Batteries: Describe the chemical reactions occurring during the discharge of a Lead storage battery

Chapter: Electrochemistry – Batteries

Topic: Lead Storage Battery – Chemical Reactions During Discharge

A lead storage battery, also called a lead-acid battery, is a rechargeable secondary cell. It is commonly used in automobiles and backup power systems.

It consists of:

• Anode (negative plate): Lead, Pb
• Cathode (positive plate): Lead dioxide, PbO₂
• Electrolyte: Sulphuric acid, H₂SO₄

Electrolyte = H₂SO₄(aq)

During discharge, the battery converts chemical energy into electrical energy.

Negative Plate: Pb

Lead undergoes oxidation and reacts with sulphate ions from the electrolyte.

Pb + SO₄²⁻ → PbSO₄ + 2e⁻

Thus, lead is converted into lead sulphate and electrons are released.

Oxidation occurs at the negative electrode.

Positive Plate: PbO₂

Lead dioxide is reduced in the presence of sulphuric acid and forms lead sulphate and water.

PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O

Thus, PbO₂ is converted into PbSO₄.

Reduction occurs at the positive electrode.

Combining the Two Half-Reactions

Negative electrode:

Pb + SO₄²⁻ → PbSO₄ + 2e⁻

Positive electrode:

PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O

Adding the two equations:

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

Overall discharge reaction:

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

Pb → PbSO_4>

⬇️

PbO₂ → PbSO₄

⬇️

H₂SO₄ is consumed

⬇️

Water concentration increases

As discharge continues, both plates gradually become coated with PbSO₄.

At the same time, sulphuric acid is consumed, so the concentration and specific gravity of the electrolyte decrease.

Therefore:
During discharge, Pb and PbO₂ are converted into PbSO₄ and the H₂SO₄ concentration decreases.

At the negative electrode, lead undergoes oxidation and releases electrons:

Pb → PbSO₄ + 2e⁻

These electrons travel through the external circuit from the negative plate to the positive plate.

Pb
Negative → e⁻ → PbO₂
Positive

Discharge	Charging
Chemical energy → Electrical energy	Electrical energy → Chemical energy
Pb and PbO₂ → PbSO₄	PbSO₄ → Pb and PbO₂
H₂SO₄ is consumed	H₂SO₄ is regenerated
Spontaneous reaction	Non-spontaneous reaction driven by external electricity

Q. Write the overall reaction occurring during the discharge of a lead storage battery.

Answer:

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

During discharge, sulphuric acid is consumed and both electrodes form lead sulphate.

Q. Write the electrode reactions occurring during discharge of a lead storage battery.

At negative electrode (oxidation):

Pb + SO₄²⁻ → PbSO₄ + 2e⁻

At positive electrode (reduction):

PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O

Q. Explain the working of a lead storage battery during discharge.

1. Pb acts as the negative electrode.
2. PbO₂ acts as the positive electrode.
3. H₂SO₄ acts as the electrolyte.
4. At the negative electrode, Pb is oxidised to PbSO₄.
5. At the positive electrode, PbO₂ is reduced to PbSO₄.
6. Electrons flow through the external circuit from Pb to PbO₂.
7. H₂SO₄ is consumed during the overall reaction.

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

Q. Describe the chemical reactions occurring during the discharge of a lead storage battery.

Negative Electrode

Lead undergoes oxidation:

Pb + SO₄²⁻ → PbSO₄ + 2e⁻

Positive Electrode

Lead dioxide undergoes reduction:

PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O

Overall Reaction

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

Thus, both electrodes are converted into PbSO₄ while sulphuric acid is consumed and water is produced.

Q. Describe in detail the chemical reactions occurring during the discharge of a lead storage battery. Explain the changes taking place at both electrodes and in the electrolyte.

Construction

• Negative electrode: Pb
• Positive electrode: PbO₂
• Electrolyte: H₂SO₄

1. Reaction at Negative Electrode

The Pb electrode undergoes oxidation and releases two electrons.

Pb + SO₄²⁻ → PbSO₄ + 2e⁻

Therefore, Pb is converted into PbSO₄.

2. Reaction at Positive Electrode

PbO₂ undergoes reduction by accepting electrons:

PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O

Therefore, PbO₂ is also converted into PbSO₄.

3. Overall Reaction

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

4. Changes During Discharge

• Both Pb and PbO₂ electrodes are converted into PbSO₄.
• H₂SO₄ is consumed.
• Water is formed.
• The concentration of H₂SO₄ decreases.
• The specific gravity of the electrolyte decreases.
• Electrons flow through the external circuit from Pb to PbO₂.

