Conductance and Electrolysis Electrolytic and metallic conduction, conductance in electrolytic solutions, molar conductivities and their variation with concentration – Kohlrausch’s law and its applications: NEET MCQs with Answers

15 free questions on Conductance and Electrolysis Electrolytic and metallic conduction, conductance in electrolytic solutions, molar conductivities and their variation with concentration – Kohlrausch’s law and its applications for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Kohlrausch’s law and its applications.

Practise all 121 Conductance and Electrolysis Electrolytic and metallic conduction, conductance in electrolytic solutions, molar conductivities and their variation with concentration – Kohlrausch’s law and its applications questions free Take a free timed test

Question 1

Previous-year question
On electrolysis of dilute sulphuric acid using a platinum (Pt) electrode, the product obtained at the anode will be [NEET 2020]
  1. oxygen gas
  2. H₂S gas
  3. SO₂ gas
  4. hydrogen gas
Show answer
Answer: (A) oxygen gas
Explanation: Sulphate ion is not oxidised; instead water is oxidised at the anode to give O₂ gas.

Question 2

Previous-year question
The number of faradays (F) required to produce 20 g of calcium from molten CaCl₂ (atomic mass of Ca = 40 g mol⁻¹) is [NEET 2020]
  1. 2
  2. 3
  3. 4
  4. 1
Show answer
Answer: (D) 1
Explanation: 20 g Ca = 0.5 mol; Ca²⁺ + 2e⁻ → Ca requires 2 F per mole, so 0.5×2 = 1 F.

Question 3

Previous-year question
The ionic mobility of which of the following alkali metal ions is lowest when an aqueous solution of their salts is placed under an electric field? [NEET 2017]
  1. K
  2. Rb
  3. Li
  4. Na
Show answer
Answer: (C) Li
Explanation: Li⁺ has the smallest crystal radius and hence the highest charge density, so it is the most heavily hydrated and moves most slowly of all the alkali metal ions.

Question 4

Previous-year question
An aqueous solution of which compound is the best conductor of electric current? [AIPMT 2015]
  1. Acetic acid, C₂H₄O₂
  2. Hydrochloric acid, HCl
  3. Ammonia, NH₃
  4. Fructose, C₆H₁₂O₆
Show answer
Answer: (B) Hydrochloric acid, HCl
Explanation: HCl is a strong electrolyte and ionises completely, giving the largest number of ions; acetic acid and ammonia are weak electrolytes and fructose is a non-electrolyte.

Question 5

Previous-year question
The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O₂ at STP will be [AIPMT 2014]
  1. 5.4 g
  2. 10.8 g
  3. 54.9 g
  4. 108.0 g
Show answer
Answer: (D) 108.0 g
Explanation: 5600 mL O₂ = 0.25 mol, requiring 4×0.25 = 1 F. One faraday deposits one equivalent of Ag = 108 g.

Question 6

Previous-year question
When 0.1 mole MnO₄²⁻ is oxidised, the quantity of electricity required to completely oxidise MnO₄²⁻ to MnO₄⁻ is [AIPMT 2014]
  1. 96500 C
  2. 2 × 96500 C
  3. 9650 C
  4. 96.50 C
Show answer
Answer: (C) 9650 C
Explanation: Mn goes from +6 to +7, i.e. one electron per ion. Q = 0.1×1×96500 = 9650 C.

Question 7

For strong electrolytes, molar conductivity varies with concentration as Λm = Λ°m − A c^(1/2). In the same solvent and at the same temperature, which pair of electrolytes has the same value of the constant A?
  1. CaCl₂ and MgSO₄
  2. MgSO₄ and KCl
  3. NaCl and CaCl₂
  4. KCl and NaNO₃
Show answer
Answer: (D) KCl and NaNO₃
Explanation: For a given solvent and temperature, A depends on the type of electrolyte, that is, on the charges on the cation and anion it produces. NaCl is a 1-1 electrolyte, CaCl₂ is 2-1 and MgSO₄ is 2-2. All electrolytes of a particular type have the same A, and KCl and NaNO₃ are both 1-1 electrolytes.
Doing well? Create a free account to take timed tests on this topic and see your weak areas.

Question 8

An unknown resistance is usually measured on a Wheatstone bridge, but measuring the resistance of an ionic solution this way presents difficulties. Which pairing of difficulty and remedy is correct?
  1. Alternating current changes the composition of the solution; a steady direct current source is used
  2. A solution cannot be connected like a metal wire; the bridge is replaced by a voltmeter placed across the solution
  3. Direct current changes the composition of the solution; an alternating current source is used
  4. Direct current boils the solution; the bridge is run from a low-voltage dry cell
Show answer
Answer: (C) Direct current changes the composition of the solution; an alternating current source is used
Explanation: Passing direct current through an ionic solution causes electrochemical reactions at the electrodes and so changes its composition; an alternating current source avoids this. The second difficulty is that a solution cannot be connected to the bridge like a metallic wire. This is solved by a specially designed conductivity cell, which has two platinum electrodes coated with platinum black.

Question 9

A current of 1.93 A is passed for 500 s through 1.0 L of aqueous NaCl using inert electrodes, and the only products released are H₂ and Cl₂. Assuming the volume stays constant and the solution started neutral, what is the approximate pH at 298 K after electrolysis? (1 F = 96500 C mol⁻¹)
  1. 2
  2. 13
  3. 7
  4. 12
Show answer
Answer: (D) 12
Explanation: The charge passed is 1.93 × 500 = 965 C, which is 0.01 F. At the cathode, 2H₂O + 2e⁻ → H₂ + 2OH⁻, so one mole of OH⁻ is produced per faraday, giving 0.01 mol OH⁻ in 1.0 L. Chloride is removed at the anode without any H⁺ being formed, so [OH⁻] = 0.01 M. Then pOH = 2 and pH = 12.

