Chapter 2: Electrochemistry

i. Which of the following elements is reduced in the reaction? 2Na + Cl₂ [source symbol unavailable] 2NaCl

Answer: b) Cl

Explanation: Chlorine (Cl₂) gains electrons (reduction) to form chloride ions (Cl⁻), going from an oxidation state of 0 to -1.

ii. In the reaction Fe + CuSO₄ [source symbol unavailable] FeSO₄ + Cu, what is the reducing agent?

Answer: a) Fe

Explanation: Iron (Fe) loses electrons (oxidation) and causes the reduction of Cu²⁺ to Cu. The species that is oxidized is the reducing agent.

iii. What is the oxidation number of sulfur in H₂SO₄?

Answer: d) +6

Explanation: In H₂SO₄, hydrogen is +1 and oxygen is -2. Let x be the oxidation number of sulfur: 2(+1) + x + 4(-2) = 0 [source symbol unavailable] 2 + x - 8 = 0 [source symbol unavailable] x

= +6.

iv. Which of the following best describes the process of oxidation?

Answer: b) Loss of electrons

Explanation: Oxidation is defined as the loss of electrons or an increase in oxidation state. A mnemonic is "OIL RIG" (Oxidation Is Loss, Reduction Is Gain).

v. In the electrolysis of water, which substance is liberated at the cathode?

Answer: a) H₂

Explanation: At the cathode (negative electrode), reduction occurs. Water is reduced to hydrogen gas (H₂) and hydroxide ions.

vi. Which of the following metals would not react with HCl to produce H₂ gas?

Answer: c) Cu

Explanation: Copper (Cu) is below hydrogen in the electrochemical series, so it cannot displace hydrogen from an acid. Zn, Fe, and Mg are above hydrogen and can react.

vii. The Winkler method measures the concentration of which substance to determine BOD?

Answer: a) O₂

Explanation: The Winkler method is a titration technique used to determine dissolved oxygen concentration in water samples, which is used to calculate Biochemical Oxygen Demand (BOD).

viii. In a redox reaction, the oxidizing agent:

Answer: b) Gains electrons

Explanation: The oxidizing agent is the species that accepts electrons (is reduced) and causes the oxidation of another species.

ix. The standard electrode potential of the hydrogen electrode is defined as:

Answer: a) 0 V

Explanation: By convention, the standard hydrogen electrode (SHE) is assigned a potential of exactly 0.00 V at 298 K, 1 atm, and 1 M H⁺ concentration.

x. What is the relationship between the Faraday constant (F), Avogadro's number (NA), and the charge of an electron (e)?

Answer: a) F = NA × e

Explanation: The Faraday constant (F) is the charge of one mole of electrons. It is equal to Avogadro's number (NA) multiplied by the elementary charge (e).

i. Explain the following with reasons.

Answer: Oxidation potential is the tendency of a species to lose electrons (be oxidized). Reduction potential is the tendency to gain electrons (be reduced). They are equal in magnitude but opposite in sign. For zinc, the standard reduction potential (Zn²⁺/Zn) is -0.76 V, meaning it has a low tendency to be reduced and a high tendency to be oxidized. Therefore, its oxidation potential is +0.76 V.

Answer: A salt bridge contains an inert electrolyte (e.g., KCl). It connects the two half-cells and allows the flow of ions. It prevents the accumulation of charge in either half-cell. Without it, the build-up of positive and negative charges would stop the reaction. It maintains electrical neutrality.

Answer: Na and K are highly reactive metals with very negative standard reduction potentials (strong reducing agents). They readily lose electrons and displace H⁺ from acids. Cu and Pt have positive standard reduction potentials (weak reducing agents). They cannot displace H⁺ from acids because they are less reactive than hydrogen.

ii. Define oxidation in terms of electron transfer.

Answer: Oxidation is defined as the loss of electrons by a chemical species. When a substance is oxidized, its oxidation state increases. The electrons lost are gained by another species, which is reduced. For example, in the reaction Zn [source symbol unavailable] Zn²⁺ + 2e⁻, zinc is oxidized. This electron transfer is the basis of redox reactions. Oxidation and reduction always occur simultaneously.

iii. What is the oxidation number of oxygen in H₂O₂?