Final Answer:

During discharge, the lead storage battery converts chemical energy into electrical energy. Pb at the negative electrode is oxidised to PbSO₄, while PbO₂ at the positive electrode is reduced to PbSO₄. Sulphuric acid is consumed and water is formed.

1. The negative electrode of a lead storage battery is made of:

A) Cu

B) Pb

C) Zn

D) Ag

✅ Answer

B) Pb

---

2. The positive electrode consists of:

A) Pb

B) Zn

C) PbO₂

D) CuO

✅ Answer

C) PbO₂

---

3. The electrolyte used is:

A) HCl

B) NaOH

C) H₂SO₄

D) KOH

✅ Answer

C) H₂SO₄

---

4. A lead storage battery is a:

A) Primary cell

B) Secondary cell

C) Fuel cell only

D) Dry cell only

✅ Answer

B) Secondary cell

---

5. During discharge, Pb undergoes:

A) Reduction

B) Oxidation

C) Neutralisation

D) Hydrolysis

✅ Answer

B) Oxidation

---

6. During discharge, PbO₂ undergoes:

A) Oxidation

B) Reduction

C) Sublimation

D) Evaporation

✅ Answer

B) Reduction

---

7. The product formed at both electrodes during discharge is:

A) PbO

B) PbSO₄

C) PbCl₂

D) Pb(NO₃)₂

✅ Answer

B) PbSO₄

---

8. During discharge, sulphuric acid is:

A) Consumed

B) Produced

C) Unchanged

D) Converted to HCl

✅ Answer

A) Consumed

---

9. The overall reaction is:

A) Pb + PbO₂ → PbO

B) Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

C) PbSO₄ → PbO₂

D) Pb + HCl → PbCl₂

✅ Answer

B

---

10. Electrons flow during discharge from:

A) PbO₂ to Pb

B) Pb to PbO₂ through the external circuit

C) Electrolyte to Pb

D) Cathode to electrolyte

✅ Answer

B

---

11. Oxidation takes place at the:

A) Negative electrode during discharge

B) Positive electrode only

C) Electrolyte

D) Salt bridge

✅ Answer

A

---

12. Reduction occurs at the:

A) Negative electrode

B) Positive electrode during discharge

C) Electrolyte

D) Separator only

✅ Answer

B

---

13. The concentration of H₂SO₄ during discharge:

A) Increases

B) Decreases

C) Becomes zero immediately

D) Remains exactly constant

✅ Answer

B

---

14. The specific gravity of the electrolyte during discharge:

A) Increases

B) Decreases

C) Is unchanged

D) Becomes infinite

✅ Answer

B

---

15. A lead storage battery is commonly used in:

A) Automobile starting systems

B) Thermometers

C) Glass manufacture only

D) Water purification only

✅ Answer

A

---

16. During discharge, chemical energy is converted into:

A) Nuclear energy

B) Electrical energy

C) Sound energy only

D) Light energy only

✅ Answer

B

---

17. The negative electrode reaction releases:

A) 1 electron

B) 2 electrons

C) 3 electrons

D) 4 electrons

✅ Answer

B) 2 electrons

---

18. Water is formed at the:

A) Negative electrode

B) Positive electrode

C) Salt bridge

D) External wire

✅ Answer

B) Positive electrode

---

19. During charging, the discharge reaction is:

A) Reversed

B) Stopped permanently

C) Converted into combustion

D) Unrelated

✅ Answer

A) Reversed

---

20. Which statement is correct?

A) Both plates form PbSO₄ during discharge

B) Only Pb forms PbSO₄

C) Only PbO₂ forms PbSO₄

D) No sulphate participates

✅ Answer

A

🎯 Quick Revision

Negative electrode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻
Positive electrode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
During discharge: H₂SO₄ ↓   |   PbSO₄ ↑   |   Chemical energy → Electrical energy

Conductance: How does molar conductivity vary with concentration for strong versus weak electrolytes

Chapter: Electrochemistry – Conductance

Topic: Variation of Molar Conductivity with Concentration — Strong vs Weak Electrolytes

Molar conductivity (Λ_m) is the conductance of the volume of solution containing one mole of an electrolyte, placed between two electrodes one unit distance apart and large enough to contain the solution.

Λ_m = κ × (1000/C)

where:

• Λ_m = molar conductivity
• κ = conductivity
• C = concentration in mol L⁻¹

The SI unit of molar conductivity is S m² mol⁻¹. It is also commonly expressed as S cm² mol⁻¹.

