Understanding current efficiency and electrolysis mcqs is extremely important for mastering numerical problems in electrochemistry. In many competitive examinations, aspirants are tested on how effectively electric charge is utilized during electrolysis. The concept behind current efficiency and electrolysis mcqs revolves around the practical application of Faraday’s laws and the difference between theoretical and actual yield of products at the electrodes.

When studying current efficiency and electrolysis mcqs, it is essential to first understand what current efficiency means. Current efficiency refers to the percentage of total electrical current that is actually used for the desired electrochemical reaction. In real systems, not all the current contributes to product formation because side reactions may occur. Therefore, in current efficiency and electrolysis mcqs, aspirants are often required to calculate actual mass deposited or gas evolved when efficiency is less than 100%.

A strong grasp of Faraday’s first and second laws is necessary to solve current efficiency and electrolysis mcqs accurately. According to Faraday’s first law, the mass of a substance deposited is directly proportional to the quantity of electricity passed. However, in practical situations highlighted in current efficiency and electrolysis mcqs, this proportionality must be adjusted using the efficiency factor.

In numerical problems based on current efficiency and electrolysis mcqs, the general formula used is:

Actual mass = (Theoretical mass × Efficiency) / 100

This adjustment becomes crucial when calculating the mass of metals like copper, silver, aluminum, or the volume of gases such as hydrogen and oxygen. Many current efficiency and electrolysis mcqs focus on comparing theoretical deposition with actual deposition under given conditions.

Another key idea explored in current efficiency and electrolysis mcqs is the concept of competing electrode reactions. For example, during electrolysis of aqueous solutions, hydrogen evolution may compete with metal deposition. In such cases, part of the current is wasted in hydrogen formation, reducing the efficiency of metal deposition. Questions framed under current efficiency and electrolysis mcqs frequently test this conceptual clarity.

Temperature, electrode material, electrolyte concentration, and overvoltage also influence outcomes in current efficiency and electrolysis mcqs. For instance, high overpotential may suppress unwanted reactions and improve efficiency. Understanding these parameters allows aspirants to approach current efficiency and electrolysis mcqs with better analytical reasoning.

Industrial applications also form an important context for current efficiency and electrolysis mcqs. In processes like electrorefining of copper, electroplating, and extraction of aluminum through the Hall–Héroult process, efficiency plays a crucial role in cost and energy consumption. Therefore, many current efficiency and electrolysis mcqs are framed around industrial scenarios.

Aspirants preparing for board exams and competitive tests should practice a wide variety of current efficiency and electrolysis mcqs to improve speed and accuracy. Often, the trick lies in identifying the number of electrons involved in the reaction and correctly applying Faraday’s constant. A small calculation mistake can lead to incorrect answers in current efficiency and electrolysis mcqs.

Conceptual understanding is equally important while solving current efficiency and electrolysis mcqs. Rather than memorizing formulas, aspirants should understand why efficiency is less than 100% and how multiple reactions share the total current. This deeper insight helps solve higher-level current efficiency and electrolysis mcqs confidently.

In advanced-level current efficiency and electrolysis mcqs, problems may involve series-connected electrolytic cells, percentage efficiency comparisons, or multi-step redox processes. Such problems demand clarity in mole–electron relationships and proportional reasoning.

Current Efficiency and Electrolysis MCQs with Answers

1.  In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under STP condition. The current to be passed is
   a) 1.93 A
   b) 9.65 A
   c) 19.3 A
   d) 0.965 A
   Answer: b

2. How many moles of platinum will be deposited on the cathode when 0.40 F of electricity is passed through a 1.0 M solution of Pt⁴⁺?
   a) 0.40 mol
   b) 1.0 mol
   c) 0.10 mol
   d) 0.45 mol
   Answer: c

3. 1 C electricity deposits:
   a) half of electrochemical equivalent of Ag
   b) electrochemical equivalent of Ag
   c) 96500 g of Ag
   d) 10.8 g of Ag
   Answer: b

4. The number of moles of electrons passed when a current of 2A is passed through a solution of electrolyte for 20 minutes is
   a) 4.1 × 10⁻² mol e⁻
   b) 1.24 × 10⁻² mol e⁻
   c) 2.487 × 10⁻¹ mol e⁻
   d) 2.487 × 10⁻² mol e⁻
   Answer: d