Answer: In hydrogen peroxide (H₂O₂), the oxidation number of oxygen is -1. This is an exception to the usual -2 oxidation state of

oxygen. It can be calculated: 2(+1) + 2(x) = 0 [source symbol unavailable] 2 + 2x = 0 [source symbol unavailable] 2x = -2 [source symbol unavailable] x = -1. The peroxide linkage (O-O) is responsible for this unusual oxidation state.

iv. Identify the reducing agent in the reaction Zn + Cu²⁺ [source symbol unavailable] Zn²⁺ + Cu

Answer: The reducing agent is Zinc (Zn). Zinc loses electrons (is oxidized) from 0 to +2, causing the reduction of Cu²⁺ to Cu. The species that is oxidized is the reducing agent. It donates electrons to the copper ions. Therefore, Zn is the reducing agent.

v. State the purpose of the Winkler method.

Answer: The Winkler method is used to determine the concentration of dissolved oxygen (DO) in water samples. This is crucial for assessing water quality and calculating the Biochemical Oxygen Demand (BOD). BOD measures the amount of oxygen consumed by microorganisms in decomposing organic matter. High BOD indicates water pollution. The method involves a series of titrations.

vi. Explain what happens at the anode during the electrolysis of aqueous sodium chloride.

Answer: During the electrolysis of aqueous sodium chloride (brine), oxidation occurs at the anode (positive electrode). Chloride ions (Cl⁻) are oxidized to chlorine gas (Cl₂): 2Cl⁻ [source symbol unavailable] Cl₂ + 2e⁻. Water can also be oxidized to oxygen, but chloride is preferentially oxidized due to its lower oxidation potential. The chlorine gas bubbles out at the anode. This is a key industrial process.

vii. What does a positive standard electrode potential indicate about a substance's tendency to gain electrons?

Answer: A positive standard electrode potential (E°) indicates that the substance has a high tendency to gain electrons (be reduced). The more positive the E°, the stronger the oxidizing agent. For example, fluorine (F₂) has a very positive E° (+2.87 V), meaning it is readily reduced. Conversely, a negative E° indicates a low tendency to gain electrons.

viii. Calculate the oxidation number of chromium in K₂Cr₂O₇.

Answer: Let x be the oxidation number of chromium. Potassium (K) is +1, and oxygen (O) is -2. The compound is neutral, so:

2(+1) + 2(x) + 7(-2) = 0

2 + 2x - 14 = 0

2x - 12 = 0

2x = +12

x = +6 The oxidation number of chromium in K₂Cr₂O₇ is +6.

ix. Describe the role of the standard hydrogen electrode in electrochemistry.

Answer: The Standard Hydrogen Electrode (SHE) is the reference electrode against which all other electrode potentials are measured. By convention, its standard electrode potential is assigned a value of 0.00 V. It consists of a platinum electrode in a 1 M H⁺ solution with H₂ gas at 1 atm and 298 K. It allows for the determination of standard reduction potentials of other half-cells.

x. What is the significance of the activity series of metals?

Answer: The activity series (or electrochemical series) is a list of metals arranged in order of decreasing reactivity. It is significant because it predicts whether a metal can displace another metal from its salt solution. Metals higher in the series are more reactive and can displace metals lower in the series. It also predicts whether a metal can displace hydrogen from acids.

xi. How can you deduce the feasibility of a redox reaction using electrode potentials?

Answer: A redox reaction is feasible (spontaneous) if the standard cell potential (E°cell) is positive. E°cell is calculated as E°cathode - E°anode. The cathode is the half-cell with the higher reduction potential (where reduction occurs). The anode is the half-cell with the lower reduction potential (where oxidation occurs). If E°cell > 0, the reaction is spontaneous. If E°cell < 0, the reaction is non- spontaneous.

xii. Bauxite ore is used for the commercial preparation of Al. For this purpose, bauxite ore is first purified to produce pure alumina, Al₂O₃. Alumina is then electrolyzed. The following reaction occurs: 2Al₂O₃(l) [source symbol unavailable] 4Al(l) + 3O₂(g). Calculate the mass of Al that collects at the cathode and the volume of oxygen that collects at the anode when Al₂O₃ is electrolyzed for 10 hours with a 15 ampere current at 1 atm and 25°C. (Ans: mass of Al: 50.36g, volume of oxygen: 34.20dm³)

Answer: Step 1: Calculate charge (Q).

