Chapter 2: Electrochemistry Application Questions | chemistry Class 12 CBSE

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A student wants to electroplate a copper ornament with silver. She sets up a cell with a silver anode, copper cathode, and silver nitrate solution as electrolyte. Explain the electrode processes and calculate the mass of silver deposited if 2 amperes current is passed for 30 minutes. (Atomic mass of Ag = 108 g mol−1, F = 96500 C mol−1)

At cathode (copper ornament): Ag+(aq) + e− → Ag(s); silver ions are reduced and deposited on the copper surface. At anode (silver): Ag(s) → Ag+(aq) + e−; silver dissolves, replenishing Ag+ ions in solution.
Charge passed Q = I × t = 2 A × 1800 s = 3600 C. For Ag+ + e− → Ag, 1F (96500 C) deposits 108 g Ag.
Mass of Ag deposited = (108 × 3600) / 96500 = 4.03 g. The silver anode ensures the Ag+ concentration in solution remains constant, making this process efficient for electroplating.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A zinc pot is suggested for storing copper sulphate solution in a chemistry laboratory. Using standard electrode potentials (E°Zn2+/Zn = −0.76 V, E°Cu2+/Cu = +0.34 V), evaluate whether this is safe and explain the chemistry.

It is not safe. Since E°Cu2+/Cu (+0.34 V) > E°Zn2+/Zn (−0.76 V), Cu2+ ions have a greater tendency to get reduced than Zn2+. Therefore, Cu2+ ions will oxidise Zn metal: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), E°cell = 0.34 − (−0.76) = 1.10 V (positive, spontaneous).
The zinc pot will dissolve over time, the CuSO4 solution will become contaminated with ZnSO4, and copper will deposit on the zinc surface. This makes storage in a zinc pot impractical and hazardous.
A suitable container would be made of a material with a more positive electrode potential than Cu2+/Cu, such as glass or plastic, which would not undergo any reaction with CuSO4 solution.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

With reference to the production of aluminium industrially, explain why electrolysis must be carried out on molten aluminium oxide (in the presence of cryolite) rather than aqueous Al3+ solution.

Al3+ has E° = −1.66 V. In aqueous solution, H2O is also present, and H+ has E° = 0.00 V >> −1.66 V. Therefore, H+ would be preferentially reduced at the cathode instead of Al3+, producing H2 gas rather than Al metal.
Electrolysis of molten Al2O3 (in cryolite to lower the melting point and improve conductivity) eliminates water and H+ ions. Al3+ is the only cation and is reduced at the cathode: Al3+ + 3e− → Al.
Cryolite (Na3AlF6) serves as a solvent for Al2O3, reducing the melting point from ~2000°C to ~950°C, making the industrial process energy-efficient. Similarly, Na and Mg are produced from their molten chlorides.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

A galvanic cell is set up: Cu(s) | Cu2+(0.1 M) || Ag+(0.01 M) | Ag(s). Given E°cell = 0.46 V, calculate E(cell) using the Nernst equation at 298 K. (n = 2)

The cell reaction is Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s). The Nernst equation is: E(cell) = E°(cell) − (0.059/n) log Q.
Q = [Cu2+]/[Ag+]2 = 0.1/(0.01)2 = 0.1/10−4 = 1000. Then E(cell) = 0.46 − (0.059/2) log 1000 = 0.46 − (0.0295 × 3) = 0.46 − 0.0885 = 0.3715 V.
The cell potential decreases from 0.46 V (standard) to 0.37 V at these non-standard concentrations, showing that lower [Ag+] and higher [Cu2+] shift the reaction quotient Q upward, reducing the driving force of the cell.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

A marine engineer notices that the hull of a steel ship corrodes rapidly near the propeller (made of a different metal alloy). Explain the electrochemical basis of this observation and suggest how to prevent it.

Different metals in electrical contact in seawater (an electrolyte) form a galvanic cell. The more reactive metal (lower E°) acts as the anode and is oxidised (corrodes), while the less reactive metal acts as the cathode and is protected. The higher the potential difference between the metals, the faster the corrosion.
Prevention: Cathodic protection (sacrificial anode) — attach a block of zinc or magnesium (more reactive metals) to the ship's hull. These metals have lower E° and will act as the anode, corroding preferentially while the steel hull (cathode) is protected.
Other methods include coating the hull with paint or anti-corrosion chemicals, or using impressed current cathodic protection (an external current source makes the hull cathodic). Replacing dissimilar metals with similar alloys also reduces galvanic corrosion.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

In a conductance experiment, a conductivity cell filled with 0.1 mol L−1 KCl (conductivity = 1.29 S m−1) shows resistance 100 Ω. When filled with an unknown electrolyte, the resistance is 520 Ω. Calculate the conductivity and molar conductivity if the electrolyte concentration is 0.02 mol L−1.

Cell constant G* = conductivity × resistance = 1.29 S m−1 × 100 Ω = 129 m−1 = 1.29 cm−1.
Conductivity of unknown solution κ = G*/R = 129 m−1 / 520 Ω = 0.248 S m−1.
Molar conductivity Λm = κ/c = 0.248 S m−1 / (0.02 × 1000 mol m−3) = 0.248/20 = 124 × 10−4 S m2 mol−1 = 124 S cm2 mol−1.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A chemist needs to determine whether Ag+(aq) can oxidise Fe2+(aq) to Fe3+(aq) under standard conditions. Using E°(Fe3+/Fe2+) = 0.77 V and E°(Ag+/Ag) = 0.80 V, carry out the analysis.