Strong Electrolytes

Strong electrolytes such as HCl, KCl, NaCl and HNO₃ are almost completely ionised in solution even at relatively higher concentrations.

When the solution is diluted, the ions experience less interionic attraction and their mobility increases.

Concentration ↓ → Molar Conductivity ↑

For strong electrolytes, the increase in molar conductivity on dilution is relatively small and gradual.

At infinite dilution, molar conductivity reaches its maximum value, called Λ°_m.

Λ_m = Λ°_m − K√C

This is the approximate relation given by Kohlrausch for strong electrolytes.

Why does Λ_m increase?

1. Dilution decreases interionic attraction.
2. Ion mobility increases.
3. Therefore, the molar conductivity increases.

Weak Electrolytes

Weak electrolytes such as CH₃COOH, NH₄OH and other weak acids and bases are only partially ionised in solution.

When the solution is diluted, the degree of ionisation increases significantly.

Dilution ↑ → Ionisation ↑ → Number of ions ↑

Therefore, molar conductivity increases very sharply on dilution.

The increase is much greater than that observed for strong electrolytes.

Why is the increase so large?

1. Weak electrolytes are incompletely ionised.
2. Dilution shifts the ionisation equilibrium towards ions.
3. The number of ions increases greatly.
4. Ion mobility also increases.
5. Therefore, Λ_m rises sharply.

Property	Strong Electrolyte	Weak Electrolyte
Ionisation	Almost complete	Partial
Examples	HCl, NaCl, KCl	CH₃COOH, NH₄OH
Effect of dilution	Small increase	Large increase
Ionisation on dilution	Changes little	Increases greatly
Graph	Gradual increase	Steep increase
At infinite dilution	Λ°m	Λ°m

Key Point:
Molar conductivity increases on dilution for both strong and weak electrolytes, but the increase is small for strong electrolytes and very large for weak electrolytes.

Variation of Molar Conductivity with √Concentration

Λ_m

√C

Strong electrolyte

Weak electrolyte

Strong electrolyte: nearly linear decrease of Λ_m with √C.

Weak electrolyte: highly curved graph because dilution causes a large increase in ionisation.

Q. How does molar conductivity change with dilution?

Answer:

Molar conductivity increases with dilution for both strong and weak electrolytes. The increase is small for strong electrolytes but large for weak electrolytes because dilution greatly increases their degree of ionisation.

Q. Explain the variation of molar conductivity of strong electrolytes with concentration.

Answer:

1. Strong electrolytes are almost completely ionised.
2. On dilution, interionic attraction decreases.
3. Ion mobility increases, so molar conductivity increases.
4. The increase is relatively small and gradual.

Q. Compare the variation of molar conductivity with concentration for strong and weak electrolytes.

Strong Electrolytes	Weak Electrolytes
Almost completely ionised.	Partially ionised.
Dilution causes a small increase in Λm.	Dilution causes a large increase in Λm.
Ionisation changes little.	Ionisation increases greatly.
Graph is approximately linear with √C.	Graph is strongly curved.

Q. Explain the variation of molar conductivity with concentration for strong and weak electrolytes with suitable examples.

Strong Electrolytes

Strong electrolytes such as HCl, NaCl and KCl are almost completely ionised. On dilution, the interionic attraction decreases and the mobility of ions increases. Hence molar conductivity increases gradually.

Λ_m = Λ°_m − K√C

Weak Electrolytes

Weak electrolytes such as CH₃COOH and NH₄OH are only partially ionised. Dilution increases their degree of ionisation considerably. Consequently, the number of ions and their mobility increase, producing a sharp rise in molar conductivity.

Conclusion:

Λ_m increases on dilution for both types, but the increase is much greater for weak electrolytes.

Q. Explain in detail how molar conductivity varies with concentration for strong and weak electrolytes. Give the reason for the difference.

1. Strong Electrolytes

Strong electrolytes such as HCl, HNO₃, NaCl and KCl are almost completely ionised even at appreciable concentrations.

When concentration is decreased, the ions become farther apart and interionic attraction decreases. Consequently, ionic mobility increases and molar conductivity increases.

Concentration ↓ → Interionic attraction ↓ → Ion mobility ↑ → Λ_m ↑

The increase is relatively small because the electrolyte is already almost completely ionised.

2. Weak Electrolytes

Weak electrolytes such as CH₃COOH and NH₄OH are only partially ionised. When the solution is diluted, their degree of ionisation increases substantially.