Q = I × t = 15 A × (10 × 3600 s) = 15 × 36000 = 540,000 C.

Step 2: Calculate moles of electrons (n).

n = QF = 540,00096,500 = 5.595 moles of electrons.

Step 3: Calculate mass of Al deposited. The half-reaction for Al is: Al³⁺ + 3e⁻ [source symbol unavailable] Al. 3 moles of electrons produce 1 mole of Al.

Moles of Al = 5.5953 = 1.865 moles.

Mass of Al = 1.865 mol × 27 g/mol = 50.36 g.

Step 4: Calculate volume of O₂ at anode. The half-reaction is: 2O²⁻ [source symbol unavailable] O₂ + 4e⁻.

4 moles of electrons produce 1 mole of O₂.

Moles of O₂ = 5.5954 = 1.399 moles.

Volume of O₂ at 1 atm and 25°C (298 K): V = nRT/P = 1.399 × 0.0821 × 2981 = 34.20 dm³.

xiii. Which of the following compounds will give more mass of metal, when 15 ampere current is passed through a molten mass of these salts for 1 hr. (a) NaCl (b) CaCl₂. (Ans: NaCl will give more Na)

Answer:

Q = I × t = 15 A × 3600 s = 54,000 C.

Moles of electrons = 54,00096,500 = 0.5596 moles.

For NaCl: Na⁺ + e⁻ [source symbol unavailable] Na. Moles of Na = 0.5596. Mass of Na = 0.5596 × 23 = 12.87 g.

For CaCl₂: Ca²⁺ + 2e⁻ [source symbol unavailable] Ca. Moles of Ca = 0.55962 = 0.2798. Mass of Ca = 0.2798 × 40 = 11.19 g.

Therefore, NaCl will give more mass of metal (Na).

xiv. How many hours would electroplating have to be continued at the rate of 5 amperes if 75g of copper is to be deposited from CuSO₄ solution?

Answer: Half-reaction: Cu²⁺ + 2e⁻ [source symbol unavailable] Cu.

Moles of Cu = 75 g / 63.5 g/mol = 1.181 moles.

Moles of electrons required = 2 × 1.181 = 2.362 moles.

Charge required (Q) = 2.362 × 96,500 = 227,933 C.

Time (t) = QI = 227,9335 = 45,586.6 s.

Time in hours = 45,586.63600 = 12.66 hours.

xv. Differentiate between the following: (a) a Galvanic and electrolytic cell (b) Oxidation half-reaction and reduction half-reaction

Answer:

Galvanic Cell: Converts chemical energy into electrical energy. The redox reaction is spontaneous. Anode is negative, cathode is positive. Example: Daniell cell. Electrolytic Cell: Converts electrical energy into chemical energy. The redox reaction is non-spontaneous. Anode is positive, cathode is negative. Example: Electrolysis of water.

Oxidation Half-Reaction: Involves the loss of electrons. The oxidation state increases. Occurs at the anode. Reduction Half-Reaction: Involves the gain of electrons. The oxidation state decreases. Occurs at the cathode.

xvi. An electroplating apparatus is used to coat jewellery with gold. What mass of gold can be deposited from a solution that contains [Au(CN)₄]⁻ ion if a current of 5.0 amperes flows for 30 minutes? The following half-reaction occurs: [Au(CN)₄]⁻ + 3e⁻ [source symbol unavailable] Au(s) + 4CN⁻ (Ans: mass of gold [source symbol unavailable] 6.12g)

Answer:

Q = I × t = 5.0 A × (30 × 60 s) = 9000 C.

Moles of electrons = 900096,500 = 0.09326 moles. From the half-reaction, 3 moles of electrons deposit 1 mole of Au.

Moles of Au = 0.093263 = 0.03109 moles.

Mass of Au = 0.03109 mol × 197 g/mol = 6.12 g.

xvii. Construct redox equations using the following half equations.