If Ag+ oxidises Fe2+: Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s). Here Ag+ is reduced (cathode, E°cathode = +0.80 V) and Fe2+ is oxidised (anode, E°anode = +0.77 V).
E°cell = E°cathode − E°anode = 0.80 − 0.77 = +0.03 V. Since E°cell > 0, ΔrG° = −nFE° < 0, the reaction is feasible and Ag+ can indeed oxidise Fe2+ under standard conditions.
ΔrG° = −1 × 96487 × 0.03 = −2894.6 J mol−1 ≈ −2.89 kJ mol−1. Although feasible, the small positive E°cell indicates only a modest thermodynamic driving force; the equilibrium constant would be moderate: log Kc = 0.03/0.059 = 0.508, Kc ≈ 3.2.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

The molar conductivity of 0.025 mol L−1 methanoic acid (HCOOH) is 46.1 S cm2 mol−1. Given λ°(H+) = 349.6 and λ°(HCOO−) = 54.6 S cm2 mol−1, calculate the degree of dissociation and dissociation constant.

Λ°m(HCOOH) = λ°H+ + λ°HCOO− = 349.6 + 54.6 = 404.2 S cm2 mol−1. Degree of dissociation α = Λm/Λ°m = 46.1/404.2 = 0.114 (11.4%).
Dissociation constant Ka = cα2/(1 − α) = (0.025 × 0.1142) / (1 − 0.114) = (0.025 × 0.013) / 0.886 = 3.67 × 10−4 mol L−1.
This shows methanoic acid is a stronger acid than acetic acid (Ka ≈ 1.8 × 10−5) at this concentration. Only 11.4% dissociation at 0.025 M confirms it is a weak acid, consistent with its classification.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

During electrolysis of an aqueous AgNO3 solution (a) with silver electrodes and (b) with platinum electrodes, what products are formed at each electrode? Explain the differences.

(a) With silver electrodes: At cathode (Ag): Ag+(aq) + e− → Ag(s); silver is deposited. At anode (reactive Ag): Ag(s) → Ag+(aq) + e−; silver dissolves. The Ag+ concentration remains constant — this is the basis of silver electroplating.
(b) With platinum electrodes (inert): At cathode (Pt): Ag+(aq) + e− → Ag(s); silver is deposited. At anode (inert Pt): 2H2O(l) → O2(g) + 4H+(aq) + 4e−; water is oxidised, releasing O2 gas.
The key difference: reactive Ag anode replenishes Ag+ ions, maintaining constant concentration. Inert Pt anode oxidises water, decreasing AgNO3 concentration over time and producing O2. Inert electrodes do not participate in reactions, allowing other species in solution to be oxidised.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

A student claims that increasing the concentration of Zn2+ ions in the Daniell cell increases the cell potential. Using the Nernst equation, evaluate this claim.

The Nernst equation for the Daniell cell is: E(cell) = E°(cell) − (0.059/2) log([Zn2+]/[Cu2+]). The reaction quotient Q = [Zn2+]/[Cu2+] appears in the log term.
Increasing [Zn2+] increases Q = [Zn2+]/[Cu2+], which increases the log term, making the subtracted term larger and thus decreasing E(cell). The student's claim is incorrect.
Cell potential increases when [Zn2+] decreases or [Cu2+] increases. Since Zn2+ is the product of the cell reaction, its increasing concentration shifts the equilibrium backward (Le Chatelier's principle), reducing the cell EMF. This is consistent with the cell approaching equilibrium.

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Chapter 2: Electrochemistry | Application Questions

A zinc pot is suggested for storing copper sulphate solution in a chemistry laboratory. Using standard electrode potentials (E°Zn2+/Zn = −0.76 V, E°Cu2+/Cu = +0.34 V), evaluate whether this is safe and explain the chemistry.

With reference to the production of aluminium industrially, explain why electrolysis must be carried out on molten aluminium oxide (in the presence of cryolite) rather than aqueous Al3+ solution.

A galvanic cell is set up: Cu(s) | Cu2+(0.1 M) || Ag+(0.01 M) | Ag(s). Given E°cell = 0.46 V, calculate E(cell) using the Nernst equation at 298 K. (n = 2)

A marine engineer notices that the hull of a steel ship corrodes rapidly near the propeller (made of a different metal alloy). Explain the electrochemical basis of this observation and suggest how to prevent it.

A chemist needs to determine whether Ag+(aq) can oxidise Fe2+(aq) to Fe3+(aq) under standard conditions. Using E°(Fe3+/Fe2+) = 0.77 V and E°(Ag+/Ag) = 0.80 V, carry out the analysis.

The molar conductivity of 0.025 mol L−1 methanoic acid (HCOOH) is 46.1 S cm2 mol−1. Given λ°(H+) = 349.6 and λ°(HCOO−) = 54.6 S cm2 mol−1, calculate the degree of dissociation and dissociation constant.

During electrolysis of an aqueous AgNO3 solution (a) with silver electrodes and (b) with platinum electrodes, what products are formed at each electrode? Explain the differences.

A student claims that increasing the concentration of Zn2+ ions in the Daniell cell increases the cell potential. Using the Nernst equation, evaluate this claim.

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