Chapter 2 Electrochemistry – 2nd Year Chemistry | Federal Board New Book
MCQs
i. Which of the following elements is reduced in the reaction?
Reaction: 2Na + Cl₂ → 2NaCl
Options:
a) Na
b) Cl
c) Both Na and Cl
d) Neither Na nor Cl
✅ Correct Answer: b) Cl
Explanation:
Chlorine (Cl₂) gains electrons to form Cl⁻ ions (reduction = gain of electrons).
Tip:
Remember OIL RIG – Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).
ii. In the reaction Fe + CuSO₄ → FeSO₄ + Cu, what is the reducing agent?
Options:
a) Fe
b) Cu
c) SO₄²⁻
d) CuSO₄
✅ Correct Answer: a) Fe
Explanation:
Fe donates electrons to Cu²⁺, reducing it to Cu. Thus, Fe is oxidized and acts as the reducing agent.
Tip:
The reducing agent gets oxidized itself while reducing another substance.
iii. What is the oxidation number of sulphur in H₂SO₄?
Options:
a) -2
b) 0
c) +2
d) +6
✅ Correct Answer: d) +6
Explanation:
Let oxidation number of S be x:
2(+1) + x + 4(-2) = 0 → x = +6
Tip:
Use the algebraic method: Sum of oxidation states = 0 for neutral compounds.
iv. Which of the following best describes the process of oxidation?
Options:
a) Gain of electrons
b) Loss of electrons
c) Gain of protons
d) Loss of protons
✅ Correct Answer: b) Loss of electrons
Explanation:
Oxidation is defined as the loss of electrons.
Tip:
LEO says GER – Loss of Electrons = Oxidation, Gain of Electrons = Reduction
v. In the electrolysis of water, which substance is liberated at the cathode?
Options:
a) H₂
b) O₂
c) H₂O
d) OH⁻
✅ Correct Answer: a) H₂
Explanation:
At the cathode, reduction takes place:
2H⁺ + 2e⁻ → H₂(g)
Tip:
Cathode = Reduction = Gain of electrons = Hydrogen gas (H₂) in water electrolysis.
vi. Which of the following metals would not react with HCl to produce H₂ gas?
Options:
a) Zn
b) Fe
c) Cu
d) Mg
✅ Correct Answer: c) Cu
Explanation:
Copper is less reactive than hydrogen and does not displace it from HCl.
Tip:
Use the reactivity series: Only metals above hydrogen can displace H₂ from acids.
vii. The Winkler method measures the concentration of which substance to determine BOD?
Options:
a) O₂
b) CO₂
c) N₂
d) H₂
✅ Correct Answer: a) O₂
Explanation:
Winkler method measures dissolved oxygen (O₂) to calculate Biological Oxygen Demand (BOD).
Tip:
BOD = the amount of O₂ needed by microorganisms to decompose organic matter.
viii. In a redox reaction, the oxidizing agent:
Options:
a) Loses electrons
b) Gains electrons
c) Loses protons
d) Gains protons
✅ Correct Answer: b) Gains electrons
Explanation:
Oxidizing agent gains electrons and gets reduced itself.
Tip:
Oxidizing agent = Electron acceptor = Gets reduced
ix. The standard electrode potential of the hydrogen electrode is defined as:
Options:
a) 0 V
b) 1 V
c) -1 V
d) 0.5 V
✅ Correct Answer: a) 0 V
Explanation:
By international convention, the standard hydrogen electrode (SHE) is assigned 0 volts.
Tip:
SHE acts as a reference electrode for all standard potentials.
x. What is the relationship between the Faraday constant (F), Avogadro’s number (NA), and charge of an electron (e)?
Options:
a) F = NA × e
b) F = e / NA
c) F = NA / e
d) F = 2 × NA
✅ Correct Answer: a) F = NA × e
Explanation:
Faraday constant is total charge per mole of electrons:
F = (6.022 × 10²³ mol⁻¹) × (1.6 × 10⁻¹⁹ C) ≈ 96500 C/mol
Tip:
Use this formula for mole-electron charge conversions.
Short Questions
i. The oxidation potential of Zn is +0.76 V and its reduction potential is -0.76 V.
Answer:
This means zinc loses electrons easily and is a good reducing agent. A positive oxidation potential means it gets oxidized readily, while the reduction potential is just the opposite in sign.
Key Concept:
Oxidation potential = how easily a substance loses electrons (oxidizes).
Reduction potential = how easily a substance gains electrons (reduces).