Answer: Oxidation: SO₂ + 2H₂O [source symbol unavailable] HSO₄⁻ + 3H⁺ + 2e⁻ Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ [source symbol unavailable] Mn²⁺ + 4H₂O Balance electrons (×5 for oxidation, ×2 for reduction): 5SO₂ + 10H₂O [source symbol unavailable] 5HSO₄⁻ + 15H⁺ + 10e⁻ 2MnO₄⁻ + 16H⁺ + 10e⁻ [source symbol unavailable] 2Mn²⁺ + 8H₂O Overall: 2MnO₄⁻ + 5SO₂ + 2H₂O [source symbol unavailable] 2Mn²⁺ + 5HSO₄⁻ + H⁺

Answer: Oxidation: 2Cl⁻ [source symbol unavailable] Cl₂ + 2e⁻ Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ [source symbol unavailable] 2Cr³⁺ + 7H₂O Balance electrons (×3 for oxidation): 6Cl⁻ [source symbol unavailable] 3Cl₂ + 6e⁻ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ [source symbol unavailable] 2Cr³⁺ + 7H₂O Overall: Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ [source symbol unavailable] 2Cr³⁺ + 3Cl₂ + 7H₂O

Answer: Oxidation: H₂O₂ [source symbol unavailable] O₂ + 2H⁺ + 2e⁻ Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ [source symbol unavailable] Mn²⁺ + 4H₂O Balance electrons (×5 for oxidation, ×2 for reduction): 5H₂O₂ [source symbol unavailable] 5O₂ + 10H⁺ + 10e⁻ 2MnO₄⁻ + 16H⁺ + 10e⁻ [source symbol unavailable] 2Mn²⁺ + 8H₂O Overall: 2MnO₄⁻ + 5H₂O₂ + 6H⁺ [source symbol unavailable] 2Mn²⁺ + 5O₂ + 8H₂O

Answer:

Oxidation: Sn²⁺ [source symbol unavailable] Sn⁴⁺ + 2e⁻ Reduction: Fe³⁺ + e⁻ [source symbol unavailable] Fe²⁺ Balance electrons (×2 for reduction): Sn²⁺ [source symbol unavailable] Sn⁴⁺ + 2e⁻ 2Fe³⁺ + 2e⁻ [source symbol unavailable] 2Fe²⁺ Overall: Sn²⁺ + 2Fe³⁺ [source symbol unavailable] Sn⁴⁺ + 2Fe²⁺

(e) IO₃⁻ [source symbol unavailable] I⁻ and AsO₂⁻ [source symbol unavailable] AsO₄³⁻

Answer: Oxidation: AsO₂⁻ + 2H₂O [source symbol unavailable] AsO₄³⁻ + 4H⁺ + 2e⁻ Reduction: IO₃⁻ + 6H⁺ + 6e⁻ [source symbol unavailable] I⁻ + 3H₂O Balance electrons (×3 for oxidation): 3AsO₂⁻ + 6H₂O [source symbol unavailable] 3AsO₄³⁻ + 12H⁺ + 6e⁻ IO₃⁻ + 6H⁺ + 6e⁻ [source symbol unavailable] I⁻ + 3H₂O Overall: IO₃⁻ + 3AsO₂⁻ + 3H₂O [source symbol unavailable] I⁻ + 3AsO₄³⁻ + 6H⁺

xviii. In the reaction between potassium permanganate and iron(II) sulphate in an acidic solution: MnO₄⁻ + Fe²⁺ [source symbol unavailable] Mn²⁺ + Fe³⁺

Answer: Mn in MnO₄⁻: +7. Mn in Mn²⁺: +2. Fe in Fe²⁺: +2. Fe in Fe³⁺: +3.

Answer: Fe²⁺ is oxidized (loses an electron, +2 to +3). MnO₄⁻ is reduced (gains electrons, +7 to +2).

Answer: Mn: +7 [source symbol unavailable] +2 (gain of 5 electrons) Fe: +2 [source symbol unavailable] +3 (loss of 1 electron) To balance, multiply Fe by 5: MnO₄⁻ + 5Fe²⁺ + 8H⁺ [source symbol unavailable] Mn²⁺ + 5Fe³⁺ + 4H₂O

Answer: Oxidizing agent: MnO₄⁻. Reducing agent: Fe²⁺.

xix. Explain why magnesium can displace zinc from zinc sulphate solution but copper cannot. Predict the outcome of placing a zinc strip in a copper(II) sulphate solution, including the balanced chemical equation.

