JEE Physics Chemical Effects of Current MCQs Set 01

Multiple Choice Questions (MCQs) for JEE Physics: Chemical Effects of Current

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Multiple Choice Questions

Question 1: The amount of substance liberated at the electrodes during electrolysis when 1 coulomb of electricity is passed is
(a) chemical equivalent
(b) electrochemical equivalent
(c) equivalent weight
(d) one mole
Show Answer & Explanation

Answer: (b) electrochemical equivalent

Explanation:
1. By definition, the mass liberated by 1 coulomb is the electrochemical equivalent \( Z \).

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 2: For gold plating on a copper chain, the substance required in the form of solution is
(a) copper sulphate
(b) copper chloride
(c) potassium cyanide
(d) potassium aurocyanide
Show Answer & Explanation

Answer: (d) potassium aurocyanide

Explanation:
1. The electrolyte must contain gold ions, supplied by potassium aurocyanide, so that gold deposits on the chain (cathode).

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 3: On passing current in a water voltameter, the hydrogen is
(a) liberated at the anode
(b) liberated at the cathode
(c) not liberated
(d) remains in the solution
Show Answer & Explanation

Answer: (b) liberated at the cathode

Explanation:
1. \( H^+ \) ions are positive, so they move to the cathode and are liberated there as hydrogen gas.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 4: For depositing 1 g of copper in a copper voltameter by passing 2 A of current, the time required will be (for copper \( Z = 0.00033 \) g/C)
(a) approximately 20 minutes
(b) approximately 25 minutes
(c) approximately 30 minutes
(d) approximately 35 minutes
Show Answer & Explanation

Answer: (b) approximately 25 minutes

Explanation:
1. \( t = \dfrac{m}{ZI} = \dfrac{1}{0.00033\times2} = 1515 \) s \( \approx 25 \) minutes.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 5: A battery of e.m.f. 3 V and internal resistance 1.0 \( \Omega \) is connected in series with a copper voltameter. The current flowing in the circuit is 1.5 A. The resistance of the voltameter will be
(a) zero
(b) 1.0 \( \Omega \)
(c) 1.5 \( \Omega \)
(d) 2.0 \( \Omega \)
Show Answer & Explanation

Answer: (b) 1.0 \( \Omega \)

Explanation:
1. Total resistance \( = \dfrac{3}{1.5} = 2\ \Omega \).

Teacher's Note:
1. Voltameter resistance \( = 2-1 = 1.0\ \Omega \).

Question 6: According to Faraday's laws of electrolysis, the amount of decomposition is proportional to the
(a) 1/(time for which current passes)
(b) electrochemical equivalent of the substance
(c) 1/current
(d) 1/(electrochemical equivalent)
Show Answer & Explanation

Answer: (b) electrochemical equivalent of the substance

Explanation:
1. \( m = ZIt \): the mass is directly proportional to Z, the current and the time.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 7: If in a voltaic cell 5 g of zinc is consumed, how many ampere-hours do we get? (E.C.E. of Zn is \( 3.387\times10^{-7} \) kg/C)
(a) 2.05
(b) 8.2
(c) 4.1
(d) \( 5\times3.387\times10^{-7} \)
Show Answer & Explanation

Answer: (c) 4.1

Explanation:
1. \( q = \dfrac{m}{Z} = \dfrac{5\times10^{-3}}{3.387\times10^{-7}} = 1.476\times10^4 \) C.

Teacher's Note:
1. In ampere-hours: \( \dfrac{1.476\times10^4}{3600} = 4.1 \).

Question 8: The current flowing in a copper voltameter is 3.2 A. The number of copper ions (\( Cu^{2+} \)) deposited at the cathode per minute is
(a) \( 0.5\times10^{20} \)
(b) \( 1.5\times10^{20} \)
(c) \( 3\times10^{20} \)
(d) \( 6\times10^{20} \)
Show Answer & Explanation

Answer: (d) \( 6\times10^{20} \)

Explanation:
1. Charge in 1 minute \( = 3.2\times60 = 192 \) C.

Teacher's Note:
1. Each ion carries \( 2e \): \( n = \dfrac{192}{2\times1.6\times10^{-19}} = 6\times10^{20} \).

Question 9: The current flowing in a copper voltameter is 1.6 A. The number of \( Cu^{2+} \) ions deposited at the cathode per minute is
(a) \( 1.5\times10^{20} \)
(b) \( 3\times10^{20} \)
(c) \( 6\times10^{20} \)
(d) \( 1\times10^{19} \)
Show Answer & Explanation

Answer: (b) \( 3\times10^{20} \)

Explanation:
1. Charge \( = 1.6\times60 = 96 \) C.