Dilution ↑ → Degree of ionisation ↑ → Number of ions ↑ → Λ_m ↑↑

Therefore, molar conductivity increases sharply on dilution.

3. Mathematical Relation

For strong electrolytes, Kohlrausch's limiting-law form is approximately:

Λ_m = Λ°_m − K√C

where Λ°_m is the molar conductivity at infinite dilution and K is a constant for a particular electrolyte at a given temperature.

4. Main Difference

Strong	Weak
Already almost completely ionised	Partially ionised
Small increase on dilution	Large increase on dilution
Mainly due to increased ionic mobility	Due to both increased ionisation and mobility

Final Conclusion:

For both strong and weak electrolytes, molar conductivity increases as concentration decreases. However, the increase is gradual for strong electrolytes and very steep for weak electrolytes because dilution greatly increases the ionisation of weak electrolytes.

1. Molar conductivity is represented by:

A) κ

B) Λ_m

C) ρ

D) R

✅ Answer

B) Λ_m

---

2. The relation between conductivity and molar conductivity is:

A) Λ_m = κC/1000

B) Λ_m = κ × 1000/C

C) Λ_m = C/κ

D) Λ_m = κ + C

✅ Answer

B

---

3. On dilution, molar conductivity generally:

A) Decreases

B) Increases

C) Becomes zero

D) Remains constant

✅ Answer

B) Increases

---

4. Strong electrolytes are:

A) Completely unionised

B) Almost completely ionised

C) Always gases

D) Insoluble

✅ Answer

B

---

5. Which is a strong electrolyte?

A) CH₃COOH

B) NH₄OH

C) HCl

D) H₂CO₃

✅ Answer

C) HCl

---

6. Which is a weak electrolyte?

A) NaCl

B) KCl

C) HCl

D) CH₃COOH

✅ Answer

D) CH₃COOH

---

7. For strong electrolytes, Λ_m increases on dilution:

A) Very sharply

B) Gradually

C) Not at all

D) Randomly

✅ Answer

B

---

8. For weak electrolytes, Λ_m increases on dilution:

A) Very sharply

B) Very slightly

C) Not at all

D) It decreases

✅ Answer

A

---

9. The large increase in Λ_m for weak electrolytes is mainly due to:

A) Decrease in ionisation

B) Increase in ionisation

C) Decrease in temperature

D) Formation of solids

✅ Answer

B

---

10. At infinite dilution, molar conductivity is called:

A) κ

B) Λ°_m

C) R

D) G

✅ Answer

B

---

11. Kohlrausch's relation for strong electrolytes involves:

A) √C

B) C²

C) 1/C²

D) T² only

✅ Answer

A) √C

---

12. On dilution, interionic attraction generally:

A) Increases

B) Decreases

C) Becomes infinite

D) Remains unchanged

✅ Answer

B

---

13. Increased ionic mobility causes molar conductivity to:

A) Increase

B) Decrease

C) Become zero

D) Remain zero

✅ Answer

A

---

14. Which electrolyte shows the steepest increase in Λ_m on dilution?

A) HCl

B) NaCl

C) CH₃COOH

D) KCl

✅ Answer

C) CH₃COOH

---

15. Weak electrolytes have:

A) High degree of ionisation at all concentrations

B) Partial ionisation

C) No ions

D) Only electrons

✅ Answer

B

---

16. Which statement is correct?

A) Λ_m decreases on dilution

B) Λ_m increases on dilution

C) Λ_m is independent of concentration

D) Λ_m is always zero

✅ Answer

B

---

17. The main reason for gradual increase in strong electrolytes is:

A) Large increase in ionisation

B) Increased ionic mobility and reduced interionic attraction

C) Formation of molecules

D) Precipitation

✅ Answer

B

---

18. The unit S cm² mol⁻¹ is commonly used for:

A) Conductivity

B) Molar conductivity

C) Resistance

D) Cell constant

✅ Answer

B

---

19. Which pair contains a strong and a weak electrolyte respectively?

A) HCl and CH₃COOH

B) CH₃COOH and HCl

C) NH₄OH and NaCl

D) CH₃COOH and KCl

✅ Answer

A

---

20. The maximum molar conductivity is obtained at:

A) High concentration

B) Infinite dilution

C) Zero temperature

D) Boiling point

✅ Answer

B) Infinite dilution

🎯 Quick Revision

Strong electrolyte: Almost completely ionised → dilution causes a small increase in Λ_m
Weak electrolyte: Partially ionised → dilution greatly increases ionisation → sharp increase in Λ_m
At infinite dilution: Λ°_m is maximum
Strong electrolyte relation: Λ_m = Λ°_m − K√C

Electrochemistry: Derive the Nernst Equation for a galvanic cell under non-standard conditions

Chapter: Electrochemistry – Nernst Equation

📘 What is the Nernst Equation?