Answer:

Magnesium vs. Zinc: Magnesium (Mg) is higher in the activity series than zinc (Zn). Its standard reduction potential is more negative (-2.37 V vs. -0.76 V). Therefore, Mg is a stronger reducing agent and can displace Zn from ZnSO₄ solution. Copper (Cu) is lower in the activity series than zinc (+0.34 V vs. -0.76 V). Cu is a weaker reducing agent and cannot displace Zn from ZnSO₄. Zinc in Copper(II) Sulphate: Zinc is higher in the activity series than copper. Zn will displace Cu from CuSO₄ solution. The balanced equation is: Zn(s) + CuSO₄(aq) [source symbol unavailable] ZnSO₄(aq) + Cu(s). This is a spontaneous redox reaction.

i. Explain the process of balancing a redox equation using the oxidation number method. Include an example.

Answer: The oxidation number method is a systematic way to balance redox equations by tracking the changes in oxidation states of the elements involved.

Steps:

1. Write the Skeleton Equation: Write the unbalanced equation.

2. Assign Oxidation Numbers: Assign oxidation numbers to all elements in the reactants and products.

3. Identify Changes: Identify which elements undergo oxidation (increase in oxidation number) and which undergo reduction (decrease in oxidation number).

4. Calculate Electron Change: Determine the total increase and decrease in oxidation numbers per atom. Multiply by the number of atoms to get the total electron change.

5. Balance Electron Loss and Gain: Adjust coefficients so that the total increase in oxidation number equals the total decrease.

6. Balance Other Atoms: Balance the remaining atoms (other than H and O) by inspection.

7. Balance Oxygen and Hydrogen: In acidic solution, add H₂O to balance O and H⁺ to balance H. In basic solution, add OH⁻ and H₂O.

8. Check: Verify that all atoms and charges are balanced.

Example: Balance MnO₄⁻ + Fe²⁺ [source symbol unavailable] Mn²⁺ + Fe³⁺ in acidic solution.

1. Skeleton: MnO₄⁻ + Fe²⁺ [source symbol unavailable] Mn²⁺ + Fe³⁺

2. Oxidation numbers: Mn in MnO₄⁻ is +7, Mn²⁺ is +2. Fe²⁺ is +2, Fe³⁺ is +3.

3. Changes: Mn decreases by 5 (reduction). Fe increases by 1 (oxidation).

4. Electron change: Mn gains 5e⁻, Fe loses 1e⁻.

5. Balance: Multiply Fe by 5 to balance electrons: MnO₄⁻ + 5Fe²⁺ [source symbol unavailable] Mn²⁺ + 5Fe³⁺.

6. Balance O and H: Add 8H⁺ to left and 4H₂O to right: MnO₄⁻ + 5Fe²⁺ + 8H⁺ [source symbol unavailable] Mn²⁺ + 5Fe³⁺ + 4H₂O.

7. Check: Mn(1), Fe(5), O(4), H(8), charge (+11 on both sides). Balanced.

In summary, the oxidation number method is a reliable technique for balancing complex redox equations.

ii. Describe the Winkler method for measuring biochemical oxygen demand (BOD) and its importance in assessing water quality.

Answer: The Winkler method is a standard titration technique used to determine the concentration of dissolved oxygen (DO) in water samples. This is essential for calculating the Biochemical Oxygen Demand (BOD), a key indicator of water quality.

Principle: The method is based on the oxidation of iodide ions (I⁻) to iodine (I₂) by dissolved oxygen in an alkaline solution. The amount of iodine produced is equivalent to the amount of dissolved oxygen present.

Procedure:

1. Sample Collection: Collect a water sample in a BOD bottle, avoiding aeration.

2. Fixation: Add manganese(II) sulfate (MnSO₄) and alkaline potassium iodide (KI-NaOH) solutions. A white precipitate of manganese(II) hydroxide, Mn(OH)₂, forms.

3. Oxidation: Dissolved oxygen oxidizes Mn²⁺ to Mn⁴⁺, forming a brown precipitate of manganese(IV) hydroxide, Mn(OH)₄.

4. Acidification: Add concentrated sulfuric acid (H₂SO₄) to dissolve the precipitate. This releases iodine (I₂) equivalent to the DO.

5. Titration: Titrate the liberated iodine with a standard sodium thiosulfate (Na₂S₂O₃) solution using starch as an indicator.

6. Calculation: Calculate the DO concentration from the volume of thiosulfate used.

BOD Calculation: BOD is the difference in DO between the initial sample and a sample incubated for 5 days at 20°C.