Teacher's Note:
1. \( n = \dfrac{96}{2\times1.6\times10^{-19}} = 3\times10^{20} \).

Question 10: In a copper voltameter experiment, the current is decreased to one-fourth of the initial value but is passed for four times the earlier duration. The amount of copper deposited will be
(a) the same
(b) one-fourth the previous value
(c) four times the previous value
(d) \( \dfrac{1}{16} \)th of the previous value
Show Answer & Explanation

Answer: (a) the same

Explanation:
1. \( m\propto It \): \( \dfrac14\times4 = 1 \), so the mass is unchanged.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 11: A certain charge liberates 0.8 g of oxygen. The same charge will liberate how many grams of silver?
(a) 108 g
(b) 10.8 g
(c) 0.8 g
(d) \( \dfrac{108}{0.8} \) g
Show Answer & Explanation

Answer: (b) 10.8 g

Explanation:
1. Equivalents of oxygen \( = 0.8/8 = 0.1 \).

Teacher's Note:
1. Silver (E = 108): \( 0.1\times108 = 10.8 \) g.

Question 12: In charging a battery of a motor-car, the following effect of electric current is used
(a) magnetic
(b) heating
(c) chemical
(d) induction
Show Answer & Explanation

Answer: (c) chemical

Explanation:
1. Charging reverses the chemical reaction inside the battery, which is the chemical effect of current.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 13: Avogadro's number is \( 6\times10^{23} \) per gram mole and the electronic charge is \( 1.6\times10^{-19} \) C. The Faraday's number is
(a) \( 6\times10^{23}\times1.6\times10^{-19} \)
(b) \( \dfrac{6\times10^{23}}{1.6\times10^{-19}} \)
(c) \( \dfrac{1.6\times10^{-19}}{6\times10^{23}} \)
(d) \( 6\times10^{23}+1.6\times10^{-19} \)
Show Answer & Explanation

Answer: (a) \( 6\times10^{23}\times1.6\times10^{-19} \)

Explanation:
1. \( F = N_Ae = 6\times10^{23}\times1.6\times10^{-19} \approx 96000 \) C.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 14: In \( CuSO_4 \) solution, when an electric current equal to 2.5 faraday is passed, the gram equivalents deposited on the cathode are
(a) 1
(b) 1.5
(c) 2
(d) 2.5
Show Answer & Explanation

Answer: (d) 2.5

Explanation:
1. 1 faraday deposits 1 gram equivalent of any substance, so 2.5 faraday deposits 2.5 gram equivalents.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 15: The atomic weights of silver and copper are 108 and 64. A silver voltameter and a copper voltameter are connected in series, and when current is passed 10.8 g of silver is deposited. The mass of copper deposited will be
(a) 6.4 g
(b) 12.8 g
(c) 3.2 g
(d) 10.8 g
Show Answer & Explanation

Answer: (c) 3.2 g

Explanation:
1. Equivalents deposited are equal: silver \( 10.8/108 = 0.1 \).

Teacher's Note:
1. Copper \( E = 64/2 = 32 \): mass \( = 0.1\times32 = 3.2 \) g.

Question 16: Faraday's laws of electrolysis are related to
(a) the atomic number of the positive ion
(b) the equivalent weight of the electrolyte
(c) the atomic number of the negative ion
(d) the velocity of the positive ion
Show Answer & Explanation

Answer: (b) the equivalent weight of the electrolyte

Explanation:
1. Faraday's second law links the mass liberated to the chemical equivalent (equivalent weight).

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 17: In the process of electrolysis, the current is carried inside the electrolyte by
(a) electrons
(b) atoms
(c) positive and negative ions
(d) all of the above
Show Answer & Explanation

Answer: (c) positive and negative ions

Explanation:
1. The electrolyte conducts through the motion of positive and negative ions.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 18: The mass of ions deposited during a given interval of time in the process of electrolysis depends on
(a) the current
(b) the resistance
(c) the temperature
(d) the electric power
Show Answer & Explanation

Answer: (a) the current

Explanation:
1. \( m = ZIt \); for a given time the mass depends on the current.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 19: The amount of charge required to liberate 9 g of aluminium (atomic weight 27, valency 3) in electrolysis is (Faraday's number = 96500 C per gram equivalent)
(a) 321660 C
(b) 69500 C
(c) 289500 C
(d) 96500 C
Show Answer & Explanation

Answer: (d) 96500 C

Explanation:
1. Equivalent weight \( = 27/3 = 9 \) g, so 9 g is 1 gram equivalent, requiring 1 faraday \( = 96500 \) C.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 20: In an electroplating experiment, \( m \) g of silver is deposited when 4 A of current flows for 2 minutes. The amount (in g) of silver deposited by 6 A of current for 40 seconds will be
(a) \( 4m \)
(b) \( m/2 \)
(c) \( m/4 \)
(d) \( 2m \)
Show Answer & Explanation

Answer: (b) \( m/2 \)

Explanation:
1. \( m\propto It \): first \( 4\times120 = 480 \); second \( 6\times40 = 240 \).

Teacher's Note:
1. The second mass is \( m/2 \).

Question 21: In electrolysis, if the duration of the passage of current is doubled, the mass liberated is
(a) doubled
(b) halved
(c) increased four times
(d) unchanged
Show Answer & Explanation

Answer: (a) doubled

Explanation:
1. \( m = ZIt\propto t \) for constant current, so it doubles.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 22: A current of 16 A flows through molten NaCl for 10 minutes. The amount of metallic sodium that appears at the negative electrode would be (atomic weight of Na = 23)
(a) 0.23 g
(b) 1.15 g
(c) 2.3 g
(d) 11.5 g
Show Answer & Explanation

Answer: (c) 2.3 g

Explanation:
1. \( q = 16\times600 = 9600 \) C.

Teacher's Note:
1. \( m = \dfrac{23\times9600}{96500} = 2.29 \approx 2.3 \) g.

Question 23: The mass of a substance liberated when a charge \( q \) flows through an electrolyte is proportional to
(a) \( q \)
(b) \( 1/q \)
(c) \( q^2 \)
(d) \( 1/q^2 \)
Show Answer & Explanation

Answer: (a) \( q \)

Explanation:
1. Faraday's first law: \( m = Zq \).

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 24: A steady current of 5 A is maintained for 45 minutes. During this time it deposits 4.572 g of zinc at the cathode of a voltameter. The E.C.E. of zinc is
(a) \( 3.387\times10^{-4} \) g/C
(b) \( 3.387\times10^{-4} \) C/g
(c) \( 3.384\times10^{-3} \) g/C
(d) \( 3.394\times10^{-3} \) C/g
Show Answer & Explanation

Answer: (a) \( 3.387\times10^{-4} \) g/C

Explanation:
1. \( q = 5\times2700 = 13500 \) C.

Teacher's Note:
1. \( Z = \dfrac{4.572}{13500} = 3.387\times10^{-4} \) g/C.

Question 25: The relation between the Faraday constant \( F \), the electron charge \( e \) and Avogadro's number \( N \) is
(a) \( F = N/e \)
(b) \( F = Ne \)
(c) \( N = F^2 \)
(d) \( F = N^2e \)
Show Answer & Explanation

Answer: (b) \( F = Ne \)

Explanation:
1. One mole of electrons carries \( N \) electrons of charge \( e \): \( F = Ne \).

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 26: The electrochemical equivalent of magnesium is 0.126 mg/C. A current of 5 A is passed in a suitable solution for 1 hour. The mass of magnesium deposited will be
(a) 0.0378 g
(b) 0.227 g
(c) 0.378 g
(d) 2.27 g
Show Answer & Explanation

Answer: (d) 2.27 g

Explanation:
1. \( m = ZIt = 0.126\times10^{-3}\times5\times3600 = 2.27 \) g.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

Question 27: Two electrolytic cells containing \( CuSO_4 \) and \( AgNO_3 \) respectively are connected in series and a current is passed until 1 mg of copper is deposited in the first cell. The amount of silver deposited in the second cell is approximately (atomic weights of copper and silver are 63.57 and 107.88)
(a) 1.7 mg
(b) 3.4 mg
(c) 5.1 mg
(d) 6.8 mg
Show Answer & Explanation

Answer: (b) 3.4 mg

Explanation:
1. Same charge: \( \dfrac{m_{Ag}}{m_{Cu}} = \dfrac{107.88}{31.79} = 3.39 \).

Teacher's Note:
1. Silver \( = 3.4 \) mg.

Question 28: A current \( I \) is passed for a time \( t \) through a number of voltameters. If \( m \) is the mass of a substance deposited on an electrode and \( z \) is its electrochemical equivalent, then
(a) \( \dfrac{zIt}{m} = \) constant
(b) \( \dfrac{z}{mIt} = \) constant
(c) \( \dfrac{I}{zmt} = \) constant
(d) \( \dfrac{It}{zm} = \) constant
Show Answer & Explanation

Answer: (a) \( \dfrac{zIt}{m} = \) constant

Explanation:
1. \( m = zIt \Rightarrow \dfrac{zIt}{m} = 1 \), a constant.

Teacher's Note:
1. Faraday: \( m = Zq = Z\,I\,t \), and \( Z = E/F \) with \( F = 96500\ \text{C} \) per gram equivalent.

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