Chapter 2: Electrochemistry Case Study Questions | chemistry Class 12 CBSE

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Chapter 2: Electrochemistry Case Study Questions | chemistry Class 12 CBSE | ByteNotes.in

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A chemistry student sets up a Daniell cell using zinc and copper electrodes dipped in ZnSO₄ and CuSO₄ solutions respectively, connected by a salt bridge. The student observes that as the cell operates, the zinc electrode gradually dissolves while copper deposits on the copper electrode. The voltmeter reads 1.1 V initially. The student then connects an external battery opposing the cell potential and slowly increases the opposing voltage. At exactly 1.1 V, the voltmeter reads zero current. On increasing the voltage beyond 1.1 V, the student notices a reversal in electrode behavior — zinc begins to deposit and copper dissolves. This experiment demonstrates the concept of electrochemical and electrolytic cells.

Question 1: In the Daniell cell, which electrode acts as the anode and which as the cathode? State the sign of each.

Zinc electrode acts as the anode (negative electrode) — oxidation occurs: Zn → Zn²⁺ + 2e⁻
Copper electrode acts as the cathode (positive electrode) — reduction occurs: Cu²⁺ + 2e⁻ → Cu

Question 2: What happens when the external opposing voltage equals exactly 1.1 V? Explain the significance of this condition.

When the external opposing voltage equals 1.1 V, no current flows through the cell (I = 0)
This condition represents equilibrium — the cell EMF is exactly balanced by the external potential, so no net chemical reaction occurs

Question 3: When the external voltage exceeds 1.1 V, the cell behaves as an electrolytic cell. Explain the change in electrode behavior and write the new electrode reactions.

When Eext > 1.1 V, the cell functions as an electrolytic cell — electrical energy drives a non-spontaneous reaction
The roles reverse: copper electrode now becomes the anode (oxidation: Cu → Cu²⁺ + 2e⁻) and zinc electrode becomes the cathode (reduction: Zn²⁺ + 2e⁻ → Zn)
Electrons now flow from Cu to Zn (opposite to galvanic mode), current flows from Zn to Cu

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The standard hydrogen electrode (SHE) is used as a reference electrode to measure the standard electrode potential of other half-cells. It consists of a platinum electrode coated with platinum black, dipped in 1 M H⁺ solution with hydrogen gas bubbled at 1 bar pressure. Using SHE as the reference (E° = 0.00 V), the standard electrode potential of Cu²⁺/Cu is found to be +0.34 V and that of Zn²⁺/Zn is –0.76 V. A table of standard electrode potentials (Table 2.1) allows prediction of the feasibility of redox reactions and relative strengths of oxidizing and reducing agents.

Question 1: What do the positive and negative values of standard electrode potential indicate about the nature of the species?

A positive E° indicates the reduced form is more stable than H₂ — the species is a stronger oxidizing agent than H⁺
A negative E° indicates H₂ is more stable than the reduced form — the species is a stronger reducing agent (weaker oxidizing agent) than H⁺

Question 2: Based on E° values of Cu²⁺/Cu (+0.34 V) and Zn²⁺/Zn (–0.76 V), calculate the standard EMF of the Daniell cell and state whether Cu dissolves in HCl.

E°cell = E°cathode – E°anode = 0.34 – (–0.76) = +1.10 V
Cu does not dissolve in HCl because E° of Cu²⁺/Cu is positive, meaning Cu²⁺ is reduced more easily than H⁺; hydrogen ions cannot oxidize Cu under standard conditions

Question 3: From Table 2.1, fluorine has E° = +2.87 V and lithium has E° = –3.05 V. Interpret these values in terms of oxidizing and reducing power, and explain the trend in the table from top to bottom.