Importance:

Water Quality: High BOD indicates high organic pollution (e.g., sewage, industrial waste). Microorganisms consume oxygen to decompose organic matter. Aquatic Life: Low DO levels (high BOD) can lead to hypoxia, killing fish and other aquatic life. Wastewater Treatment: BOD is used to measure the efficiency of wastewater treatment plants.

In summary, the Winkler method is a crucial analytical technique for monitoring water quality and assessing the impact of organic pollution on aquatic ecosystems.

iii. Explain how the activity series of metals can be used to predict the outcome of displacement reactions.

Answer: The activity series (or electrochemical series) is a list of metals arranged in order of decreasing reactivity. It is a powerful tool for predicting whether a displacement reaction will occur.

Principle: A more reactive metal (higher in the series) can displace a less reactive metal (lower in the series) from its salt solution. The more reactive metal loses electrons (is oxidized) and the less reactive metal ion gains electrons (is reduced).

Predicting Outcomes:

1. Metal + Salt Solution: If the solid metal is above the metal ion in the solution, a reaction occurs. If it is below, no reaction occurs.

Example: Zinc (Zn) is above Copper (Cu). Zn + CuSO₄ [source symbol unavailable] ZnSO₄ + Cu (spontaneous). Example: Copper (Cu) is below Zinc (Zn). Cu + ZnSO₄ [source symbol unavailable] No reaction.

2. Metal + Acid: Metals above hydrogen in the series can displace hydrogen from acids (e.g., HCl, H₂SO₄) to produce hydrogen gas.

Example: Mg + 2HCl [source symbol unavailable] MgCl₂ + H₂. Example: Cu + HCl [source symbol unavailable] No reaction.

3. Metal + Water: Highly reactive metals (Group 1, Ca, etc.) react with cold water. Less reactive metals (Mg, Fe) react with steam. Metals below hydrogen do not react.

Explanation using Standard Reduction Potentials:

A metal with a more negative E° (higher in the series) has a greater tendency to be oxidized. A metal ion with a more positive E° (lower in the series) has a greater tendency to be reduced.

The reaction is spontaneous if E°cell = E°cathode - E°anode > 0.

In summary, the activity series is a simple yet effective guide for predicting displacement reactions based on the relative reactivity of metals.

iv. Describe the construction and function of an electrolytic cell.

Answer: An electrolytic cell is an electrochemical cell that uses electrical energy to drive a non-spontaneous redox reaction.

Construction:

1. External Power Source: A battery or DC power supply provides the electrical energy.

2. Electrodes: Two electrodes (conductors) are immersed in the electrolyte.

Anode: The positive electrode, connected to the positive terminal of the power source. Cathode: The negative electrode, connected to the negative terminal.

3. Electrolyte: A molten ionic compound or an aqueous solution of an ionic compound that conducts electricity.

4. Container: A suitable vessel to hold the electrolyte.

Function:

Oxidation at Anode: Anions (negative ions) migrate to the anode and lose electrons (oxidation). Reduction at Cathode: Cations (positive ions) migrate to the cathode and gain electrons (reduction). Electron Flow: Electrons flow from the anode to the cathode through the external circuit. Ion Flow: Ions move through the electrolyte to complete the circuit.

Examples:

Electrolysis of Molten NaCl: Anode: 2Cl⁻ [source symbol unavailable] Cl₂ + 2e⁻. Cathode: Na⁺ + e⁻ [source symbol unavailable] Na. Electrolysis of Water: Anode: 2H₂O [source symbol unavailable] O₂ + 4H⁺ + 4e⁻. Cathode: 2H₂O + 2e⁻ [source symbol unavailable] H₂ + 2OH⁻.

Applications:

Electroplating, extraction of metals (e.g., aluminum), production of chlorine and hydrogen.

In summary, an electrolytic cell uses electrical energy to drive non-spontaneous chemical reactions, with oxidation at the anode and reduction at the cathode.

v. Calculate the mass of copper deposited at the cathode during the electrolysis of CuSO₄ solution when a current of 2 A flows for 30

minutes. (Faraday's constant = 96500 C/mol, Molar mass of Cu = 63.5 g/mol. Ans: 1.18g)

Answer: Step 1: Calculate charge (Q).

Q = I × t = 2 A × (30 × 60 s) = 2 × 1800 = 3600 C.

