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Electrochemistry: Class 12 Chemistry Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based3 marks

A team of environmental engineers is monitoring an industrial effluent treatment plant. They set up the following electrochemical cell to test the redox potential of the treated water sample containing Cr₂O₇²⁻ and Cr³⁺ ions. The dichromate/chromium(III) couple acts as the cathode and the iron(III)/iron couple acts as the anode.

A team of environmental engineers is monitoring an industrial effluent treatment plant. They set up the following electrochemical cell to test the redox potential of the treated water sample containing Cr₂O₇²⁻ and Cr³⁺ ions:

Pt(s) | Cr³⁺(0.1 M), Cr₂O₇²⁻(0.01 M), H⁺(0.1 M) || Fe³⁺(0.001 M) | Fe(s)

The cell uses the dichromate/chromium(III) half-cell as cathode and the iron(III)/iron half-cell as anode.

[Given: E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V, E°(Fe³⁺/Fe) = −0.04 V, 2.303RT/F = 0.0591 V at 298 K, log 2 = 0.301, log 3 = 0.477]

(i) Write the balanced half-reactions at the cathode and anode, and calculate E°cell. (2 marks)
(ii) Using the Nernst equation, calculate Ecell at 298 K. (1 mark)
(iii) Calculate ΔrG for this cell reaction. (1 mark)

Show answer
(i) Balanced half-reactions and E°cell:

Cathode (Reduction):
Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l)

Anode (Oxidation):
2Fe(s) → 2Fe³⁺(aq) + 6e⁻ [×1, already 6e⁻ when multiplied by 3: Fe(s) → Fe³⁺(aq) + 3e⁻, ×2]

Wait — balancing electrons: anode half-reaction multiplied by 2:
2Fe(s) → 2Fe³⁺(aq) + 6e⁻

Overall cell reaction (n = 6):
Cr₂O₇²⁻(aq) + 14H⁺(aq) + 2Fe(s) → 2Cr³⁺(aq) + 7H₂O(l) + 2Fe³⁺(aq)

E°cell = E°cathode − E°anode
E°cell = (+1.33) − (−0.04)
∴ E°cell = +1.37 V

(ii) Ecell using Nernst equation (n = 6):

Nernst equation:
Ecell = E°cell − (0.0591/n) log Q

Reaction quotient Q:
Q = [Cr³⁺]² [Fe³⁺]² / ([Cr₂O₇²⁻][H⁺]¹⁴)

Substituting concentrations:
[Cr³⁺] = 0.1 M, [Cr₂O₇²⁻] = 0.01 M, [H⁺] = 0.1 M, [Fe³⁺] = 0.001 M

Q = (0.1)² × (0.001)² / [(0.01) × (0.1)¹⁴]
= (10⁻²) × (10⁻⁶) / [(10⁻²) × (10⁻¹⁴)]
= 10⁻⁸ / 10⁻¹⁶
= 10⁸

log Q = 8

Ecell = 1.37 − (0.0591/6) × 8
= 1.37 − (0.00985 × 8)
= 1.37 − 0.0788
∴ Ecell = +1.2912 V ≈ +1.29 V

(iii) ΔrG for the cell reaction:

ΔrG = −nFEcell
ΔrG = −6 × 96500 × 1.29
ΔrG = −6 × 96500 × 1.29
= −746,370 J mol⁻¹
∴ ΔrG ≈ −746.4 kJ mol⁻¹

(The large negative value of ΔrG confirms that the cell reaction is spontaneous under the given non-standard conditions, indicating that the dichromate-based redox couple can effectively oxidise iron — a useful diagnostic for the engineers.)
Q2Case-based4 marks

A school science club sets up an electrochemical cell to investigate corrosion of iron pipes in a water supply system. They construct the following cell at 298 K:

Fe(s) | Fe²⁺(aq, 0.001 M) ‖ Cu²⁺(aq, 0.1 M) | Cu(s)

Given: E°(Fe²⁺/Fe) = −0.44 V, E°(Cu²⁺/Cu) = +0.34 V, F = 96500 C mol⁻¹

A school science club sets up an electrochemical cell to investigate corrosion of iron pipes in a water supply system. They construct the following cell at 298 K:

Fe(s) | Fe²⁺(aq, 0.001 M) ‖ Cu²⁺(aq, 0.1 M) | Cu(s)

Given: E°(Fe²⁺/Fe) = −0.44 V, E°(Cu²⁺/Cu) = +0.34 V, F = 96500 C mol⁻¹

(a) Identify the anode and cathode of the cell. Write the overall balanced cell reaction. (2 marks)
(b) Calculate the cell potential (E_cell) at the given concentrations using the Nernst equation. (1 mark)
(c) The students observe that iron corrodes faster when it is in electrical contact with copper in the presence of an electrolyte. Using the concept of electrode potential, justify this observation. (1 mark)

Show answer
MARKING SCHEME — Case-Based Question (4 marks)

──────────────────────────────────────────
Part (a) — Anode, Cathode and Cell Reaction [2 marks]
──────────────────────────────────────────

Anode (oxidation): Fe(s) → Fe²⁺(aq) + 2e⁻
Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

Overall balanced cell reaction:
Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)

Value points:
• Correct identification of anode as Fe and cathode as Cu [½ mark]
• Correct half-reactions written and balanced [½ mark]
• Correct overall balanced equation [1 mark]

──────────────────────────────────────────
Part (b) — Cell Potential using Nernst Equation [1 mark]
──────────────────────────────────────────

Step 1 — Standard cell potential:
E°_cell = E°_cathode − E°_anode
E°_cell = (+0.34) − (−0.44) = +0.78 V

Step 2 — Nernst equation (n = 2, T = 298 K):
E_cell = E°_cell − (0.0591/n) × log Q

where Q = [Fe²⁺] / [Cu²⁺] = 0.001 / 0.1 = 0.01 = 10⁻²

Step 3 — Substitution:
E_cell = 0.78 − (0.0591/2) × log(10⁻²)
E_cell = 0.78 − (0.02955) × (−2)
E_cell = 0.78 + 0.0591

∴ E_cell = 0.8391 V ≈ 0.84 V

Value point: Correct application of Nernst equation with correct Q and final answer with unit [1 mark]

──────────────────────────────────────────
Part (c) — Justification of faster corrosion [1 mark]
──────────────────────────────────────────

Due to the large difference in standard electrode potentials of iron (−0.44 V) and copper (+0.34 V), when they are in electrical contact in the presence of an electrolyte, a galvanic cell is set up. Iron acts as the anode (lower electrode potential) and undergoes oxidation, i.e., it corrodes preferentially and at a faster rate, while copper (higher electrode potential) acts as the cathode and is protected.

Value point: Correct reasoning — iron has lower electrode potential → acts as anode → oxidised faster (galvanic corrosion) [1 mark]
Q3Case-based4 marks

A school science club sets up a simple galvanic cell for a demonstration. They dip a zinc rod into 100 mL of 0.10 M ZnSO₄ solution and a copper rod into 100 mL of 0.10 M CuSO₄ solution, connected by a salt bridge. Given: E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V, R = 8.314 J K⁻¹ mol⁻¹, F = 96500 C mol⁻¹, T = 298 K, 2.303RT/F = 0.0591 V, log 2 = 0.3010.

A school science club sets up a simple galvanic cell for a demonstration. They dip a zinc rod into 100 mL of 0.10 M ZnSO₄ solution and a copper rod into 100 mL of 0.10 M CuSO₄ solution, connected by a salt bridge.

Given:
E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V
R = 8.314 J K⁻¹ mol⁻¹, F = 96500 C mol⁻¹, T = 298 K
2.303RT/F = 0.0591 V, log 2 = 0.3010

(a) Write the cell representation (cell notation) for this galvanic cell, identifying the anode and cathode. (1 mark)
(b) Calculate E°cell for this cell. (1 mark)
(c) The students accidentally add 0.10 mol of CuSO₄ to the copper half-cell, making [Cu²⁺] = 0.20 M, while [Zn²⁺] remains 0.10 M. Calculate the new Ecell using the Nernst equation. (2 marks)

Show answer
(a) Cell Notation (1 mark)

At the anode, zinc undergoes oxidation; at the cathode, copper ions are reduced.