🔁 They are numerically equal but opposite in sign.
ii. A salt bridge maintains electrical neutrality in the cell.
Answer:
In a galvanic cell, when electrons flow from one electrode to another, ions also need to move to balance the charge. The salt bridge allows this movement of ions and prevents charge buildup, so the cell keeps working.
Key Concept:
⚖️ Salt bridge = balance of charges = completes the circuit in electrochemical cells.
iii. Na and K can displace hydrogen from acids but Cu and Pt cannot.
Answer:
This is because Na and K are more reactive than hydrogen, while Cu and Pt are less reactive. According to the reactivity series, only metals above hydrogen can displace it from acids.
Key Concept:
🔁 Reactivity series → metals higher than H₂ can displace H⁺ from acids.
iv. Define oxidation in terms of electron transfer.
Answer:
Oxidation means loss of electrons.
For example, when Zn → Zn²⁺ + 2e⁻, zinc loses electrons and is oxidized.
Key Concept:
🔁 OIL RIG – Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons)
v. What is the oxidation number of oxygen in H₂O₂?
Answer:
In hydrogen peroxide (H₂O₂), oxygen has an oxidation number of -1 (not -2 like in water).
That’s because in peroxides, oxygen atoms are paired and each shares one electron.
Key Concept:
⚛️ Peroxides = oxygen oxidation number is -1.
(H₂O₂ is a peroxide.)
vi. Identify the reducing agent in the reaction: Zn + Cu²⁺ → Zn²⁺ + Cu
Answer:
Zinc is the reducing agent because it loses electrons (oxidized) and reduces Cu²⁺ to Cu.
Key Concept:
⚡ Reducing agent = the one that gives electrons (gets oxidized itself).
Unit 2: Electrochemistry
vii. State the purpose of the Winkler method.
Answer:
The Winkler method measures the amount of dissolved oxygen in water, which helps determine water quality and biological oxygen demand (BOD).
Key Concept:
🌊 Dissolved O₂ = how clean or polluted water is → BOD test.
viii. Explain what happens at the anode during the electrolysis of aqueous sodium chloride.
Answer:
At the anode, Cl⁻ ions lose electrons (oxidized) and form Cl₂ gas.
So, chlorine is liberated at the anode.
Key Concept:
Anode = Oxidation = Cl⁻ → Cl₂
💨 Gas forms at electrode.
ix. What does a positive standard electrode potential indicate about a substance’s tendency to gain electrons?
Answer:
A positive E° (standard reduction potential) means the substance easily gains electrons, so it’s a strong oxidizing agent.
Key Concept:
📈 More positive E° → More likely to get reduced (gain e⁻).
x. Calculate the oxidation number of chromium in K₂Cr₂O₇.
Answer:
Let the oxidation number of Cr be x:
2(+1) + 2(x) + 7(-2) = 0
→ 2 + 2x – 14 = 0 → 2x = +12 → x = +6
Answer: Oxidation number of Cr = +6
Key Concept:
🧮 Total charge = 0 in neutral compound → Solve like an equation.
xi. Describe the role of the standard hydrogen electrode in electrochemistry.
Answer:
The standard hydrogen electrode (SHE) is used as a reference electrode with a potential of 0 V. It helps compare the electrode potentials of other half-cells.
Key Concept:
⚖️ SHE = universal standard to measure all electrode potentials.
xii. What is the significance of the activity series of metals?
Answer:
It shows which metals are more reactive. A metal higher in the series can displace a metal lower in the series or hydrogen from acid.
Key Concept:
🔄 Activity series helps predict displacement reactions.
xiii. How can you deduce the feasibility of a redox reaction using electrode potentials?
Answer:
A redox reaction is feasible if the net EMF is positive:
E°cell = E°cathode – E°anode > 0
That means the reaction can spontaneously occur.
