Electrochemistry — Previous Year Questions (Class 12 Chemistry)
Electrochemistry covers galvanic cells, electrode potentials, Faraday's laws, and electrolysis. CBSE tests cell reactions, Nernst equation, and calculation
TL;DR: Electrochemistry covers galvanic cells, electrode potentials, Faraday's laws, and electrolysis. CBSE tests cell reactions, Nernst equation, and calcul…
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Electrochemistry covers galvanic cells, electrode potentials, Faraday's laws, and electrolysis. CBSE tests cell reactions, Nernst equation, and calculation
Electrochemistry — Previous Year Questions with Solutions
Q (2023, 3 marks): Define a galvanic cell and explain how a Daniell cell works with a cell diagram.
Answer: Galvanic Cell (Voltaic Cell): A spontaneous redox reaction is used to generate electrical energy.
Daniell Cell:
Composition:
Anode (oxidation): Zn(s) → Zn^2+(aq) + 2e^-
Cathode (reduction): Cu^2+(aq) + 2e^- → Cu(s)
Cell diagram: Zn(s) | Zn^2+(aq) || Cu^2+(aq) | Cu(s)
Electrolyte: ZnSO4 (anode compartment) and CuSO4 (cathode compartment)
Salt bridge: Contains KNO3 or KCl to complete the circuit and maintain charge neutrality.
Working:
1. Zn is oxidized (loses electrons) at the anode, creating a negative terminal.
2. Cu^2+ is reduced (gains electrons) at the cathode, creating a positive terminal.
3. Electrons flow from Zn to Cu through external circuit.
4. Anions from salt bridge move to anode compartment, cations move to cathode compartment.
5. Cell potential E = E_cathode - E_anode = 0.34 - (-0.76) = 1.10 V
Final Answer: Galvanic cell converts chemical energy to electrical energy. Daniell cell has Zn anode and Cu cathode with E_cell = 1.10 V.
Q (2022, 2 marks): How much copper will be deposited when a current of 2 A is passed through a CuSO4 solution for 10 minutes?
Answer: Current I = 2 A, Time t = 10 min = 600 s
Reaction: Cu^2+ + 2e^- → Cu
Charge Q = I × t = 2 × 600 = 1200 coulombs
Faraday's constant F = 96500 C/mol
Electrons transferred = Q / F = 1200 / 96500 ≈ 0.01244 mol
From reaction, 2 electrons deposit 1 Cu atom:
Moles of Cu = 0.01244 / 2 = 0.00622 mol
Mass of Cu = 0.00622 × 64 = 0.398 g ≈ 0.4 g
Final Answer: Mass of Cu deposited ≈ 0.4 g or 398 mg
Q (2021, 3 marks): State and mathematically represent Faraday's laws of electrolysis.
Answer: Faraday's First Law:
The amount of substance deposited (or dissolved) is directly proportional to the charge passed.
m = Z × Q or m = Z × I × t
Where m = mass, Z = electrochemical equivalent, Q = charge, I = current, t = time.
Alternatively: n = Q / (n × F)
Where n = moles of substance, n = number of electrons, F = Faraday constant (96500 C/mol)
Faraday's Second Law:
When the same charge is passed through different electrolytes, the amounts of different substances deposited are in the ratio of their molar masses divided by the number of electrons involved.
m1 : m2 = (M1/n1) : (M2/n2)
Where M = molar mass, n = number of electrons transferred.
Example:
For Cu^2+ (M = 64, n = 2) and Ag^+ (M = 108, n = 1):
Ratio = (64/2) : (108/1) = 32 : 108
So same charge deposits 1 part Cu but 3.375 parts Ag.
Final Answer: m ∝ Q; same Q deposits different amounts proportional to M/n.
Q (2023, 2 marks): Calculate the standard cell potential for the reaction: Zn + Cu^2+ → Zn^2+ + Cu. (Given: E°(Zn^2+/Zn) = -0.76 V, E°(Cu^2+/Cu) = +0.34 V)
Answer: Standard cell potential: E°_cell = E°_cathode - E°_anode
In the reaction Zn + Cu^2+ → Zn^2+ + Cu:
Zn is oxidized (anode): E°_anode = -0.76 V
Cu^2+ is reduced (cathode): E°_cathode = +0.34 V
E°_cell = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V
Final Answer: E°_cell = 1.10 V (positive, so reaction is spontaneous)
Q (2022, 2 marks): What is passivation? Give an example.
Answer: Passivation: A highly reactive metal becomes unreactive (passive) due to the formation of a protective oxide layer on its surface.
Example:
Iron becomes passive in concentrated nitric acid (HNO3):
When Iron metal is dipped in concentrated HNO3, a thin layer of Fe3O4 (or Fe2O3) forms on the surface.
This oxide layer prevents further reaction of Fe with HNO3, so Fe remains unreactive (passive).
Another example: Aluminum (Al) is passivated in concentrated HNO3, forming a protective Al2O3 layer.
Importance:
Passivation prevents corrosion of metals in certain environments.
It is the basis for stainless steel protection (formed by alloying steel with chromium, which forms a passive Cr2O3 layer).
Final Answer: Passivation = formation of protective oxide layer making metal unreactive. Example: Fe in conc. HNO3 forms Fe3O4 layer.
Q (2021, 3 marks): Explain the difference between electrochemical cell and electrolytic cell.
Answer: Electrochemical (Galvanic) Cell:
1. Uses spontaneous redox reaction to generate electrical energy.
2. Anode is negative (-), cathode is positive (+).
3. Oxidation occurs at anode, reduction at cathode.
4. Electrons flow from anode to cathode through external circuit.
5. Cell potential is positive (E_cell > 0).
6. Ions flow from anode to cathode inside the cell.
7. Example: Daniell cell, battery.
Electrolytic Cell:
1. Requires external electrical energy to drive non-spontaneous redox reaction.
2. Anode is positive (+), cathode is negative (-).
3. Oxidation occurs at anode, reduction at cathode.
4. Electrons flow from cathode to anode through external source.
5. Cell potential is negative (E_cell < 0); external EMF > E_cell required.
6. Ions flow from anode to cathode inside the cell.
7. Example: Electroplating, electrolysis of water, CuSO4 solution.
Key difference: Galvanic generates electricity; electrolytic consumes electricity.
Final Answer: Galvanic cell is spontaneous (E > 0) and generates electricity. Electrolytic cell is non-spontaneous (E < 0) and requires external electrical energy.
Frequently Asked Questions
What is the Nernst equation and when is it used?
Nernst equation: E_cell = E°_cell - (RT/nF) ln Q. It gives the cell potential under non-standard conditions. Used when ion concentrations are not 1 M (standard state) or temperature is not 25°C. Q is the reaction quotient.
Why is salt bridge used in a galvanic cell?
Salt bridge allows ionic current to flow between anode and cathode compartments, maintaining electrical neutrality. Without it, the anode compartment would become positively charged and the cathode negatively charged, stopping the reaction.
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