Cell representation:
Zn(s) | Zn²⁺(0.10 M) ‖ Cu²⁺(0.10 M) | Cu(s)

Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

(b) Calculation of E°cell (1 mark)

E°cell = E°cathode − E°anode

E°cell = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn)

E°cell = (+0.34) − (−0.76)

∴ E°cell = +1.10 V

(c) Calculation of Ecell using the Nernst equation (2 marks)

The overall cell reaction is:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

For this reaction, n = 2.

The reaction quotient Q is:
Q = [Zn²⁺] / [Cu²⁺] = 0.10 / 0.20 = 1/2 = 0.5

Applying the Nernst equation at 298 K:
Ecell = E°cell − (0.0591/n) × log Q

Ecell = 1.10 − (0.0591/2) × log(0.5)

log(0.5) = log(1/2) = −log 2 = −0.3010

Ecell = 1.10 − (0.02955) × (−0.3010)

Ecell = 1.10 + 0.02955 × 0.3010

Ecell = 1.10 + 0.00889

∴ Ecell ≈ 1.109 V

Since [Cu²⁺] has increased, Q decreases below 1, so Ecell > E°cell — the cell emf increases slightly, which the students can observe as a marginally higher voltage reading.
Q4Case-based4 marks

A technician in an electrochemical laboratory sets up the following cell at 298 K:

Ni(s) | Ni²⁺ (0.01 M) || Ag⁺ (0.001 M) | Ag(s)

Given: E°(Ni²⁺/Ni) = −0.25 V, E°(Ag⁺/Ag) = +0.80 V
[log 2 = 0.3010, log 3 = 0.4771, log 10 = 1]

A technician in an electrochemical laboratory sets up the following cell at 298 K:

Ni(s) | Ni²⁺ (0.01 M) || Ag⁺ (0.001 M) | Ag(s)

Given: E°(Ni²⁺/Ni) = −0.25 V, E°(Ag⁺/Ag) = +0.80 V
[log 2 = 0.3010, log 3 = 0.4771, log 10 = 1]

(a) Identify the anode and cathode of the cell. Write the overall balanced cell reaction. (2 marks)
(b) Calculate the e.m.f. of the cell at 298 K using the Nernst equation. (1 mark)
(c) The technician observes that as the cell operates, the e.m.f. gradually decreases. Justify this observation in terms of changes in ion concentrations. (1 mark)

Show answer
(a) Identification of anode and cathode, and overall cell reaction:

At the anode (oxidation — lower reduction potential):
Ni(s) → Ni²⁺(aq) + 2e⁻

At the cathode (reduction — higher reduction potential):
2Ag⁺(aq) + 2e⁻ → 2Ag(s)

Overall balanced cell reaction:
Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s)

∴ Anode: Ni electrode; Cathode: Ag electrode. [1 mark for correct identification with reasoning; 1 mark for balanced overall equation]

(b) Calculation of e.m.f. at 298 K:

Step 1 — Calculate E°<sub>cell</sub>:
E°<sub>cell</sub> = E°<sub>cathode</sub> − E°<sub>anode</sub>
E°<sub>cell</sub> = (+0.80) − (−0.25) = +1.05 V

Step 2 — Apply the Nernst equation (n = 2 electrons transferred):
E<sub>cell</sub> = E°<sub>cell</sub> − (0.0591/n) log Q

where Q = [Ni²⁺] / [Ag⁺]²

Substituting values:
Q = (0.01) / (0.001)² = (0.01) / (10<sup>−6</sup>) = 10<sup>4</sup>

E<sub>cell</sub> = 1.05 − (0.0591/2) × log(10<sup>4</sup>)
E<sub>cell</sub> = 1.05 − (0.02955) × 4
E<sub>cell</sub> = 1.05 − 0.1182

∴ E<sub>cell</sub> = 0.9318 V ≈ 0.93 V [1 mark]

(c) Justification for gradual decrease in e.m.f.:

As the cell operates, Ni is oxidised at the anode, so [Ni²⁺] increases, while Ag⁺ is consumed at the cathode, so [Ag⁺] decreases. This causes the reaction quotient Q = [Ni²⁺]/[Ag⁺]² to increase continuously. By the Nernst equation, E<sub>cell</sub> = E°<sub>cell</sub> − (0.0591/n) log Q, an increasing Q increases the log Q term, thereby decreasing E<sub>cell</sub>. The cell e.m.f. falls until Q = K<sub>c</sub> (equilibrium), at which point E<sub>cell</sub> = 0. [1 mark]
Q5Case-based4 marks

A research group is developing a portable electrochemical device for monitoring water quality. The device uses a galvanic cell based on the half-reactions for Mn²⁺/Mn and Cl₂/Cl⁻ couples at 298 K under specified non-standard conditions.

A research group is developing a portable electrochemical device for monitoring water quality. The device uses a galvanic cell based on the following half-reactions at 298 K:

Mn²⁺(aq) + 2e⁻ → Mn(s) E° = −1.18 V
Cl₂(g) + 2e⁻ → 2Cl⁻(aq) E° = +1.36 V

During a field test, the cell operates under non-standard conditions: [Mn²⁺] = 0.001 mol L⁻¹, P(Cl₂) = 2 atm, and [Cl⁻] = 0.1 mol L⁻¹.

(a) Identify the anode and cathode of this cell and write the overall balanced cell reaction. Also calculate E°cell. (2 marks)

(b) Using the Nernst equation, calculate the cell potential Ecell under the given non-standard conditions. (1 mark)

(c) The research group observes that beyond a certain point, even though Cl₂ is still being produced upstream, the device gives zero reading. Give ONE electrochemical reason for this observation. (1 mark)

Show answer
(a) Identifying anode, cathode and calculating E°cell:

At the anode (oxidation), the electrode with lower reduction potential undergoes oxidation:
Anode: Mn(s) → Mn²⁺(aq) + 2e⁻ (E°oxidation = +1.18 V)

At the cathode (reduction), the electrode with higher reduction potential undergoes reduction:
Cathode: Cl₂(g) + 2e⁻ → 2Cl⁻(aq) (E°reduction = +1.36 V)

Overall balanced cell reaction (electrons cancel directly, n = 2):
Mn(s) + Cl₂(g) → Mn²⁺(aq) + 2Cl⁻(aq)

Cell notation: Mn(s) | Mn²⁺(aq) ‖ Cl₂(g) | Cl⁻(aq) | Pt(s)

E°cell = E°cathode − E°anode
E°cell = (+1.36) − (−1.18)
∴ E°cell = +2.54 V

(b) Calculating Ecell using the Nernst equation:

The Nernst equation at 298 K is:
Ecell = E°cell − (0.0591/n) × log Q

Here n = 2. The reaction quotient Q is:
Q = [Mn²⁺][Cl⁻]² / P(Cl₂)

Substituting the given values:
[Mn²⁺] = 0.001 mol L⁻¹, [Cl⁻] = 0.1 mol L⁻¹, P(Cl₂) = 2 atm

Q = (0.001 × (0.1)²) / 2
Q = (0.001 × 0.01) / 2
Q = 1 × 10⁻⁵ / 2
Q = 5 × 10⁻⁶

log Q = log(5 × 10⁻⁶) = log 5 + log 10⁻⁶ = 0.699 + (−6) = −5.301

Ecell = 2.54 − (0.0591/2) × (−5.301)
Ecell = 2.54 − (0.02955) × (−5.301)
Ecell = 2.54 + 0.1567
∴ Ecell ≈ +2.70 V

(c) Electrochemical reason for zero reading (Ecell = 0):

When Ecell = 0, the cell has reached electrochemical equilibrium — the reaction quotient Q has become equal to the equilibrium constant K. At this point, the forward and reverse electrode reactions proceed at equal rates, so there is no net electron flow (no net current), and the device records zero potential even though Cl₂ may still be present in the surroundings.
Q6Case-based4 marks

A school science club sets up a simple electrochemical demonstration. They connect a zinc plate and a copper plate into a beaker containing 1.0 mol L⁻¹ CuSO₄ solution. The standard electrode potentials are given as: E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V.

A school science club sets up a simple electrochemical demonstration. They connect a zinc plate and a copper plate into a beaker containing 1.0 mol L⁻¹ CuSO₄ solution. The standard electrode potentials are given as:
E°(Cu²⁺/Cu) = +0.34 V
E°(Zn²⁺/Zn) = −0.76 V

(a) Write the cell notation for this electrochemical cell. Identify the anode and the cathode. (2 marks)
(b) Calculate the standard cell potential (E°cell) for this cell. (1 mark)
(c) The students observe that the blue colour of the CuSO₄ solution gradually fades. Give one reason for this observation. (1 mark)

Show answer
(a) Cell Notation and Electrode Identification (2 marks)

In a galvanic cell, oxidation occurs at the anode and reduction occurs at the cathode.

Zinc has a lower (more negative) standard electrode potential (−0.76 V) compared to copper (+0.34 V), so zinc undergoes oxidation → zinc is the ANODE.
Copper undergoes reduction → copper is the CATHODE.

Cell notation (anode on left, cathode on right):

Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)

∴ Anode: Zinc plate (oxidation takes place)
∴ Cathode: Copper plate (reduction takes place)

[1 mark for correct cell notation with state symbols and double salt bridge ‖]
[1 mark for correct identification of anode and cathode with reason]

─────────────────────────────────────────
(b) Standard Cell Potential (1 mark)

The formula for standard cell potential is:

E°cell = E°cathode − E°anode

Substituting the given values:

E°cell = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn)
E°cell = (+0.34) − (−0.76)
E°cell = +0.34 + 0.76

∴ E°cell = +1.10 V

[1 mark for correct substitution and answer with unit]

─────────────────────────────────────────
(c) Reason for Fading of Blue Colour (1 mark)

Due to the reduction of Cu²⁺ ions at the cathode:

Cu²⁺(aq) + 2e⁻ → Cu(s)

the concentration of Cu²⁺ ions in the solution decreases steadily. Since the blue colour of CuSO₄ solution is due to the presence of Cu²⁺(aq) ions, as these ions are consumed (deposited as copper metal on the cathode), the intensity of the blue colour gradually fades.

[1 mark for correct reason: Cu²⁺ ions are consumed by reduction at the cathode, reducing their concentration]
Q7Case-based4 marks

Riya is a Class 12 student investigating electrochemical cells in her school laboratory. She sets up a galvanic cell using zinc and copper electrodes dipped in their respective 1.0 M sulphate solutions at 25°C and records E°cell = +1.10 V. Her teacher then asks her to explore how the cell potential changes when concentrations are altered, and how the cell can be used to determine thermodynamic quantities.

Riya is a Class 12 student investigating electrochemical cells in her school laboratory. She sets up a galvanic cell using zinc and copper electrodes dipped in their respective 1.0 M sulphate solutions at 25°C and records E°cell = +1.10 V. Her teacher then asks her to explore how the cell potential changes when concentrations are altered, and how the cell can be used to determine thermodynamic quantities.

Based on the above scenario, answer the following:

(a) Write the cell notation, the anode reaction, the cathode reaction, and the overall cell reaction for the Zn–Cu galvanic cell described above. (2 marks)

(b) Calculate the cell potential (E_cell) when [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.01 M at 25°C. (Given: E°cell = +1.10 V; log 10 = 1) (1 mark)

(c) Calculate the Gibbs energy change (ΔG°) for the cell reaction under standard conditions. (Given: F = 96500 C mol⁻¹) (1 mark)

Show answer
(a) Cell notation, electrode reactions and overall cell reaction: [2 marks]

Cell notation:
Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)

(Anode on left — oxidation; Cathode on right — reduction.)

Anode reaction (oxidation):
Zn(s) → Zn²⁺(aq) + 2e⁻

Cathode reaction (reduction):
Cu²⁺(aq) + 2e⁻ → Cu(s)

Overall cell reaction:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

(1 mark for correct cell notation + anode and cathode reactions; 1 mark for balanced overall cell reaction with state symbols.)

(b) Calculation of E_cell using the Nernst equation: [1 mark]

The Nernst equation at 298 K is:
E_cell = E°_cell − (0.0591/n) × log Q

Here, n = 2 (2 electrons transferred).

The reaction quotient Q is:
Q = [Zn²⁺] / [Cu²⁺] = 0.1 / 0.01 = 10

Substituting:
E_cell = 1.10 − (0.0591/2) × log 10
E_cell = 1.10 − (0.02955) × 1
E_cell = 1.10 − 0.0296

∴ E_cell = 1.0704 V ≈ 1.07 V

(c) Calculation of ΔG°: [1 mark]

The relation between standard Gibbs energy change and standard cell potential is:
ΔG° = −nFE°_cell

Substituting:
ΔG° = −2 × 96500 × 1.10
ΔG° = −2 × 96500 × 1.10
ΔG° = −212300 J mol⁻¹

∴ ΔG° = −212300 J mol⁻¹ = −212.3 kJ mol⁻¹

(The negative value confirms the cell reaction is spontaneous under standard conditions.)
Q8Case-based4 marks

A portable water-quality monitoring device uses an electrochemical cell to detect dissolved zinc ions in a river sample. The technician sets up the following cell at 298 K:

Zn(s) | Zn²⁺ (0.0001 M) || Cu²⁺ (0.1 M) | Cu(s)

Given: E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V, log 10 = 1

A portable water-quality monitoring device uses an electrochemical cell to detect dissolved zinc ions in a river sample. The technician sets up the following cell at 298 K:

Zn(s) | Zn²⁺ (0.0001 M) || Cu²⁺ (0.1 M) | Cu(s)

Given: E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V, log 10 = 1

(a) Write the cell reaction and calculate E°cell. [1]
(b) Using the Nernst equation, calculate the emf of the cell (Ecell) at 298 K. [2]
(c) The technician notices that when the concentration of Cu²⁺ is increased further, the cell potential rises. Justify this observation on the basis of the Nernst equation. [1]

Show answer
(a) Cell reaction and E°cell:

At anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
At cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

Overall cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

E°cell = E°cathode − E°anode
E°cell = (+0.34) − (−0.76)
∴ E°cell = +1.10 V [1]

(b) Nernst equation (n = 2, T = 298 K):

Ecell = E°cell − (0.0591/n) log Q

Here Q = [Zn²⁺] / [Cu²⁺] = (0.0001) / (0.1) = 10⁻³

Ecell = 1.10 − (0.0591/2) log(10⁻³)
Ecell = 1.10 − (0.02955) × (−3)
Ecell = 1.10 + 0.0887
∴ Ecell = 1.1887 V ≈ 1.19 V [2]

(Award 1 mark for correct Nernst substitution; 1 mark for correct final value with unit.)

(c) Justification using Nernst equation:

From the Nernst equation: Ecell = E°cell − (0.0591/n) log([Zn²⁺]/[Cu²⁺])

When [Cu²⁺] increases, the reaction quotient Q = [Zn²⁺]/[Cu²⁺] decreases. A smaller Q means log Q becomes more negative, so the term −(0.0591/n) log Q becomes more positive. Therefore, Ecell increases.

∴ A higher [Cu²⁺] shifts the equilibrium further towards products, increasing the driving force of the cell reaction and hence the cell potential rises. [1]
Q9Short Answer1 mark

Assertion (A) : The molar conductivity of a strong electrolyte increases on dilution.
Reason (R) : On dilution, the degree of dissociation of a strong electrolyte increases significantly.

Show answer
(C) A is true, but R is false.

