At the cathode, H⁺ ions are still discharged, producing hydrogen gas.
10. How would you test and identify the products of different substances at the anode and cathode during the electrolysis of dilute copper(II) sulphate solution using copper electrodes?
Answer: During the electrolysis of dilute copper(II) sulphate solution using copper electrodes, the products are:
At Cathode: Copper metal is deposited. Test: Observe the cathode; a reddish-brown coating of copper metal will form on the electrode. At Anode: The copper anode dissolves. Test: Observe the anode; it will decrease in mass. The blue color of the solution remains constant because Cu²⁺ ions are removed at the cathode and added at the anode at the same rate. There is no gas evolution at the anode if copper electrodes are used.
11. Compare and contrast the products formed at the electrodes during the electrolysis of concentrated aqueous sodium chloride and dilute sulphuric acid using inert electrodes.
Answer:
| Feature | Concentrated NaCl (Brine) | Dilute H₂SO₄ |
| At Cathode | Hydrogen gas (H₂) | Hydrogen gas (H₂) |
| At Anode | Chlorine gas (Cl₂) | Oxygen gas (O₂) |
| Reason | Cl⁻ ions are discharged in preference to OH⁻ due to high concentration. | OH⁻ ions are discharged because SO₄²⁻ is very stable. |
| Overall | 2NaCl + 2H₂O [source symbol unavailable] 2NaOH + Cl₂ + H₂ | 2H₂O [source symbol unavailable] 2H₂ + O₂ |
12. Assess the environmental impact of using hydrogen-oxygen fuel cells compared to gasoline engines in vehicles. Provide arguments for and against each technology.
Answer:
Hydrogen-Oxygen Fuel Cells (For): Zero emission of greenhouse gases (only water vapor). High efficiency. Quiet
operation. (Against): Production of hydrogen often requires fossil fuels (unless using green hydrogen). Storage and transportation of hydrogen are difficult and expensive. Gasoline Engines (For): Established infrastructure. High energy density. (Against): Emit CO₂, CO, NOx, and particulate matter. Contribute to climate change and air pollution. Lower efficiency.
13. Compare the ionic half-equations for the electrolysis of a dilute solution and a concentrated solution of sodium chloride. How do the products differ?
Answer:
Dilute NaCl:
Cathode: 2H₂O + 2e⁻ [source symbol unavailable] H₂ + 2OH⁻ Anode: 2H₂O [source symbol unavailable] O₂ + 4H⁺ + 4e⁻ Products: Hydrogen gas and Oxygen gas.
Concentrated NaCl:
Cathode: 2H₂O + 2e⁻ [source symbol unavailable] H₂ + 2OH⁻ Anode: 2Cl⁻ [source symbol unavailable] Cl₂ + 2e⁻ Products: Hydrogen gas and Chlorine gas.
1. Sketch an electrolytic cell for the electrolysis of fused KCl.
Answer: (Diagram Description)
Container: A crucible or cell containing molten KCl. Electrodes: Two inert electrodes (e.g., graphite) dipped into the molten KCl. Power Source: A battery connected to the electrodes. At Cathode: K⁺ ions are reduced to K metal. 2K⁺ + 2e⁻ [source symbol unavailable] 2K. At Anode: Cl⁻ ions are oxidized to Cl₂ gas. 2Cl⁻ [source symbol unavailable] Cl₂ + 2e⁻.
2. Sketch a simple representation of electrolytic cell for the electrolysis of fused sodium chloride. Indicate the direction of the flow of electrons. Identify anode and cathode.
Answer: (Diagram Description)
Cell: Container with molten NaCl. Electrodes:
Anode (+): Connected to the positive terminal of the battery. Electrons flow from the anode to the battery. Cathode (-): Connected to the negative terminal. Electrons flow from the battery to the cathode.
Electron Flow: Anode [source symbol unavailable] External Circuit [source symbol unavailable] Battery [source symbol unavailable] Cathode. Reactions:
Anode: 2Cl⁻ [source symbol unavailable] Cl₂ + 2e⁻ Cathode: 2Na⁺ + 2e⁻ [source symbol unavailable] 2Na
Sketch a voltaic cell labeling the cathode, the anode and the direction of flow of the electrons. Use the following chemicals: Silver, Zinc, Silver Nitrate and zinc sulphate (ZnSO₄). (Hint: Zn is more active than Ag)
Answer: (Diagram Description)
Anode: Zinc (Zn) electrode immersed in ZnSO₄ solution. It is the negative electrode where oxidation occurs: Zn [source symbol unavailable] Zn²⁺ + 2e⁻. Cathode: Silver (Ag) electrode immersed in AgNO₃ solution. It is the positive electrode where reduction occurs: Ag⁺ + e⁻ [source symbol unavailable] Ag. Electron Flow: From the Zinc anode to the Silver cathode through the external wire. Salt Bridge: Connects the two solutions to maintain charge balance.
1. Which of the following metals is more likely to undergo reduction?
Al³⁺(aq) + 3e⁻ [source symbol unavailable] Al(s) E° = -1.66 V
Fe³⁺(aq) + 3e⁻ [source symbol unavailable] Fe(s) E° = -0.44 V
Answer: Fe³⁺ is more likely to undergo reduction.
Explanation: A more positive (or less negative) standard reduction potential indicates a greater tendency to be reduced (gain electrons). Since -0.44 V is more positive than -1.66 V, Fe³⁺ has a higher reduction potential and is more easily reduced.
2. Arrange the following metals in order of increasing ability to get reduce.
Na⁺(aq) + e⁻ [source symbol unavailable] Na(s) E° = -2.71 V
Zn²⁺(aq) + 2e⁻ [source symbol unavailable] Zn(s) E° = -0.76 V
Cu²⁺(aq) + 2e⁻ [source symbol unavailable] Cu(s) E° = +0.34 V
Answer: Increasing ability to get reduced means moving from the most negative reduction potential to the most positive. Order: Na⁺ < Zn²⁺ < Cu²⁺
Explanation: Na⁺ has the most negative potential (-2.71 V) so it is the hardest to reduce. Cu²⁺ has the most positive potential (+0.34
V) so it is the easiest to reduce.
3. Identify the half-cell in which oxidation occurs and the half-cell in which reduction occurs in the given voltaic cell (Figure 4.8). (Diagram of a voltaic cell with H₂H⁺ half-cell and Zn/Zn²⁺ half-cell)
Answer:
Oxidation: Occurs at the Zinc (Zn) half-cell. Zinc is the more active metal and acts as the anode. Zn(s) [source symbol unavailable] Zn²⁺(aq) + 2e⁻. Reduction: Occurs at the Hydrogen (H₂) half-cell. H⁺ ions are reduced to hydrogen gas. 2H⁺(aq) + 2e⁻ [source symbol unavailable] H₂(g).