F₂ has the highest E° (+2.87 V) → it has the maximum tendency to get reduced → F₂ is the strongest oxidizing agent; F⁻ is the weakest reducing agent
Li has the lowest E° (–3.05 V) → Li⁺ is the weakest oxidizing agent; Li metal is the most powerful reducing agent in aqueous solution
As we go from top to bottom in Table 2.1, E° decreases: oxidizing power of species on the left decreases, reducing power of species on the right increases

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Nernst equation accounts for the dependence of electrode potential on ion concentrations. For the Daniell cell, the cell potential changes with the concentrations of Zn²⁺ and Cu²⁺ ions. When the circuit is closed, Zn²⁺ concentration increases and Cu²⁺ concentration decreases over time. The voltage on the voltmeter gradually decreases until equilibrium is reached, at which point E(cell) = 0. At this stage, using the Nernst equation, one can derive the relationship between the standard cell potential and the equilibrium constant of the cell reaction. At 298 K, for the Daniell cell with E°cell = 1.1 V, this equilibrium constant was calculated to be approximately 2 × 10³⁷.

Question 1: Write the Nernst equation for a general cell reaction: aA + bB → cC + dD involving n electrons, and identify each term.

E(cell) = E°(cell) – (RT/nF) ln Q, where Q = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
R = gas constant (8.314 J K⁻¹ mol⁻¹), T = temperature in Kelvin, n = moles of electrons transferred, F = Faraday constant (96487 C mol⁻¹)

Question 2: At 298 K, what simplification of the Nernst equation for the Daniell cell gives E(cell) in terms of log? How does E(cell) change with increasing Cu²⁺ concentration?

At 298 K: E(cell) = E°(cell) – (0.059/2) log([Zn²⁺]/[Cu²⁺])
As Cu²⁺ concentration increases, the log term decreases, so E(cell) increases — higher Cu²⁺ concentration gives a higher cell potential

Question 3: At equilibrium in the Daniell cell, E(cell) = 0. Using E°cell = 1.1 V and n = 2, derive the relationship between E°cell and Kc, and calculate the value of Kc at 298 K.

At equilibrium, E(cell) = 0, so: 0 = E°(cell) – (2.303RT/nF) log Kc
Therefore: E°(cell) = (2.303RT/nF) log Kc — this is the general relationship
At T = 298 K: 1.1 = (0.059/2) log Kc → log Kc = (1.1 × 2)/0.059 = 37.28 → Kc = 2 × 10³⁷

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The relationship between standard Gibbs energy (ΔrG°) and the standard EMF of a cell is fundamental to electrochemistry. The reversible electrical work done by a galvanic cell equals the decrease in Gibbs energy. This relationship allows calculation of thermodynamic quantities from measurable electrical parameters. For the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), the standard cell potential is 1.1 V and n = 2. Using Faraday's constant F = 96487 C mol⁻¹, the standard Gibbs energy of this reaction can be calculated. Furthermore, since ΔrG° = –RT ln K, these relationships connect all three quantities: E°cell, ΔrG°, and Kc.

Question 1: State the relationship between ΔrG° and E°cell. Why is E(cell) called an intensive property while ΔrG is extensive?

ΔrG° = –nFE°(cell), where n = moles of electrons, F = 96487 C mol⁻¹
E(cell) is intensive because it does not depend on how much of the reaction is written; ΔrG is extensive because it depends on n — doubling the reaction doubles ΔrG but E(cell) remains the same

Question 2: For the reaction Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), E°cell = 0.46 V. Calculate the equilibrium constant Kc at 298 K.

Using E°(cell) = (0.059/n) log Kc at 298 K, with n = 2:
0.46 = (0.059/2) log Kc → log Kc = (0.46 × 2)/0.059 = 15.6 → Kc = 3.92 × 10¹⁵

Question 3: For the Daniell cell reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), E°cell = 1.1 V. Calculate ΔrG°. If the equation is doubled, what will be the new ΔrG° and new E°cell? Explain.

ΔrG° = –nFE°(cell) = –2 × 96487 × 1.1 = –212,271 J mol⁻¹ = –212.27 kJ mol⁻¹
For doubled equation: 2Zn + 2Cu²⁺ → 2Zn²⁺ + 2Cu; n becomes 4; ΔrG° = –4 × 96487 × 1.1 = –424.54 kJ mol⁻¹ (doubled)
E°cell remains 1.1 V (unchanged) because it is an intensive property — independent of how the equation is written

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Case Set 5

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A student is studying the electrical conductance of ionic solutions in the laboratory. The student uses a conductivity cell with two platinum electrodes (platinized) to measure resistance using an AC source and a Wheatstone bridge arrangement. The student dissolves KCl in water at different concentrations and measures conductivity. It is observed that as concentration decreases, conductivity also decreases, but molar conductivity increases. For 0.1 mol L⁻¹ KCl, conductivity is 0.0129 S cm⁻¹ and molar conductivity is 129.0 S cm² mol⁻¹. The student also notes that pure water has very low conductivity (~3.5 × 10⁻⁵ S m⁻¹).

Question 1: Why is an AC source used instead of a DC source for measuring conductivity of ionic solutions?

Passing DC current through an ionic solution over a prolonged period changes the composition of the solution due to electrochemical reactions (electrolysis)
AC source prevents this composition change, ensuring accurate and stable conductivity measurements

Question 2: Define cell constant (G*). A cell filled with 0.1 mol L⁻¹ KCl (κ = 1.29 S m⁻¹) shows resistance 100 Ω. Find the cell constant.

Cell constant G* = l/A (ratio of distance between electrodes to their cross-sectional area); SI units: m⁻¹ or cm⁻¹
G* = κ × R = 1.29 S m⁻¹ × 100 Ω = 129 m⁻¹ = 1.29 cm⁻¹

Question 3: Explain why conductivity decreases but molar conductivity increases with dilution for a strong electrolyte. Write the formula for molar conductivity and relate it to conductivity.

Conductivity decreases with dilution because fewer ions are present per unit volume to carry current — dilution reduces ion concentration
Molar conductivity Λm = κ/c, where c is molar concentration (mol m⁻³)
On dilution, though κ decreases, c decreases even more rapidly, so Λm = κ/c increases overall; the increase in volume V (containing 1 mol) more than compensates for the decrease in κ

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Chapter 2: Electrochemistry | Case Study Questions

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Question 1: In the Daniell cell, which electrode acts as the anode and which as the cathode? State the sign of each.

Question 2: What happens when the external opposing voltage equals exactly 1.1 V? Explain the significance of this condition.

Question 3: When the external voltage exceeds 1.1 V, the cell behaves as an electrolytic cell. Explain the change in electrode behavior and write the new electrode reactions.

Passage

Question 1: What do the positive and negative values of standard electrode potential indicate about the nature of the species?

Question 2: Based on E° values of Cu²⁺/Cu (+0.34 V) and Zn²⁺/Zn (–0.76 V), calculate the standard EMF of the Daniell cell and state whether Cu dissolves in HCl.

Question 3: From Table 2.1, fluorine has E° = +2.87 V and lithium has E° = –3.05 V. Interpret these values in terms of oxidizing and reducing power, and explain the trend in the table from top to bottom.

Passage

Question 1: Write the Nernst equation for a general cell reaction: aA + bB → cC + dD involving n electrons, and identify each term.

Question 2: At 298 K, what simplification of the Nernst equation for the Daniell cell gives E(cell) in terms of log? How does E(cell) change with increasing Cu²⁺ concentration?

Question 3: At equilibrium in the Daniell cell, E(cell) = 0. Using E°cell = 1.1 V and n = 2, derive the relationship between E°cell and Kc, and calculate the value of Kc at 298 K.

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Question 1: State the relationship between ΔrG° and E°cell. Why is E(cell) called an intensive property while ΔrG is extensive?

Question 2: For the reaction Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), E°cell = 0.46 V. Calculate the equilibrium constant Kc at 298 K.

Question 3: For the Daniell cell reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), E°cell = 1.1 V. Calculate ΔrG°. If the equation is doubled, what will be the new ΔrG° and new E°cell? Explain.

Passage

Question 1: Why is an AC source used instead of a DC source for measuring conductivity of ionic solutions?

Question 2: Define cell constant (G*). A cell filled with 0.1 mol L⁻¹ KCl (κ = 1.29 S m⁻¹) shows resistance 100 Ω. Find the cell constant.

Question 3: Explain why conductivity decreases but molar conductivity increases with dilution for a strong electrolyte. Write the formula for molar conductivity and relate it to conductivity.

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