Explanation: Strong electrolytes are completely dissociated at all concentrations — their degree of dissociation does not change appreciably with dilution. Therefore R is false. However, A is true: on dilution, interionic attractions decrease, so each ion moves more freely. The molar conductivity Λm = κ × 1000 / M rises with decreasing concentration and approaches a limiting value Λ°m (obtained by extrapolating the Kohlrausch plot Λm vs √C to C = 0).
Q10MCQ1 mark

Which of the following expressions correctly represents the relationship between molar conductivity (Λ<sub>m</sub>), conductivity (κ) and concentration (C) in mol L<sup>−1</sup>?

(A) Λ<sub>m</sub> = κ × C / 1000
(B) Λ<sub>m</sub> = κ × 1000 / C
(C) Λ<sub>m</sub> = C / (κ × 1000)
(D) Λ<sub>m</sub> = 1000 / (κ × C)

Show answer
(B) Λ<sub>m</sub> = κ × 1000 / C

Molar conductivity is related to conductivity by:

Λ<sub>m</sub> = κ × 1000 / C

where κ is in S cm⁻¹ and C is in mol L⁻¹, giving Λ<sub>m</sub> in S cm² mol⁻¹.
Q11MCQ1 mark

Which of the following is the correct expression for the relationship between standard Gibbs energy change (ΔG°) and the standard EMF (E°cell) of a cell?

Show answer
(B) ΔG° = −nFE°cell

Explanation: For a spontaneous electrochemical cell reaction, the standard Gibbs energy change is related to the standard cell potential by ΔG° = −nFE°cell, where n is the number of moles of electrons transferred per mole of reaction and F = 96500 C mol⁻¹. Since a spontaneous cell has E°cell > 0, ΔG° is negative, consistent with a thermodynamically favourable process. Option (A) is incorrect because the negative sign is missing. Options (C) and (D) incorrectly introduce RT in place of F.
Q12MCQ1 mark

The molar ionic conductivities of Na⁺ and SO₄²⁻ are 50.1 and 160.0 S cm² mol⁻¹ respectively. The value of limiting molar conductivity (Λ°m) of Na₂SO₄ will be:

Show answer
Answer: (B) 260.2 S cm² mol⁻¹

Explanation: By Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte equals the sum of the limiting molar conductivities of its constituent ions, each multiplied by the number of ions furnished per formula unit. Na₂SO₄ dissociates as Na₂SO₄ → 2Na⁺ + SO₄²⁻, giving two Na⁺ ions and one SO₄²⁻ ion per formula unit.

∴ Λ°m(Na₂SO₄) = 2 × λ°(Na⁺) + 1 × λ°(SO₄²⁻)
= 2 × 50.1 + 160.0
= 100.2 + 160.0
= 260.2 S cm² mol⁻¹
Q13Short Answer2 marks

The conductivity of a 0.02 M KCl solution at 298 K is 2.48 × 10⁻³ S cm⁻¹. Calculate the molar conductivity of the solution.

Show answer
Given:
Conductivity, κ = 2.48 × 10⁻³ S cm⁻¹
Concentration, M = 0.02 mol L⁻¹

Formula:
Λ<sub>m</sub> = (κ × 1000) / M

Substituting:
Λ<sub>m</sub> = (2.48 × 10⁻³ × 1000) / 0.02
Λ<sub>m</sub> = 2.48 / 0.02

∴ Λ<sub>m</sub> = 124 S cm² mol⁻¹
Q14Short Answer2 marks

Calculate the emf of the following cell at 298 K:

Zn(s) | Zn²⁺(0·001 M) ‖ Cu²⁺(0·1 M) | Cu(s)

Given: E°cell = 1·10 V, log 10 = 1, log 100 = 2

Show answer
Using the Nernst equation at 298 K:

E_cell = E°_cell − (0·0591/n) log Q

For the cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), n = 2

Q = [Zn²⁺] / [Cu²⁺] = 0·001 / 0·1 = 0·01 = 10⁻²

E_cell = 1·10 − (0·0591/2) log (10⁻²)

E_cell = 1·10 − (0·02955) × (−2)

E_cell = 1·10 + 0·0591

∴ E_cell = 1·1591 V ≈ 1·16 V
Q15Short Answer2 marks

The conductivity of a 0.02 M KCl solution at 298 K is 2.48 × 10⁻³ S cm⁻¹. Calculate its molar conductivity.

Show answer
Using the relation:

Λm = (κ × 1000) / M

where κ = 2.48 × 10⁻³ S cm⁻¹, M = 0.02 mol L⁻¹

Λm = (2.48 × 10⁻³ × 1000) / 0.02

Λm = 2.48 / 0.02

∴ Λm = 124 S cm² mol⁻¹
Q16Short Answer2 marks

The following data is given for acetic acid (CH₃COOH) at 298 K:

Λ°m(CH₃COOH) = 390.5 S cm² mol⁻¹
Λm(CH₃COOH) at 0.01 M = 16.2 S cm² mol⁻¹

(a) Calculate the degree of dissociation (α) of acetic acid at 0.01 M.
(b) Give one reason why Λm of acetic acid increases sharply as concentration approaches zero.

Show answer
(a) The degree of dissociation is given by:

α = Λm / Λ°m

Substituting the given values:

α = 16.2 / 390.5

∴ α = 0.0415 (or 4.15%)

(b) Due to weak electrolyte behaviour — at very low concentrations, acetic acid dissociates almost completely into CH₃COO⁻ and H⁺ ions, increasing the number of charge carriers drastically. Since Λm accounts for the conductivity per mole of electrolyte, it rises sharply and approaches Λ°m as concentration → 0.
Q17Short Answer3 marks

A battery manufacturer is testing a new electrochemical cell based on the following half-reactions at 298 K:

Cathode: Fe³⁺(aq) + e⁻ → Fe²⁺(aq), E° = +0.77 V
Anode: Sn(s) → Sn²⁺(aq) + 2e⁻, E° = −(−0.14 V) = +0.14 V (oxidation)

The electrolyte solution contains Fe³⁺ at 0.010 M, Fe²⁺ at 0.100 M, and Sn²⁺ at 0.050 M.

(a) Write the balanced overall cell reaction and calculate E°cell. (2 marks)
(b) Using the Nernst equation, calculate Ecell at 298 K for the given ion concentrations. (1 mark)
(c) Calculate ΔG for the cell under these non-standard conditions. State whether the cell reaction is spontaneous. (1 mark)

(Given: F = 96500 C mol⁻¹; log 2 = 0.301; log 5 = 0.699)

Show answer
(a) Balanced overall cell reaction and E°cell:

The cathode (reduction) half-reaction involves Fe³⁺/Fe²⁺ and the anode (oxidation) involves Sn/Sn²⁺. To balance electrons, multiply the cathode half-reaction by 2:

Cathode (×2): 2Fe³⁺(aq) + 2e⁻ → 2Fe²⁺(aq)
Anode: Sn(s) → Sn²⁺(aq) + 2e⁻

Overall balanced cell reaction:
Sn(s) + 2Fe³⁺(aq) → Sn²⁺(aq) + 2Fe²⁺(aq)

E°cell = E°cathode − E°anode
= (+0.77) − (−0.14)
= +0.77 + 0.14
∴ E°cell = +0.91 V

[Note: E° values are intensive — the cathode half-reaction is not multiplied when computing E°cell.]