The Nernst equation gives the electrode potential or cell potential of an electrochemical cell under non-standard conditions.

For a galvanic cell, the cell potential depends on the concentration or activity of the reacting species, in addition to the standard cell potential.

E_cell = E°_cell − (RT/nF) ln Q

At 298 K:

E_cell = E°_cell − (0.0591/n) log Q

🔬 Step 1: Consider a General Galvanic Cell

Consider the cell reaction:

aA + bB → cC + dD

For this reaction, the reaction quotient is:

Q = [C]^c[D]^d / [A]^a[B]^b

Pure solids and pure liquids are not included in Q because their activities are taken as unity.

🔬 Step 2: Gibbs Energy and Cell Potential

The Gibbs energy change for an electrochemical reaction is related to the cell potential by:

ΔG = −nFE

Under standard conditions:

ΔG° = −nFE°

Therefore:

ΔG = ΔG° + RT ln Q

🔬 Step 3: Substitute the Electrochemical Relations

We know:

ΔG = −nFE

and

ΔG° = −nFE°

Substituting these into:

ΔG = ΔG° + RT ln Q

we get:

−nFE = −nFE° + RT ln Q

🔬 Step 4: Rearrangement

Divide the entire equation by −nF:

E = E° − (RT/nF) ln Q

This is the Nernst Equation.

🌡️ Step 5: Nernst Equation at 298 K

At 298 K:

R = 8.314 J mol⁻¹ K⁻¹
F = 96485 C mol⁻¹

Changing natural logarithm to common logarithm gives:

E = E° − (0.0591/n) log Q

where n is the number of electrons transferred in the balanced cell reaction.

🔋 Application to a Zn–Cu Galvanic Cell

Consider the Daniell cell:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Overall reaction:

Zn + Cu²⁺ → Zn²⁺ + Cu

Here, two electrons are transferred:

n = 2

Therefore:

Q = [Zn²⁺]/[Cu²⁺]

Hence, at 298 K:

E_cell = E°_cell − (0.0591/2) log ([Zn²⁺]/[Cu²⁺])

✅ Final Derivation

Starting from:

ΔG = ΔG° + RT ln Q

Using:

ΔG = −nFE      and      ΔG° = −nFE°

we obtain:

E_cell = E°_cell − (RT/nF) ln Q

At 298 K:

E_cell = E°_cell − (0.0591/n) log Q

E = Cell potential under non-standard conditions

E° = Standard cell potential

R = Gas constant = 8.314 J mol⁻¹ K⁻¹

T = Absolute temperature in kelvin

n = Number of electrons transferred

F = Faraday constant ≈ 96485 C mol⁻¹

Q = Reaction quotient

Q. Write the Nernst equation for a galvanic cell.

Answer:

E_cell = E°_cell − (RT/nF) ln Q

At 298 K:

E_cell = E°_cell − (0.0591/n) log Q

Q. Derive the Nernst equation briefly.

1. For an electrochemical reaction: ΔG = ΔG° + RT ln Q.
2. Using ΔG = −nFE and ΔG° = −nFE°.
3. Substitution and rearrangement give: E = E° − (RT/nF) ln Q.

Q. Derive the Nernst equation at 298 K.

ΔG = ΔG° + RT ln Q

−nFE = −nFE° + RT ln Q

E = E° − (RT/nF) ln Q

At 298 K, converting ln to log:

E = E° − (0.0591/n) log Q

Q. Derive the Nernst equation for a galvanic cell and explain the terms involved.

For a general cell reaction:

aA + bB → cC + dD

The reaction quotient is:

Q = [C]^c[D]^d / [A]^a[B]^b

Thermodynamically:

ΔG = ΔG° + RT ln Q

For an electrochemical cell:

ΔG = −nFE
ΔG° = −nFE°

Therefore:

E = E° − (RT/nF) ln Q

At 298 K:

E = E° − (0.0591/n) log Q

Q. Derive the Nernst equation for a galvanic cell under non-standard conditions. Explain its significance.

Step 1: Gibbs Energy Relation

ΔG = ΔG° + RT ln Q

Step 2: Relation with Cell Potential

For a galvanic cell:

ΔG = −nFE

Under standard conditions:

ΔG° = −nFE°

Step 3: Substitute

−nFE = −nFE° + RT ln Q

Step 4: Rearrange

E = E° − (RT/nF) ln Q

Step 5: At 298 K

E = E° − (0.0591/n) log Q

Significance

• It calculates cell potential under non-standard conditions.
• It shows how ion concentration affects cell potential.
• It can be used to determine equilibrium conditions.
• It is useful for calculating the potential of individual electrodes.