Key Concept:
⚡ Positive E°cell = spontaneous redox reaction.
xiv. Electrolysis of alumina reaction: 2Al₂O₃ → 4Al + 3O₂
Given:
Current = 15 A, Time = 10 hours = 36000 s
Faraday’s constant = 96500 C/mol
Reaction:
4Al³⁺ + 12e⁻ → 4Al → 3 mol O₂ → 6 mol O (atoms)
Mass of Al collected:
Electrons for 4 mol Al = 12 mol e⁻
For 1 mol Al = 3 mol e⁻
Total charge = I × t = 15 × 36000 = 540000 C
Moles of e⁻ = 540000 / 96500 ≈ 5.6 mol
Moles of Al = 5.6 / 3 ≈ 1.87 mol
Mass = 1.87 × 27 = 50.36 g ✅
Volume of O₂ at STP:
From reaction: 4Al → 3O₂
Moles of O₂ = (1.87 × 3)/4 = 1.4 mol
Volume = 1.4 × 22.4 = ~34.2 dm³ ✅
Key Concept:
🔋 Use Faraday’s law:
Q = I × t, then use moles = Q / F\text{Q = I × t, then use moles = Q / F}
xv. 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 hour
Options:
a) NaCl
b) CaCl₂
✅ Correct Answer: a) NaCl
Answer:
Both compounds provide metal ions, but Na⁺ requires 1 electron, whereas Ca²⁺ needs 2 electrons to deposit.
So, less charge is required per atom of Na, hence more Na is deposited for the same current.
Key Concept:
Faraday’s law: Mass deposited ∝ Charge / n (electrons required)
xvi. How many hours would electroplating have to be continued at 5 A if 75g of copper is to be deposited from CuSO₄ solution?
Solution:
- Molar mass of Cu = 63.5 g/mol
- n (electrons) = 2 for Cu²⁺ → Cu
- Mass (m) = 75 g
- Faraday’s constant (F) = 96500 C/mol
Use formula: Q= n × F × m /M = 2 × 96500 × 75/ 63.5 ≈227559 C
Now, Time (t)=Q/I=227559/5 ≈ 45511.8 s≈12.64 hours
✅ Answer: About 12.6 hours
Key Concept:
Faraday’s Second Law: Q=n×F×m /M
xvii. Differentiate between the following:
(a) Galvanic and Electrolytic Cell
| Feature | Galvanic Cell | Electrolytic Cell |
|---|---|---|
| Energy | Converts chemical → electrical | Converts electrical → chemical |
| Spontaneity | Spontaneous | Non-spontaneous |
| Electrodes | Anode (−), Cathode (+) | Anode (+), Cathode (−) |
(b) Oxidation Half-Reaction and Reduction Half-Reaction
| Type | Description | Example |
|---|---|---|
| Oxidation Half-Reaction | Loss of electrons | Zn → Zn²⁺ + 2e⁻ |
| Reduction Half-Reaction | Gain of electrons | Cu²⁺ + 2e⁻ → Cu |
Key Concept:
🔁 Oxidation = Loss of e⁻, Reduction = Gain of e⁻ (OIL RIG)
xviii. What mass of gold is deposited when a 5 A current is passed for 30 minutes?
Half-reaction:
Au(CN)2− + 3e⁻ → Au + 4CN⁻
- Time = 30 mins = 1800 s
- Charge (Q) = 5 × 1800 = 9000 C
- n = 3
- Molar mass of Au = 197 g/mol
Mass=Q×M / n×F=9000 × 197/ 3×96500≈6.12g
✅ Answer: 6.12 g
Key Concept:
Faraday’s Law: Mass=Q×M/ n×F
xix. Construct redox equations using the following half equations
(a) SO₂ → HSO₄⁻
Half-reactions:
- Oxidation: SO₂ + 2H₂O → HSO₄⁻ + 3H⁺ + 2e⁻
- Balancing with reduction partner not given, so redox equation incomplete without full pairing.
(b) MnO₄⁻ → Mn²⁺
Half-reaction:
- MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Again, this is a reduction half-reaction; needs pairing with oxidation to form full redox.
(c) Cl⁻ → Cl₂
Oxidation half-reaction:
2Cl⁻ → Cl₂ + 2e⁻
(d) H₂O₂ → O₂
Oxidation:
H₂O₂ → O₂ + 2H⁺ + 2e⁻
(e) Mn²⁺ → MnO₄⁻
Complicated half-reaction, usually:
Mn²⁺ + 4H₂O → MnO₄⁻ + 8H⁺ + 5e⁻
(Reverse of MnO₄⁻ → Mn²⁺)
(f) Sn²⁺ → Sn⁴⁺
Oxidation:
Sn²⁺ → Sn⁴⁺ + 2e⁻
(g) Fe²⁺ → Fe³⁺
Oxidation:
Fe²⁺ → Fe³⁺ + e⁻
(h) IO₃⁻ → I₂
Reduction:
2IO₃⁻ + 12H⁺ + 10e⁻ → I₂ + 6H₂O
(i) AsO₃³⁻ → AsO₄³⁻
Oxidation:
AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻
