Electrolytic Cells and Electrolysis MCQ Quiz in தமிழ் - Objective Question with Answer for Electrolytic Cells and Electrolysis - இலவச PDF ஐப் பதிவிறக்கவும்

CONCEPT:

Electrochemical Cell vs. Electrolytic Cell

• An electrochemical cell converts chemical energy into electrical energy through spontaneous redox reactions.
• An electrolytic cell requires an external electrical energy source to drive non-spontaneous redox reactions.

EXPLANATION:

• In an electrochemical cell, the cell potential (E_cell) is positive, indicating a spontaneous reaction.
• For an electrochemical cell to behave like an electrolytic cell, an external voltage (E_ext) must be applied that exceeds the cell potential (E_cell).
• This external voltage forces the cell to undergo a non-spontaneous reaction, similar to an electrolytic cell.
• The condition for this to occur is:

  E_ext > E_cell

Therefore, an electrochemical cell can behave like an electrolytic cell when the external voltage (E_ext) is greater than the cell potential (E_cell).

On electrolysis of aq. NaCl

Cathode: 

Anode: 

Net reaction :

CONCEPT:

Molar Conductance (Λm)

• Molar conductance is the conductance of a solution containing 1 mole of electrolyte and is expressed in terms of S cm² mol⁻¹.
• It is calculated using the formula:

  Λm = (1000 × κ) / C

  where:
  • Λm = Molar conductance (S cm² mol⁻¹)
  • κ = Conductivity (S cm⁻¹)
  • C = Concentration of the solution (mol L⁻¹)

EXPLANATION:

• The resistance (R) of the solution is given as 400 ohm, and the concentration (C) is 5 × 10⁻³ mol L⁻¹.
• Conductivity (κ) is related to resistance by the formula:

  κ = 1 / R

• Substitute the given resistance (R = 400 ohm):

  κ = 1 / 400 = 0.0025 S cm⁻¹

• Now calculate the molar conductance (Λm) using the formula:

  Λm = (1000 × κ) / C

  Substitute κ = 0.0025 S cm⁻¹ and C = 5 × 10⁻³ mol L⁻¹:

  Λm = (1000 × 0.0025) / (5 × 10⁻³)

  Λm = 192.5 S cm² mol⁻¹

Therefore, the molar conductance of Na2SO4 is 192.5 S cm² mol⁻¹.

Current passed,

Time for which current is passed

So, total charge flown,

Number of moles of electrons moles

Hence, option is correct.

As the reduction potential of and are lower than hydrogen ion, hydrogen ion gets preferably reduced and hydrogen gas is produced at the cathode.

Also, as oxidation potential of is higher than sulphate ion, it will oxidise first and oxygen is evolved at the anode.

Hence, a fairly concentrated electrolytic aqueous solution of and produces oxygen and hydrogen gas.

Corrosion takes place with greater rate in the presence of electrolyte because electrolyte increases concentration of which further increases the mobility of and rate of corrosion increases.

Electroplating prevents corrosion. Oxygen is a must for corrosion to happen.

So, in vacuum no corrosion takes place.

The correct answers are - option A, C, D.

Nernst equation is written as'-

for electrolytic cells,

so at

Also, in electrolytic concentration cell,

The electrode material and the solution in both half cells are composed of the same substance.

However, the concentration of solutions of the same substance is different.

So, all options are correct.

At Cathode :

At Anode :

Increase in mass of cathode = decrease in mass of Anode =

Options A and B are correct.

CONCEPT:

Electrolysis and Gas Volume Calculation

• During electrolysis, the amount of substance discharged at the electrodes is determined by Faraday’s laws.
• The number of moles of electrons passed is given by:

• The volume of gas produced at STP can be calculated using:

EXPLANATION:

• Step 1: Calculating Moles of Electrons
  • Given current I = 20 A and time t = 28950 s:
• Step 2: Determining Discharge at Anode
  • Among Br-  and Cl- , bromide has a lower discharge potential and is discharged first.
  • 2L of 2M bromide solution contains 4 moles of Br-  , which provides 4 moles of electrons to form 2 moles of Br2 .
• Step 3: Remaining Electron Contribution
  • Remaining 2 moles of electrons will be supplied by 2 moles of Cl- to form 1 mole of Cl2 .
• Step 4: Volume Calculation
  • Volume of Br₂   = 2 x 22.4 = 44.8 L
  • Volume ofCl₂ = 1 x 22.4 = 22.4 L

Correct Options: Option 1: 44.8 L Br₂ & Option 3: 22.4 L Cl₂