(b) Nernst equation at 298 K:

For the overall reaction: Sn(s) + 2Fe³⁺(aq) → Sn²⁺(aq) + 2Fe²⁺(aq), n = 2

Ecell = E°cell − (0.0591/n) × log Q

Q = [Sn²⁺][Fe²⁺]² / [Fe³⁺]²

Substituting the given concentrations:
[Sn²⁺] = 0.050 M, [Fe²⁺] = 0.100 M, [Fe³⁺] = 0.010 M

Q = (0.050)(0.100)² / (0.010)²
= (0.050 × 0.0100) / (0.0001)
= 5.0 × 10⁻⁴ / 1.0 × 10⁻⁴
= 5.0

log Q = log 5 = 0.699

Ecell = 0.91 − (0.0591/2) × 0.699
= 0.91 − 0.02955 × 0.699
= 0.91 − 0.02066
= 0.91 − 0.0207
∴ Ecell = +0.8893 V ≈ +0.889 V

(c) ΔG under non-standard conditions:

ΔG = −nFEcell

Substituting:
n = 2, F = 96500 C mol⁻¹, Ecell = +0.889 V

ΔG = −2 × 96500 × 0.889
= −2 × 96500 × 0.889
= −171,577 J mol⁻¹
∴ ΔG ≈ −171.6 kJ mol⁻¹

Since ΔG < 0 (negative), the cell reaction is spontaneous under the given non-standard conditions.
Q18Short Answer3 marks

The conductivity of 0.001 mol L⁻¹ acetic acid solution is 4.95 × 10⁻⁵ S cm⁻¹. Calculate the molar conductivity and degree of dissociation of acetic acid at this concentration. Given: Λ°m (CH₃COOH) = 390.5 S cm² mol⁻¹.

Show answer
Step 1 — Molar Conductivity

The governing formula is:

Λm = (κ × 1000) / M

where κ = conductivity (S cm⁻¹) and M = molarity (mol L⁻¹).

Substituting:

Λm = (4.95 × 10⁻⁵ S cm⁻¹ × 1000 cm³ L⁻¹) / (0.001 mol L⁻¹)

Λm = (4.95 × 10⁻² S cm⁻¹ L) / (0.001 mol L⁻¹)

∴ Λm = 49.5 S cm² mol⁻¹ … (1 mark)

Step 2 — Degree of Dissociation

For a weak electrolyte, the degree of dissociation is given by:

α = Λm / Λ°m

Substituting:

α = 49.5 S cm² mol⁻¹ / 390.5 S cm² mol⁻¹

∴ α = 0.1268 ≈ 0.127 … (1 mark)

Conclusion:

The molar conductivity of acetic acid at 0.001 mol L⁻¹ is 49.5 S cm² mol⁻¹ and its degree of dissociation is 0.127 (i.e., approximately 12.7%). Since Λm ≪ Λ°m, acetic acid is a weak electrolyte that is only partially dissociated at this concentration. … (1 mark)
Q19Short Answer3 marks

Calculate the EMF of the following cell at 298 K:

Fe(s) | Fe²⁺(0.001 M) || Ag⁺(0.1 M) | Ag(s)

Given: E°(Ag⁺/Ag) = +0.80 V, E°(Fe²⁺/Fe) = −0.44 V
[Given: log 10 = 1, log 100 = 2]

Show answer
Step 1 — Standard cell EMF (1 mark)

E°cell = E°cathode − E°anode
= E°(Ag⁺/Ag) − E°(Fe²⁺/Fe)
= (+0.80) − (−0.44)
= +1.24 V

The cell reaction (n = 2):

Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s)

Step 2 — Nernst equation with substitution (1 mark)

Ecell = E°cell − (0.0591/n) log Q

where Q = [Fe²⁺] / [Ag⁺]²
= (0.001) / (0.1)²
= (1 × 10⁻³) / (1 × 10⁻²)
= 10⁻¹
= 0.1

∴ log Q = log(0.1) = log(10⁻¹) = −1

Ecell = 1.24 − (0.0591/2) × (−1)
= 1.24 + 0.02955

Step 3 — Final answer with unit (1 mark)

∴ Ecell = 1.2696 V ≈ 1.27 V

(Accept 1.2696 V or 1.27 V; ½ mark deducted if unit is missing.)
Q20Short Answer3 marks

Calculate the emf of the following cell at 298 K:

Ni(s) | Ni²⁺ (0·01 M) || Ag⁺ (0·1 M) | Ag(s)

Given: E°(Ag⁺/Ag) = +0·80 V, E°(Ni²⁺/Ni) = −0·25 V, log 10 = 1

Also calculate ΔG° for the cell reaction.
(Given: F = 96500 C mol⁻¹)

Show answer
Step 1 — Identify cathode and anode, and write E°cell

E°cell = E°cathode − E°anode
= E°(Ag⁺/Ag) − E°(Ni²⁺/Ni)
= (+0·80) − (−0·25)
= +1·05 V … (½ mark)

Step 2 — Write the balanced cell reaction

Oxidation (anode): Ni(s) → Ni²⁺(aq) + 2e⁻
Reduction (cathode): 2Ag⁺(aq) + 2e⁻ → 2Ag(s)
────────────────────────────────────────────────
Net cell reaction: Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s) n = 2

Step 3 — Apply the Nernst equation

Ecell = E°cell − (0·0591/n) log Q

Q = [Ni²⁺] / [Ag⁺]²
= (0·01) / (0·1)²
= 0·01 / 0·01
= 1 … (½ mark)

∴ log Q = log 1 = 0

Ecell = 1·05 − (0·0591/2) × 0
= 1·05 − 0

∴ Ecell = 1·05 V … (1 mark)

Step 4 — Calculate ΔG°

ΔG° = −nFE°cell
= −2 × 96500 × 1·05
= −202650 J mol⁻¹

∴ ΔG° = −202650 J mol⁻¹ (or −202·65 kJ mol⁻¹) … (1 mark)
Q21Short Answer3 marks

Answer the following:
(a) State Kohlrausch's law of independent migration of ions. Write the expression for the degree of dissociation (α) of a weak electrolyte in terms of its molar conductivity.
(b) Calculate the emf of the following cell at 298 K:
Cu(s) | Cu²⁺(0.01 M) ‖ Ag⁺(0.1 M) | Ag(s)
Given: E°(Ag⁺/Ag) = +0.80 V, E°(Cu²⁺/Cu) = +0.34 V
[log 10 = 1, log 2 = 0.301]

Show answer
(a) Kohlrausch's Law: At infinite dilution, the molar conductivity of an electrolyte is the sum of the individual contributions of its constituent ions (cations and anions), independent of the nature of the counter ion.

For an electrolyte A<sub>x</sub>B<sub>y</sub>:
Λ°<sub>m</sub> = x λ°<sub>+</sub> + y λ°<sub>−</sub>

Degree of dissociation:
α = Λ<sub>m</sub> / Λ°<sub>m</sub>

where Λ<sub>m</sub> = molar conductivity at the given concentration and Λ°<sub>m</sub> = limiting molar conductivity. [1 mark]

(b) Step 1 – Identify cathode and anode and write E°<sub>cell</sub>:

E°<sub>cell</sub> = E°<sub>cathode</sub> − E°<sub>anode</sub>
= E°(Ag⁺/Ag) − E°(Cu²⁺/Cu)
= +0.80 − 0.34
= +0.46 V

The cell reaction is:
Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) (n = 2)

Step 2 – Write the Nernst equation:

E<sub>cell</sub> = E°<sub>cell</sub> − (0.0591 / n) log Q

where Q = [Cu²⁺] / [Ag⁺]²

Substituting values:
Q = (0.01) / (0.1)²
= 0.01 / 0.01
= 1

log Q = log 1 = 0

Step 3 – Calculate E<sub>cell</sub>:

E<sub>cell</sub> = 0.46 − (0.0591 / 2) × 0

∴ E<sub>cell</sub> = +0.46 V [1 + 1 marks]
Q22Short Answer3 marks

Answer the following:
(a) The molar conductivity of a weak acid HA increases sharply on dilution, but that of a strong electrolyte NaCl increases only gradually. Give one reason for each behaviour.
(b) Calculate the standard Gibbs energy change (ΔG°) for the cell reaction:
Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s)
Given: E°(Ag⁺/Ag) = +0.80 V; E°(Fe²⁺/Fe) = −0.44 V; F = 96500 C mol⁻¹

Show answer
(a) Weak acid HA (1 mark):
For weak electrolyte HA, degree of dissociation (α) increases greatly on dilution. The number of ions (H⁺ and A⁻) carrying current therefore increases sharply, causing Λm to rise steeply on dilution.

Strong electrolyte NaCl (1 mark):
NaCl is completely dissociated at all dilutions, so the number of ions does not increase. The gradual rise in Λm on dilution is due to the decrease in inter-ionic attractions (interionic forces) between Na⁺ and Cl⁻, which allows the ions to move more freely.

(b) Numerical (1 mark):
Step 1 — Identify cathode and anode, then calculate E°cell:
Cathode (reduction): Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V
Anode (oxidation): Fe(s) → Fe²⁺(aq) + 2e⁻, E° = −(−0.44) = +0.44 V

E°cell = E°cathode − E°anode = 0.80 − (−0.44) = +1.24 V

Step 2 — Apply ΔG° = −nFE°cell:
Here n = 2 (2 electrons transferred per formula unit)

ΔG° = −nFE°cell
ΔG° = −2 × 96500 × 1.24
ΔG° = −239,480 J mol⁻¹

∴ ΔG° = −239480 J mol⁻¹ = −239.48 kJ mol⁻¹
Q23Short Answer3 marks

The conductivity of 0.025 mol L⁻¹ methanoic acid is 46.1 × 10⁻⁵ S cm⁻¹. Calculate: (i) the molar conductivity of the solution, and (ii) the degree of dissociation of methanoic acid at this concentration. Given: λ°(H⁺) = 349.6 S cm² mol⁻¹ and λ°(HCOO⁻) = 54.6 S cm² mol⁻¹.

Show answer
(i) Molar Conductivity of the Solution

The governing formula is:

Λm = (κ × 1000) / M

where κ is the conductivity in S cm⁻¹ and M is the molarity in mol L⁻¹.

Substituting the given values:

Λm = (46.1 × 10⁻⁵ S cm⁻¹ × 1000 cm³ L⁻¹) / (0.025 mol L⁻¹)

Λm = (46.1 × 10⁻² S cm⁻² L) / (0.025 mol L⁻¹)

Λm = 46.1 × 10⁻² / 0.025 S cm² mol⁻¹

∴ Λm = 184.4 S cm² mol⁻¹

(ii) Degree of Dissociation of Methanoic Acid

First, calculate the limiting molar conductivity using Kohlrausch's Law:

Λ°m(HCOOH) = λ°(H⁺) + λ°(HCOO⁻)

Λ°m(HCOOH) = 349.6 + 54.6 = 404.2 S cm² mol⁻¹

The degree of dissociation is given by:

α = Λm / Λ°m

Substituting:

α = 184.4 / 404.2

∴ α = 0.456 (i.e., degree of dissociation ≈ 0.46 or 45.6%)
Q24Short Answer3 marks

An electrochemical cell is set up to monitor the corrosion of iron in an industrial pipeline. The cell is represented as:

Fe(s) | Fe²⁺ (0.001 M) || Cu²⁺ (0.1 M) | Cu(s)

Given: E°(Fe²⁺/Fe) = −0.44 V, E°(Cu²⁺/Cu) = +0.34 V, 2.303RT/F at 298 K = 0.0591 V

(a) Write the cell reaction and calculate E_cell at 298 K using the Nernst equation. (2 marks)

(b) A corrosion-inhibitor company claims that coating the iron pipeline with a thin layer of zinc (sacrificial anode protection) will stop iron from corroding. Justify this claim using standard electrode potential values. (1 mark)

(c) When the above Fe–Cu cell reaches equilibrium, E_cell = 0. Calculate the equilibrium constant K_c for the cell reaction at 298 K. (1 mark)

Show answer
(a) Cell Reaction and E_cell at 298 K

At the anode (oxidation): Fe(s) → Fe²⁺(aq) + 2e⁻
At the cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

Overall cell reaction: Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)

E°_cell = E°_cathode − E°_anode
= (+0.34) − (−0.44)
= +0.78 V

Nernst equation (n = 2):

E_cell = E°_cell − (0.0591/n) log([Fe²⁺]/[Cu²⁺])

Substituting values:

E_cell = 0.78 − (0.0591/2) log(0.001/0.1)
= 0.78 − (0.02955) log(10⁻²)
= 0.78 − (0.02955)(−2)
= 0.78 + 0.0591

∴ E_cell = 0.8391 V ≈ 0.84 V

(b) Justification of Sacrificial Anode Protection

E°(Zn²⁺/Zn) = −0.76 V, which is more negative than E°(Fe²⁺/Fe) = −0.44 V.

Since zinc has a lower (more negative) standard electrode potential than iron, zinc is oxidised preferentially over iron whenever both are in electrical contact with an electrolyte. Zinc acts as the anode and is consumed, while iron acts as the cathode and is protected from oxidation (corrosion).

∴ The zinc coating acts as a sacrificial anode and prevents iron from corroding.

(c) Equilibrium Constant K_c

At equilibrium, E_cell = 0 ∴ E°_cell = (0.0591/n) log K_c

log K_c = (n × E°_cell) / 0.0591
= (2 × 0.78) / 0.0591
= 1.56 / 0.0591
= 26.40

∴ K_c = 10^(26.40) ≈ 2.51 × 10²⁶
Q25Case-based4 marks

A school science club is investigating the conductivity of solutions for a water-quality project. They measure the conductivity of three water samples:
• Sample P: tap water with conductivity κ = 2.90 × 10⁻⁴ S cm⁻¹ at a concentration of 0.010 M NaCl
• Sample Q: dilute acetic acid (CH₃COOH), concentration 0.010 M, conductivity κ = 5.20 × 10⁻⁵ S cm⁻¹
• Sample R: distilled water — negligible conductivity
Given: λ°(Na⁺) = 50.1 S cm² mol⁻¹, λ°(Cl⁻) = 76.3 S cm² mol⁻¹, λ°(CH₃COO⁻) = 40.9 S cm² mol⁻¹, λ°(H⁺) = 349.6 S cm² mol⁻¹

A school science club is investigating the conductivity of solutions for a water-quality project. They measure the conductivity of three water samples and record the following data:

• Sample P: tap water with conductivity κ = 2.90 × 10⁻⁴ S cm⁻¹ at a concentration of 0.010 M dissolved NaCl
• Sample Q: a dilute acetic acid (CH₃COOH) solution of concentration 0.010 M with conductivity κ = 5.20 × 10⁻⁵ S cm⁻¹
• Sample R: distilled water — negligible conductivity

Given: λ°(Na⁺) = 50.1 S cm² mol⁻¹, λ°(Cl⁻) = 76.3 S cm² mol⁻¹,
λ°(CH₃COO⁻) = 40.9 S cm² mol⁻¹, λ°(H⁺) = 349.6 S cm² mol⁻¹

(a) Calculate the molar conductivity (Λm) of Sample P and Sample Q. (2 marks)
(b) Using Kohlrausch's law, calculate the limiting molar conductivity (Λ°m) of acetic acid. Hence find the degree of dissociation (α) of acetic acid in Sample Q. (2 marks)

Show answer
Part (a) — Molar Conductivity of Sample P and Sample Q [2 marks]

Formula:

Λm = (κ × 1000) / M

where κ is in S cm⁻¹ and M is molarity in mol L⁻¹.

For Sample P (NaCl, κ = 2.90 × 10⁻⁴ S cm⁻¹, M = 0.010 mol L⁻¹):

Λm(P) = (2.90 × 10⁻⁴ × 1000) / 0.010

Λm(P) = 0.290 / 0.010

∴ Λm(P) = 29.0 S cm² mol⁻¹

For Sample Q (CH₃COOH, κ = 5.20 × 10⁻⁵ S cm⁻¹, M = 0.010 mol L⁻¹):

Λm(Q) = (5.20 × 10⁻⁵ × 1000) / 0.010

Λm(Q) = 0.0520 / 0.010

∴ Λm(Q) = 5.20 S cm² mol⁻¹

[Award 1 mark for Λm(P) = 29.0 S cm² mol⁻¹ with correct formula and substitution; 1 mark for Λm(Q) = 5.20 S cm² mol⁻¹ with correct formula and substitution. ECF applies if formula is correct but arithmetic error occurs — penalise once.]