5. How many Faradays of electricity are required to deposit 10 g of calcium from molten calcium chloride using inert electrodes? (Molar mass of calcium = 40 g mol⁻¹)
   a) 0.5 F
   b) 1 F
   c) 0.25 F
   d) 2 F
   Answer: a

6. According to Faraday’s first law
   a) w=96500×EI×tw = \frac {96500 \times E}{I \times t}w=I×t96500×E​
   b) w=I×E×t96500w = \frac {I \times E \times t}{96500}w=96500I×E×t​
   c) E=I×96500×twE = \frac {I \times 96500 \times t}{w}E=wI×96500×t​
   d) E=I×wt×96500E = \frac {I \times w}{t \times 96500}E=t×96500I×w​
   Answer: b

7. The amount of silver deposited on passing 2 F of electricity through aqueous solution of AgNO₃
   a) 54 g
   b) 108 g
   c) 216 g
   d) 324 g
   Answer: c

8. The number of Faradays (F) required to produce 20g of calcium from molten CaCl₂ is (Atomic mass of Ca = 40 mol⁻¹)
   a) 1
   b) 2
   c) 3
   d) 4
   Answer: a

9. During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is
   a) 55 minutes
   b) 110 minutes
   c) 220 minutes
   d) 330 minutes
   Answer: b

10. The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60 × 10⁻¹⁹ C)
    a) 6 × 10²⁰
    b) 6 × 10¹⁸
    c) 3.75 × 10²⁰
    d) 7.48 × 10²³
    Answer: c

11. When 0.1 mol MnO₄²⁻ is oxidized, the quantity of electricity required to completely oxidize MnO₄²⁻ to MnO₄⁻ is
    a) 96500 C
    b) 2 × 96500 C
    c) 96.5 C
    d) 9650 C
    Answer: d

12. The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O₂ at STP will be
    a) 54.0 g
    b) 10.88 g
    c) 54.0 g
    d) 108.0 g
    Answer: d

13. How many grams of cobalt metal will be deposited when a solution of cobalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes? (1 Faraday = 96,500 C; Atomic mass of Co = 59 u)
    a) 4.0 g
    b) 20.0 g
    c) 6.0 g
    d) 0.66 g
    Answer: b

14. Al₂O₃ is reduced by electrolysis at low potentials and high currents. If 4.0 × 10⁴ amperes of current is passed through molten Al₂O₃ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, Atomic mass of Al = 27 g/mol)
    a) 8.1 × 10⁴ g
    b) 2.4 × 10⁴ g
    c) 9.0 × 10⁴ g
    d) 8.1 × 10³ g
    Answer: a

15. For the cell reaction: 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq), Ecell0E^0_{cell}Ecell0​ = 0.24 V at 298 K. The standard Gibbs energy change is
    a) -23.16 kJ mol⁻¹
    b) 46.32 kJ mol⁻¹
    c) 23.16 kJ mol⁻¹
    d) -46.32 kJ mol⁻¹
    Answer: d

16. The equilibrium constant of the reaction: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), Ecell0E^0_{cell}Ecell0​ = 0.46V at 289 K is
    a) 2 × 10¹⁸
    b) 2 × 10¹⁶
    c) 4 × 10¹⁸
    d) 4 × 10¹⁵
    Answer: d

17. In acidic medium MnO₄⁻ is converted to Mn²⁺. The quantity of electricity in faraday required to reduce 0.5 mole of MnO₄⁻ to Mn²⁺ would be
    a) 0.25 F
    b) 0.5 F
    c) 1 F
    d) 2.5 F
    Answer: d

18. In electrochemical reaction of which set of reactants, the metal displacement will not take place?
    a) Mg + Cu²⁺
    b) Pb + Ag⁺
    c) Zn + Cu²⁺
    d) Cu + Mg²⁺
    Answer: d

19. One faraday of current was passed through electrolytic cells in series containing Ag⁺, Ni²⁺ and Cr³⁺. Ratio of amounts of Ag, Ni and Cr deposited (At. wt. Ag=108, Ni=59, Cr=52)
    a) 108 : 29.5 : 17.3
    b) 17.4 : 29.5 : 108
    c) 1 : 2 : 3
    d) 108 : 59 : 52
    Answer: a

20. On passing ‘C’ ampere current for time ‘t’ sec through 1 L of 2(M) CuSO₄ solution, the amount ‘m’ of Cu deposited is
    a) m=Ct63.5×96500m = \frac{Ct}{63.5 \times 96500}m=63.5×96500Ct​
    b) m=Ct31.25×96500m = \frac{Ct}{31.25 \times 96500}m=31.25×96500Ct​
    c) m=C×9650031.25×tm = \frac{C \times 96500}{31.25 \times t}m=31.25×tC×96500​
    d) m=31.75×C×t96500m = \frac{31.75 \times C \times t}{96500}m=9650031.75×C×t​
    Answer: d

21. The quantity of electricity needed to separately electrolyze 1 M solution of ZnSO₄, AlCl₃, and AgNO₃ completely is in the ratio of
    a) 2 : 3 : 1
    b) 2 : 1 : 1
    c) 2 : 1 : 3
    d) 2 : 2 : 1
    Answer: a

22. What amount of electricity can deposit 1 mole of Al metal at cathode when passed through molten AlCl₃?
    a) 0.3F
    b) 1F
    c) 3F
    d) 1/3F
    Answer: c

23. Assertion: One mole of silver deposits by 1 Faraday charge.
    Reason: Faraday charge required depends upon number of electrons.
    a) Both correct and Reason explains Assertion
    b) Both correct but Reason doesn’t explain
    c) Assertion correct but Reason incorrect
    d) Both incorrect
    Answer: a

24. Assertion: Galvanised iron does not rust
    Reason: Zinc has a more negative electrode potential than iron.
    a) Both correct and Reason explains Assertion
    b) Both correct but Reason doesn’t explain
    c) Assertion correct but Reason incorrect
    d) Both incorrect
    Answer: a

25. Time taken to completely (in hrs) decompose 36 g water by passing 3A current is:
    a) 35.8 hrs
    b) 40 hrs
    c) 51.8 hrs
    d) 22.5 hrs
    Answer: a

26. The cell potential for Zn | Zn²⁺(aq) | Snˣ⁺ | Sn is 0.801 V at 298 K. Reaction quotient is 10⁻¹². Number of electrons involved is:
    a) 1
    b) 2
    c) 3
    d) 4
    Answer: d

27. The resistance of 0.01 N solution is 220 Ω at 298 K using a conductivity cell with cell constant 0.88 cm⁻¹. Equivalent conductance is:
    a) 400 ohm cm² g eq⁻¹
    b) 295 ohm cm² g eq⁻¹
    c) 419 ohm cm² g eq⁻¹
    d) 425 ohm cm² g eq⁻¹
    Answer: a

28. The EMF of the cell Mg | Mg²⁺(0.01M) || Sn²⁺(0.1M) | Sn at 298K is (E°Mg²⁺/Mg=-2.34V, E°Sn²⁺/Sn=-0.14V)
    a) 2.17 V
    b) 2.5 V
    c) 2.45 V
    d) 2.23 V
    Answer: d

29. In the reversible reaction 2NO₂ ⇌ N₂O₄, the rate of disappearance of NO₂ is equal to:
    a) 2k1k2[NO2]2\frac{2k_1}{k_2}[NO_2]^2k2​2k1​​[NO2​]2
    b) 2k₁[NO₂]² − 2k₂[N₂O₄]
    c) 2k₁[NO₂]² − k₂[N₂O₄]
    d) (2k₁ − k₂)[NO₂]
    Answer: c

30. For the cell: Ag | Ag⁺ | AgCl | Cl⁻ | Cl₂,Pt
    ΔG°f(AgCl)=-109 kJ/mol, ΔG°f(Cl⁻)=-129 kJ/mol, ΔG°f(Ag⁺)=78 kJ/mol. E° of the cell is
    a) -0.60 V
    b) 6 V
    c) None of these
    d) 0.60 V
    Answer: a

Ultimately, mastering current efficiency and electrolysis mcqs strengthens a aspirant’s command over electrochemistry numericals. Since efficiency calculations are common in exam patterns, regular practice of current efficiency and electrolysis mcqs ensures conceptual clarity and computational precision. With systematic revision and focused problem-solving, aspirants can confidently tackle any question related to current efficiency and electrolysis mcqs and improve their performance significantly.