Step 2: Calculate moles of electrons (n).

n = QF = 360096500 = 0.0373 moles of electrons.

Step 3: Calculate moles of Cu deposited. The half-reaction for Cu is: Cu²⁺ + 2e⁻ [source symbol unavailable] Cu. 2 moles of electrons deposit 1 mole of Cu.

Moles of Cu = 0.03732 = 0.01865 moles.

Step 4: Calculate mass of Cu.

Mass = moles × molar mass = 0.01865 mol × 63.5 g/mol = 1.184 g [source symbol unavailable] 1.18 g.

Therefore, the mass of copper deposited is 1.18 g.

vi. Balance the following equations by the oxidation number method.

Answer: Oxidation: 2Cl⁻ [source symbol unavailable] Cl₂ + 2e⁻ (Cl: -1 to 0) Reduction: Mn⁴⁺ + 2e⁻ [source symbol unavailable] Mn²⁺ (Mn: +4 to +2) Balance electrons: 1 Mn gains 2e⁻, 2 Cl lose 2e⁻. Balanced: MnO₂ + 4HCl [source symbol unavailable] MnCl₂ + 2H₂O + Cl₂

Answer: Oxidation: 2I⁻ [source symbol unavailable] I₂ + 2e⁻ Reduction: N⁵⁺ + 3e⁻ [source symbol unavailable] N²⁺ (N: +5 to +2) Balance electrons: Multiply oxidation by 3, reduction by 2. Balanced: 2HNO₃ + 6HI [source symbol unavailable] 2NO + 3I₂ + 4H₂O

Answer: Oxidation: 4Ag [source symbol unavailable] 2Ag₂S + 4e⁻ (Ag: 0 to +1) Reduction: O₂ + 4e⁻ [source symbol unavailable] 2O²⁻ (O: 0 to -2) Balanced: 4Ag + 2H₂S + O₂ [source symbol unavailable] 2Ag₂S + 2H₂O

Answer: Oxidation: Zn [source symbol unavailable] Zn²⁺ + 2e⁻ Reduction: N⁵⁺ + 3e⁻ [source symbol unavailable] N²⁺ Balance electrons: Multiply oxidation by 3, reduction by 2. Balanced: 3Zn + 8HNO₃ [source symbol unavailable] 3Zn(NO₃)₂ + 2NO + 4H₂O

e. Cu + H₂SO₄ [source symbol unavailable] CuSO₄ + SO₂ + H₂O

Answer:

Oxidation: Cu [source symbol unavailable] Cu²⁺ + 2e⁻ Reduction: S⁶⁺ + 2e⁻ [source symbol unavailable] S⁴⁺ Balance electrons: 1 Cu loses 2e⁻, 1 S gains 2e⁻. Balanced: Cu + 2H₂SO₄ [source symbol unavailable] CuSO₄ + SO₂ + 2H₂O

vii. During the electrolysis of molten sodium chloride:

Answer: At the anode (positive electrode), chloride ions (Cl⁻) are oxidized to chlorine gas: 2Cl⁻ [source symbol unavailable] Cl₂ + 2e⁻. At the cathode (negative electrode), sodium ions (Na⁺) are reduced to sodium metal: Na⁺ + e⁻ [source symbol unavailable] Na.

Answer: Q = I × t = 3 A × (2 × 3600 s) = 3 × 7200 = 21,600 C.

viii. Consider the following half-reactions and their standard electrode potentials:

Ag⁺ + e⁻ [source symbol unavailable] Ag E° = +0.80 V

Zn²⁺ + 2e⁻ [source symbol unavailable] Zn E° = -0.76 V

Answer: E°cell = E°cathode - E°anode = +0.80 V - (-0.76 V) = 1.56 V.

Answer: Electrons flow from the anode (Zn) to the cathode (Ag).

Answer: Since E°cell is positive (+1.56 V), the reaction is spontaneous and feasible. The cell will produce electrical energy.

ix. A current of 1.0 A is passed through a solution of AgNO₃ for 965 seconds.

Answer: Moles of electrons = I × tF = 1.0 × 96596500 = 0.01 F.

Answer: Ag⁺ + e⁻ [source symbol unavailable] Ag. 1 mole of electrons deposits 1 mole of Ag. Therefore, 0.01 moles of electrons deposit 0.01 moles of Ag.