Conclusion:

The Nernst equation connects the standard cell potential with the actual cell potential at a specified temperature and composition.

1. The Nernst equation is used to calculate:

A) Cell potential under non-standard conditions

B) Atomic mass

C) Boiling point

D) Density

✅ Answer

A

---

2. Which equation is correct?

A) E = E° + RT ln Q/nF

B) E = E° − RT ln Q/nF

C) E = E°Q

D) E = E° + Q

✅ Answer

B

---

3. At 298 K, the Nernst equation contains the factor:

A) 0.0591/n

B) 0.591n

C) 59.1n

D) 9.81/n

✅ Answer

A

---

4. In the Nernst equation, n represents:

A) Number of ions

B) Number of electrons transferred

C) Number of atoms

D) Number of solutions

✅ Answer

B

---

5. F in the Nernst equation represents:

A) Force constant

B) Faraday constant

C) Free energy

D) Frequency

✅ Answer

B

---

6. The approximate value of Faraday constant is:

A) 8.314 C mol⁻¹

B) 96485 C mol⁻¹

C) 22.4 C mol⁻¹

D) 6.022 × 10²³ C mol⁻¹

✅ Answer

B

---

7. Q in the Nernst equation represents:

A) Heat

B) Reaction quotient

C) Charge only

D) Volume

✅ Answer

B

---

8. For a reaction aA + bB → cC + dD, Q is:

A) [A][B]/[C][D]

B) [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ

C) [A]ᵃ[B]ᵇ

D) [C][D]

✅ Answer

B

---

9. Pure solids are generally:

A) Included in Q

B) Not included in Q

C) Always numerator

D) Always denominator

✅ Answer

B

---

10. The relation between Gibbs energy and cell potential is:

A) ΔG = nFE

B) ΔG = −nFE

C) ΔG = RT

D) ΔG = E/nF

✅ Answer

B

---

11. Under standard conditions:

A) ΔG° = −nFE°

B) ΔG° = nFE°

C) ΔG° = RT

D) ΔG° = Q

✅ Answer

A

---

12. A galvanic cell converts:

A) Electrical energy into chemical energy spontaneously

B) Chemical energy into electrical energy

C) Heat into mass

D) Light into chemical energy only

✅ Answer

B

---

13. If Q = 1, the Nernst equation gives:

A) E = E°

B) E = 0

C) E = 2E°

D) E = Q

✅ Answer

A

---

14. At equilibrium, the cell potential of a reversible galvanic cell is:

A) Maximum

B) Zero

C) Negative infinity

D) Always 1 V

✅ Answer

B

---

15. At equilibrium, the reaction quotient is equal to:

A) 0

B) 1

C) K

D) nF

✅ Answer

C) K

---

16. For the Daniell cell Zn + Cu²⁺ → Zn²⁺ + Cu, n equals:

A) 1

B) 2

C) 3

D) 4

✅ Answer

B) 2

---

17. For the Daniell cell, Q is:

A) [Cu²⁺]/[Zn²⁺]

B) [Zn²⁺]/[Cu²⁺]

C) [Zn][Cu]

D) [Zn²⁺][Cu²⁺]

✅ Answer

B

---

18. Increasing the concentration of products generally makes Q:

A) Increase

B) Decrease

C) Zero

D) Always equal to 1

✅ Answer

A

---

19. The Nernst equation is derived using the relation:

A) ΔG = ΔG° + RT ln Q

B) PV = nRT only

C) E = mc²

D) q = mcΔT

✅ Answer

A

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20. The Nernst equation is important because it relates cell potential to:

A) Concentration/activity of reacting species

B) Colour of the solution

C) Atomic radius

D) Melting point only

✅ Answer

A

🎯 Quick Revision

General equation: E = E° − (RT/nF) ln Q
At 298 K: E = E° − (0.0591/n) log Q
ΔG: ΔG = −nFE
Standard ΔG: ΔG° = −nFE°
Q: Reaction quotient
n: Number of electrons transferred
F: Faraday constant