Electrochemistry forms a crucial part of Class 12 Chemistry, especially when numerical problems are involved. One of the most important ideas students must master is the relationship between electric charge and chemical change. This is where coulomb faraday conversion electrochemistry mcqs become highly relevant for board and competitive exam preparation.

In electrochemical reactions, the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electric charge passed through the electrolyte. This fundamental idea is tested repeatedly through coulomb Faraday conversion electrochemistry mcqs, which require a strong grasp of Faraday’s laws.

In coulomb Faraday conversion electrochemistry mcqs , a coulomb is the SI unit of electric charge, while a Faraday represents the charge carried by one mole of electrons. Numerically, 1 Faraday = 96,500 coulombs. Understanding this conversion is the backbone of solving coulomb Faraday conversion electrochemistry mcqs accurately and quickly.

Most numerical problems in electrolysis involve converting current and time into charge using the relation:
Charge (Q) = Current (I) × Time (t).
Once the charge is calculated in coulombs, it is converted into faradays. This step is unavoidable in coulomb faraday conversion electrochemistry mcqs, as the stoichiometry of electrochemical reactions is always expressed in faradays.

The importance of coulomb faraday conversion electrochemistry mcqs lies in their ability to test both conceptual clarity and calculation speed. Aspirants often make mistakes by forgetting the electron exchange involved in the electrode reaction, which leads to incorrect faraday calculations.

Another key concept emphasized in coulomb faraday conversion electrochemistry mcqs is the oxidation number of the ion being reduced or oxidized. For example, deposition of Al³⁺ requires three electrons, whereas Ag⁺ requires only one. This directly affects the number of faradays needed and is a frequent focus in coulomb faraday conversion electrochemistry mcqs.

Electrolysis problems involving gases such as hydrogen and oxygen also depend heavily on coulomb–faraday conversion. Since gas volumes are linked to moles, and moles are linked to faradays, aspirants must be comfortable converting between these quantities when tackling coulomb faraday conversion electrochemistry mcqs.

Efficiency-based problems are another common variation. In such cases, only a fraction of the total charge contributes to the desired reaction. These are advanced forms of coulomb faraday conversion electrochemistry mcqs, requiring careful adjustment of effective charge.

In competitive exams like JEE and NEET, coulomb faraday conversion electrochemistry mcqs often appear as multi-step numericals, combining Faraday’s laws with molar mass, equivalent weight, and gas laws. Practicing such problems builds confidence and improves accuracy.

Repeated exposure to coulomb faraday conversion electrochemistry mcqs helps aspirants recognize standard patterns, such as calculating deposited mass, time required, current needed, or volume of gas evolved. Over time, these calculations become almost formula-driven.

Teachers and examiners favor coulomb faraday conversion electrochemistry mcqs because they assess both theoretical understanding and numerical precision. This makes them a high-scoring area if prepared properly.

Coulomb Faraday Conversion in Electrochemistry MCQs :

Q1. On passing 0.5 Faraday of electricity through NaCl, the amount of Cl deposited on the cathode is
a) 35.5 g
b) 17.75 g
c) 71 g
d) 142 g
Answer: b

Q2. How many cc of oxygen will be liberated by 2A current flowing for 3 min 13s through acidulated water?
a) 11.2 cc
b) 33.6 cc
c) 44.8 cc
d) 22.4 cc
Answer: d

Q3. How much electricity in terms of Faraday is required for the reduction of 2 mole Cr₂O₇²⁻ into Cr³⁺ in acidic medium?
a) 2 F
b) 3 F
c) 6 F
d) 9 F
Answer: c

Q4. For how much time, 10 ampere current should be passed through a dilute aqueous NiSO₄ solution during electrolysis using an inert electrode, in order to get 5.85 gm Nickel? (At. mass of Ni = 58.5 gm)
a) 968 s
b) 3860 s
c) 1930 s
d) 9650 s
Answer: c

Q5. Two electrolytic cells containing molten solutions of Nickel chloride & Aluminum chloride are connected in series. If the same amount of electric current is passed through them, what will be the weight of Nickel obtained when 18 gm of Aluminum is obtained? (Al=27 gm/mol, Ni=58.5 gm/mol)
a) 58.5 gm
b) 117 gm
c) 29.25 gm
d) 5.85 gm
Answer: a

Q6. These are physical properties of an element: A. Sublimation enthalpy B. Ionisation enthalpy C. Hydration enthalpy D. Electron gain enthalpy. The total number of above properties that affect the reduction potential is ___
a) 1
b) 2
c) 3
d) 4
Answer: c

Q7. Consider the cell at 25°C: Zn | Zn²⁺(1M) || Fe³⁺, Fe²⁺ | Pt. The fraction of total iron present as Fe³⁺ at cell potential 1.500 V is X×10⁻². X is:
a) 24
b) 3
c) 2
d) 20
Answer: a

Q8. Cu(s) + 2Ag⁺(1×10⁻³ M) → Cu²⁺(0.250 M) + 2Ag(s); E°cell = 2.97 V. Ecell (nearest integer) is:
a) 3
b) 6
c) 9
d) 12
Answer: a

Q9. Cd(s) + Hg₂SO₄(s) + (9/5)H₂O(l) ⇌ CdSO₄·(9/5)H₂O(s) + 2Hg(l). E°cell = 4.315 V. If ΔH° = -825.2 kJ mol⁻¹, ΔS° (JK⁻¹) is:
a) 25
b) 30
c) 27
d) 37
Answer: a