---

Part (b) — Λ°m of Acetic Acid and Degree of Dissociation α [2 marks]

Step 1 — Kohlrausch's Law for CH₃COOH:

According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte equals the sum of the limiting molar conductivities of its constituent ions:

Λ°m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻)

Λ°m(CH₃COOH) = 349.6 + 40.9

∴ Λ°m(CH₃COOH) = 390.5 S cm² mol⁻¹

Step 2 — Degree of Dissociation α:

α = Λm / Λ°m

α = 5.20 / 390.5

∴ α = 0.0133 (i.e., approximately 1.33 × 10⁻²)

This low value of α confirms that acetic acid is a weak electrolyte — only about 1.33% of the molecules are dissociated into ions at this concentration, which is why Sample Q conducts electricity far less than Sample P (a strong electrolyte) even at the same concentration.

[Award 1 mark for correct application of Kohlrausch's law and Λ°m = 390.5 S cm² mol⁻¹; 1 mark for correct formula α = Λm/Λ°m and final answer α = 1.33 × 10⁻² (accept 0.0133). ECF: if Λm(Q) is wrong from part (a) but correctly used here, full marks for this step.]
Q26Case-based4 marks

A school chemistry club sets up a galvanic cell Fe(s)|Fe²⁺(0.01 M)||Ag⁺(0.1 M)|Ag(s) at 298 K to demonstrate electrochemical principles at a science exhibition. They explore how electrode reactions and ion concentrations affect the cell's EMF.

A school chemistry club is investigating electrochemical cells for a science exhibition. They set up the following cell at 298 K:

Fe(s) | Fe²⁺(0.01 M) || Ag⁺(0.1 M) | Ag(s)

Given: E°(Fe²⁺/Fe) = −0.44 V, E°(Ag⁺/Ag) = +0.80 V, 2.303RT/F = 0.0591 V at 298 K, log 2 = 0.301

(a) Identify the anode and cathode in this cell. Write the balanced cell reaction. (2 marks)

(b) Calculate the EMF of the cell using the Nernst equation. (1 mark)

(c) One student remarks: 'If we double the concentration of Ag⁺ from 0.1 M to 0.2 M, the EMF will also double.' Is this statement correct? Justify your answer with a calculation. (1 mark)

Show answer
PART (a) — Identification of Anode and Cathode + Balanced Cell Reaction [2 marks]

Anode (oxidation): Fe(s) → Fe²⁺(aq) + 2e⁻
Cathode (reduction): Ag⁺(aq) + e⁻ → Ag(s) ×2

Balancing electrons (multiply cathode half-reaction by 2):

Cathode: 2Ag⁺(aq) + 2e⁻ → 2Ag(s)
Anode: Fe(s) → Fe²⁺(aq) + 2e⁻

Overall balanced cell reaction:

Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s)

[½ mark — correct identification of anode and cathode; ½ mark — correct half-reactions; 1 mark — balanced overall cell reaction with correct charges and stoichiometry]

---

PART (b) — EMF Calculation using the Nernst Equation [1 mark]

Step 1 — Standard EMF:

E°cell = E°cathode − E°anode
E°cell = (+0.80) − (−0.44)
E°cell = +1.24 V

Step 2 — Reaction Quotient Q:

For the cell reaction Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s), n = 2

Q = [Fe²⁺] / [Ag⁺]²
Q = (0.01) / (0.1)²
Q = 0.01 / 0.01 = 1

Step 3 — Nernst Equation:

Ecell = E°cell − (0.0591/n) × log Q
Ecell = 1.24 − (0.0591/2) × log 1
Ecell = 1.24 − (0.02955) × 0
Ecell = 1.24 − 0

∴ Ecell = +1.24 V

[1 mark — correct substitution into Nernst equation and correct final answer with unit]

---

PART (c) — Evaluation of the Student's Claim [1 mark]

The student's statement is INCORRECT.

Due to the logarithmic relationship in the Nernst equation, the EMF does not change proportionally (linearly) with concentration — it changes logarithmically.

Verification by calculation when [Ag⁺] = 0.2 M:

Q' = [Fe²⁺] / [Ag⁺]²
Q' = (0.01) / (0.2)²
Q' = 0.01 / 0.04 = 0.25

log Q' = log (0.25) = log (1/4) = −log 4 = −2 × log 2 = −2 × 0.301 = −0.602

Ecell = 1.24 − (0.0591/2) × (−0.602)
Ecell = 1.24 − (0.02955) × (−0.602)
Ecell = 1.24 + 0.0178
∴ Ecell ≈ +1.258 V

The EMF increases only slightly (from 1.24 V to ~1.258 V) on doubling [Ag⁺] from 0.1 M to 0.2 M — it certainly does NOT double (which would imply 2.48 V). The claim is incorrect because EMF depends on log[Ag⁺], not directly on [Ag⁺].

[1 mark — correct rejection of the claim with a valid reason/supporting calculation]
Q27Case-based4 marks

Hydrogen–oxygen fuel cells convert chemical energy directly into electrical energy with high efficiency and water as the only by-product. Such cells are preferred in space and underwater missions because they do not produce polluting gases and have a high energy density. In a hydrogen–oxygen fuel cell, hydrogen gas is fed at the anode and oxygen gas at the cathode; both electrodes are typically made of porous platinum to allow gas diffusion and catalysis. The electrolyte used may be aqueous KOH (alkaline) or a proton-exchange membrane (acidic). The overall reaction releases energy that drives the external circuit.

A research team is developing a hydrogen–oxygen fuel cell for an unmanned underwater vehicle. The cell operates at 298 K under standard conditions. The standard electrode potentials are: O₂/H₂O: E° = +1.23 V and H⁺/H₂: E° = 0.00 V.

(a) Write the cell notation (cell representation) for this fuel cell and calculate E°cell. (2 marks)

(b) Calculate ΔG° for the cell reaction: 2H₂(g) + O₂(g) → 2H₂O(l), given that n = 4. (1 mark)

(c) The engineers observe that the actual cell potential drops below E°cell during operation. Give ONE reason for this observation in the context of fuel cell working. (1 mark)

Show answer
Part (a) — 2 marks

Cell notation:
Pt(s) | H₂(g) | H⁺(aq) ‖ O₂(g) | H₂O(l) | Pt(s)

The anode (oxidation) is placed on the left and the cathode (reduction) on the right.

Using the relation:
E°cell = E°cathode − E°anode
E°cell = (+1.23 V) − (0.00 V)
∴ E°cell = +1.23 V

Part (b) — 1 mark

Using the relation:
ΔG° = −nFE°cell

Here n = 4 (4 moles of electrons transferred for the overall reaction 2H₂(g) + O₂(g) → 2H₂O(l)), F = 96500 C mol⁻¹, E°cell = 1.23 V.

ΔG° = −4 × 96500 × 1.23
ΔG° = −4 × 96500 × 1.23
ΔG° = −474,540 J mol⁻¹

∴ ΔG° = −4.75 × 10⁵ J mol⁻¹ (or −475 kJ mol⁻¹)

The large negative value of ΔG° confirms the reaction is spontaneous under standard conditions.

Part (c) — 1 mark

During actual operation, the concentration of reactants (H₂ and O₂) at the electrode surfaces decreases as they are consumed, so the reaction quotient Q increases above 1. By the Nernst equation:
Ecell = E°cell − (0.0591/n) log Q

Since log Q > 0, Ecell < E°cell.

∴ The actual cell potential drops below E°cell because the non-standard concentrations/partial pressures of H₂ and O₂ at the electrodes during operation make Q > 1, lowering the cell potential as described by the Nernst equation.

(Alternatively accepted: Activation overpotential / internal resistance / polarisation losses cause the working voltage to be lower than the theoretical E°cell.)
Q28Case-based4 marks

A chemistry student sets up a galvanic cell using two half-cells: a zinc electrode dipped in 0.001 M ZnSO₄ solution and a copper electrode dipped in 0.1 M CuSO₄ solution, connected by a salt bridge at 298 K. Standard electrode potentials: E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V.