Question 10

For the anode reactions in the electrolysis of concentrated aqueous NaCl, E° is 1.36 V for Cl₂/Cl⁻ and 1.23 V for O₂/H₂O. On E° values alone oxygen should be released, yet chlorine is the main product. What is the reason?
  1. The oxidation of water to oxygen is kinetically slow and needs an overpotential
  2. Water molecules cannot reach the anode because it is fully covered by Na⁺ ions
  3. Chloride ions have a higher reduction potential than water at every concentration
  4. Oxygen is released first but reacts completely with sodium ions at the anode
Show answer
Answer: (A) The oxidation of water to oxygen is kinetically slow and needs an overpotential
Explanation: A reaction with the more favourable electrode potential can still be the slower one at the electrode. Oxidation of water to O₂ needs an extra voltage called overpotential, so it does not proceed at the voltage expected from E°. Oxidation of Cl⁻ therefore takes over and Cl₂ is the main anode product. Na⁺ is a cation and moves towards the cathode, not the anode.

Question 11

A current of 10 A is passed through copper sulphate solution for 30 minutes in a cell working at 90% current efficiency. If the electrochemical equivalent of copper is 3.3 × 10⁻⁴ g C⁻¹, the mass of copper deposited is nearest to
  1. 5.94 g
  2. 4.81 g
  3. 2.67 g
  4. 5.35 g
Show answer
Answer: (D) 5.35 g
Explanation: Total charge passed is 10 × 1800 = 18000 C, and only 90% of it deposits metal, which is 16200 C. By Faraday's first law m = ZQ = 3.3 × 10⁻⁴ × 16200 = 5.35 g.

Question 12

Aqueous AgNO₃ is electrolysed in two cells. Cell X has a silver anode and cell Y has a platinum anode. What happens at the anode and to the Ag⁺ concentration in the bulk solution?
  1. X: silver dissolves, Ag⁺ falls; Y: NO₂ evolved, Ag⁺ stays nearly constant
  2. X: O₂ evolved, Ag⁺ falls; Y: silver dissolves, Ag⁺ stays nearly constant
  3. X: silver dissolves, Ag⁺ stays nearly constant; Y: O₂ evolved, Ag⁺ falls
  4. X and Y: O₂ evolved, Ag⁺ falls in both
Show answer
Answer: (C) X: silver dissolves, Ag⁺ stays nearly constant; Y: O₂ evolved, Ag⁺ falls
Explanation: Silver metal is oxidised more easily than water or nitrate, so a silver anode dissolves and replaces the Ag⁺ that deposits at the cathode, keeping the concentration nearly constant. A platinum anode is inert, so water is oxidised to O₂ there. Ag⁺ is not replaced in that cell and its concentration falls as silver deposits.

Question 13

An aqueous solution of Na₂SO₄ containing a little neutral indicator is electrolysed between platinum electrodes. Which statement correctly describes the electrodes?
  1. Sodium deposits at the cathode and SO₂ is evolved at the anode
  2. H₂ forms at the cathode and the solution around it turns alkaline; O₂ forms at the anode and the solution around it turns acidic
  3. H₂ forms at the cathode and the solution around it turns acidic; O₂ forms at the anode and the solution around it turns alkaline
  4. Sodium deposits at the cathode and O₂ is evolved at the anode, with no change of pH anywhere
Show answer
Answer: (B) H₂ forms at the cathode and the solution around it turns alkaline; O₂ forms at the anode and the solution around it turns acidic
Explanation: Neither Na⁺ nor SO₄²⁻ is discharged in preference to water. At the cathode 2H₂O + 2e⁻ → H₂ + 2OH⁻, which makes that region alkaline. At the anode 2H₂O → O₂ + 4H⁺ + 4e⁻, which makes that region acidic.

Question 14

Sulphuric acid is electrolysed using platinum electrodes. As the acid is changed from dilute to fairly concentrated, how does the main oxidation product at the anode change?
  1. From S₂O₈²⁻ to O₂
  2. From O₂ to SO₃
  3. From O₂ to S₂O₈²⁻
  4. From SO₂ to O₂
Show answer
Answer: (C) From O₂ to S₂O₈²⁻
Explanation: Two anode reactions compete: oxidation of water to O₂ (E° = +1.23 V) and oxidation of sulphate to peroxodisulphate (E° = +1.96 V). In dilute acid the water reaction, having the lower potential, is preferred. At higher concentration of H₂SO₄ the formation of S₂O₈²⁻ becomes the main reaction.

Question 15

An aqueous solution of CuBr₂ is electrolysed between platinum electrodes. Which pair correctly gives the main products at the cathode and the anode?
  1. Cathode: H₂; anode: Br₂
  2. Cathode: Cu; anode: O₂
  3. Cathode: H₂; anode: O₂
  4. Cathode: Cu; anode: Br₂
Show answer
Answer: (D) Cathode: Cu; anode: Br₂
Explanation: Cu²⁺ has a positive standard reduction potential, so it is reduced far more readily than water at the cathode and copper deposits. At the anode, bromide is discharged in preference to water because Br⁻ needs a lower discharge potential than OH⁻ or water. The products are therefore copper and bromine.

Want more Conductance and Electrolysis Electrolytic and metallic conduction, conductance in electrolytic solutions, molar conductivities and their variation with concentration – Kohlrausch’s law and its applications questions?

E-GYAN NIDHI has 121 questions on this topic and 63,000+ in total, all free: timed CBT tests, NEET marking and instant analysis.

Register free Free Sunday Mock Test

More topics in REDOX REACTIONS AND ELECTROCHEMISTRY

↑