Q10. Aluminium oxide is electrolyzed at 1000°C. To prepare 5.12 kg Al, electricity required is:
a) 5.49 × 10⁷ C
b) 5.49 × 10⁴ C
c) 1.83 × 10⁷ C
d) 5.49 × 10 C
Answer: a

Q11. Two Faraday of electricity is passed through CuSO₄. Mass of Cu deposited (At. mass = 63.5 g):
a) 63.5 g
b) 127 g
c) 31.75 g
d) 15.875 g
Answer: a

Q12. KClO₃ prepared by electrolysis: 6OH⁻ + Cl⁻ → ClO₃⁻ + 3H₂O + 6e⁻. If only 60% current utilized, time to produce 10 g KClO₃ with 2 A (nearest hour) is:
a) 10 hr
b) 11 hr
c) 12 hr
d) 6 hr
Answer: b

Q13. Acidic dichromate electrolyzed 8 min using 2A: Cr₂O₇²⁻ +14H⁺+6e⁻→2Cr³⁺+7H₂O. Cr³⁺ obtained = 0.104 g. Efficiency (%) is:
a) 50%
b) 60%
c) 70%
d) 30%
Answer: b

Q14. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Faradays needed to reduce 5 moles MnO₄⁻ is:
a) 25 F
b) 20 F
c) 50 F
d) 30 F
Answer: a

Q15. Among the following, the form of water with the lowest ionic conductance at 298 K is:
a) saline water for IV injection
b) distilled water
c) water from a well
d) sea water
Answer: b

Q16. 3 Faradays passed through molten Al₂O₃, aqueous CuSO₄, molten NaCl in separate cells. Mole ratio of Al : Cu : Na deposited is:
a) 3 : 4 : 6
b) 2 : 1 : 6
c) 3 : 2 : 1
d) 2 : 3 : 6
Answer: d

Q17. Faraday’s constant is defined as:
a) charge carried by 1 electron
b) charge carried by one mole of electrons
c) charge required to deposit one mole of substance
d) charge carried by two moles of electrons
Answer: b

Q18. A certain current liberates 0.504 g H₂ in 2 h. Mass of O₂ liberated by same current in same time is:
a) 2.0 g
b) 0.4 g
c) 4.0 g
d) 8.0 g
Answer: c

Q19. Highest electrical conductivity among following 0.1 M solutions is of:
a) difluoroacetic acid
b) fluoroacetic acid
c) chloroacetic acid
d) acetic acid
Answer: a

Q20. Aluminium oxide electrolyzed at 1000°C. To prepare 5.12 kg Al, electricity required is:
a) 5.49 × 10⁷ C
b) 5.49 × 10⁸ C
c) 1.83 × 10⁷ C
d) 5.49 × 10 C
Answer: a

Q21. Which of the following does not conduct electricity?
a) Fused NaCl
b) Solid NaCl
c) Brine solution
d) Copper
Answer: b

Q22. Time required for depositing all silver in 125 mL of 1 M AgNO₃ using 241.25 A (1F = 96500 C) is:
a) 10 s
b) 50 s
c) 1000 s
d) 100 s
Answer: b

Q23. Number of Faraday required to get 1 g atom of Mg from MgCl₂ is:
a) 0.0035 F
b) 2 F
c) 1 F
d) 0.2 F
Answer: b

Q24. How many Faradays are required to reduce 1 mol of BrO₃⁻ to Br⁻?
a) 3
b) 5
c) 6
d) 4
Answer: c

Q25. Charge required for reduction of 1 mol of MnO₄⁻ to MnO₂ is:
a) 1 F
b) 3 F
c) 5 F
d) 7 F
Answer: b

Q26. Current in Faraday required for reduction of 1 mol Cr₂O₇²⁻ to Cr³⁺ is:
a) 1 F
b) 2 F
c) 6 F
d) 4 F
Answer: c

Q27. Coulombs required for oxidation of one mole of water to dioxygen is:
a) 9.65 × 10⁴ C
b) 1.93 × 10⁵ C
c) 1.93 × 10⁶ C
d) 9.65 × 10⁶ C
Answer: b

Q28. Electrolytic refining of impure Cu (Fe, Au, Ag impurities): 140 A for 482.5 s decreases anode mass by 22.26 g and increases cathode mass by 22.011 g. % of iron is:
a) 0.98
b) 0.85
c) 0.97
d) 0.9
Answer: d

Q29. Approximate time (hours) to electroplate 30 g Ca from molten CaCl₂ using 5 A (At. mass Ca = 40) is:
a) 8
b) 16
c) 6
d) 10
Answer: a

Q30. 9.65 C current passed through fused anhydrous MgCl₂. Mg obtained is converted to Grignard reagent. Moles of Grignard reagent formed is:
a) 5 × 10⁻⁴
b) 5 × 10⁻⁵
c) 1 × 10⁻⁵
d) 1 × 10³
Answer: b

In summary, mastering coulomb Faraday conversion electrochemistry mcqs is essential for success in class 12 electrochemistry. Consistent practice of coulomb Faraday conversion electrochemistry mcqs strengthens conceptual foundations, reduces calculation errors, and ensures strong performance in Class 12 board exams as well as competitive entrance tests.