A chemistry student sets up a galvanic cell using two half-cells: a zinc electrode dipped in 0.001 M ZnSO₄ solution and a copper electrode dipped in 0.1 M CuSO₄ solution, connected by a salt bridge at 298 K. The standard electrode potentials are given as: E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V.

(a) Write the cell notation (cell representation) for this galvanic cell and identify the anode and cathode. (2 marks)

(b) Calculate the EMF of the cell under the given non-standard conditions using the Nernst equation. (Given: log 10 = 1, log 100 = 2) (1 mark)

(c) If the same galvanic cell is allowed to operate until equilibrium is reached, what will be the value of Ecell at that point? Justify your answer. (1 mark)

Show answer
Part (a): [2 marks]

Cell Notation:
Zn(s) | Zn²⁺(aq, 0.001 M) ‖ Cu²⁺(aq, 0.1 M) | Cu(s)

Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

• Zinc electrode is the anode (undergoes oxidation; more negative E°).
• Copper electrode is the cathode (undergoes reduction; more positive E°).

E°cell = E°cathode − E°anode
E°cell = (+0.34) − (−0.76) = +1.10 V

[Value points: correct cell notation with anode on left ½; anode/cathode identification ½; half-reactions written correctly ½; E°cell = +1.10 V ½]

---

Part (b): [1 mark]

Applying the Nernst equation at 298 K (n = 2 electrons transferred):

Ecell = E°cell − (0.0591/n) × log([Zn²⁺]/[Cu²⁺])

Substituting values:
Ecell = 1.10 − (0.0591/2) × log(0.001/0.1)
Ecell = 1.10 − (0.02955) × log(10⁻²)
Ecell = 1.10 − (0.02955) × (−2)
Ecell = 1.10 + 0.0591

∴ Ecell = 1.1591 V ≈ 1.16 V

[Value point: correct substitution into Nernst equation and correct final value with unit ½ + ½]

---

Part (c): [1 mark]

At equilibrium, Ecell = 0 V.

Justification: At equilibrium, the concentrations of the reactants and products adjust such that the reaction quotient Q equals the equilibrium constant K. Since Ecell = E°cell − (0.0591/n) log Q, when Q = K, Ecell becomes zero. This means there is no net driving force for the cell reaction, and the cell can no longer do electrical work.

[Value points: Ecell = 0 V ½; correct justification (Q = K, no net driving force) ½]
Q29Case-based4 marks

A school science club is designing a simple galvanic cell for a project. They use a zinc rod dipped in 1 M ZnSO₄ solution and a copper rod dipped in 1 M CuSO₄ solution, connected by a salt bridge. The teacher provides the following standard electrode potential data:

Zn²⁺(aq) + 2e⁻ → Zn(s), E° = −0.76 V
Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V

A school science club is designing a simple galvanic cell for a project. They use a zinc rod dipped in 1 M ZnSO₄ solution and a copper rod dipped in 1 M CuSO₄ solution, connected by a salt bridge. The teacher provides the following standard electrode potential data:

Zn²⁺(aq) + 2e⁻ → Zn(s), E° = −0.76 V
Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V

(a) Identify the anode and the cathode in this cell. Write the cell notation (cell representation) for this galvanic cell. (2 marks)
(b) Calculate the standard EMF (E°cell) of the cell. (1 mark)
(c) The students notice that the copper rod becomes heavier during the operation of the cell. Give one reason for this observation. (1 mark)

Show answer
(a) Identification of anode and cathode, and cell notation: (2 marks)

The electrode with the lower (more negative) standard reduction potential undergoes oxidation and acts as the anode. The electrode with the higher (more positive) standard reduction potential undergoes reduction and acts as the cathode.

∴ Anode: Zinc (Zn) — oxidation occurs here.
∴ Cathode: Copper (Cu) — reduction occurs here.

Electrode reactions:

At anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
At cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

Cell notation:

Zn(s) | Zn²⁺(aq, 1 M) ‖ Cu²⁺(aq, 1 M) | Cu(s)

(Anode written on the left; cathode on the right; double vertical line represents the salt bridge.)

(b) Calculation of E°cell: (1 mark)

Using the relation:

E°cell = E°cathode − E°anode

E°cell = (+0.34 V) − (−0.76 V)

E°cell = 0.34 + 0.76

∴ E°cell = +1.10 V

(c) Reason for the copper rod becoming heavier: (1 mark)

At the cathode, Cu²⁺ ions from the CuSO₄ solution are reduced and deposited as copper metal on the copper rod:

Cu²⁺(aq) + 2e⁻ → Cu(s)

Due to this continuous deposition of copper atoms on the cathode, the mass of the copper rod increases during the operation of the cell.
Q30Case-based4 marks

A manufacturer uses an electrolytic cell to silver-plate cutlery. The cell contains aqueous AgNO₃ solution, a pure silver anode, and a steel spoon as the cathode. The circuit is run at a steady current of 2 A for 1930 seconds. Standard electrode potentials: Ag⁺(aq)/Ag(s) E° = +0.80 V; Fe²⁺(aq)/Fe(s) E° = −0.44 V.

A manufacturer uses an electrolytic cell to silver-plate cutlery. The cell contains aqueous AgNO₃ solution, a pure silver anode, and a steel spoon as the cathode. The circuit is run at a steady current of 2 A for 1930 seconds.

(a) Using Faraday's first law of electrolysis, calculate the mass of silver deposited on the spoon. (Molar mass of Ag = 108 g mol⁻¹, F = 96500 C mol⁻¹) [2 marks]

(b) In a separate quality-control experiment, the standard electrode potentials are used to calculate the EMF of the silver-plating cell at 298 K:

Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V
Fe²⁺(aq) + 2e⁻ → Fe(s), E° = −0.44 V

The cell is observed to operate spontaneously under standard conditions with Fe as the anode and Ag⁺/Ag as the cathode. Calculate the standard EMF of this cell. [1 mark]

(c) The manufacturer notices that when the AgNO₃ solution becomes very dilute, the rate of silver deposition slows down even though the current remains the same. Using the Nernst equation, explain why a decrease in [Ag⁺] reduces the electrode potential of the Ag⁺/Ag half-cell, and state the effect this has on the cell EMF. [1 mark]

Show answer
(a) Faraday's First Law: The mass of substance deposited at an electrode during electrolysis is directly proportional to the quantity of charge passed.

m = (M / nF) × Q

Step 1 — Calculate total charge:
Q = I × t = 2 × 1930 = 3860 C

Step 2 — Moles of electrons:
For Ag⁺ + e⁻ → Ag(s), n = 1

Step 3 — Substitute:
m = (108 / (1 × 96500)) × 3860
m = (108 × 3860) / 96500
m = 416880 / 96500
∴ m = 4.32 g

∴ Mass of silver deposited = 4.32 g

[Award ½ for correct formula; ½ for correct Q = 3860 C; ½ for correct substitution; ½ for final answer with unit]

(b) E°cell = E°cathode − E°anode
E°cell = E°(Ag⁺/Ag) − E°(Fe²⁺/Fe)
E°cell = (+0.80) − (−0.44)
∴ E°cell = +1.24 V

[Award 1 mark for correct application of the formula and correct answer with unit]

(c) For the Ag⁺/Ag half-cell, the Nernst equation gives:

E(Ag⁺/Ag) = E°(Ag⁺/Ag) − (0.0591 / 1) log (1 / [Ag⁺])

As [Ag⁺] decreases, log(1/[Ag⁺]) increases, so E(Ag⁺/Ag) decreases below E°.

Since E_cell = E_cathode − E_anode, a decrease in the cathode potential (Ag⁺/Ag) directly reduces the overall cell EMF.

∴ A decrease in [Ag⁺] lowers the electrode potential of the cathode and therefore reduces the cell EMF.

[Award 1 mark for correctly applying the Nernst equation to show the direction of change in E and its effect on EMF; accept 'cell EMF decreases' as the stated effect]

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