Statement I: One coulomb of electric charge deposits weight equal to the electrochemical equivalent of the substance

Statement II: One Faraday deposits one mole of the substance

Statement I is correct but Statement II is false

The correct answer is option 3. Statement I is correct but Statement II is false.

Statement I: One coulomb of electric charge deposits weight equal to the electrochemical equivalent of the substance

The electrochemical equivalent \((Z)\) of a substance is defined as the mass of the substance deposited (or liberated) by one coulomb of electric charge.

Mathematically, it is given by:

\(Z = \frac{M}{n \times F}\)

where:

\( M \) is the molar mass of the substance,

\( n \) is the number of electrons involved in the reaction,

\( F \) is Faraday's constant, approximately 96485 coulombs per mole of electrons

If one coulomb of charge is passed through the electrolytic cell, the mass \( m \) of the substance deposited is:

\(m = Z \times Q\)

where \( Q \) is the charge in coulombs. For \( Q = 1 \) coulomb:

\(m = Z \times 1 = Z\).

Thus, one coulomb of electric charge deposits a mass equal to the electrochemical equivalent of the substance. This makes Statement I correct.

Statement II: One Faraday deposits one mole of the substance

One Faraday \((F)\) of charge (approximately 96485 coulombs) corresponds to one mole of electrons. The amount of substance deposited depends on the number of electrons (\( n \)) required to deposit one mole of the substance. This is governed by Faraday's laws of electrolysis.

For a reaction where the deposition involves \( n \) electrons per ion, the relationship is:

\(1 \text{ Faraday} = 1 \text{ mole of electrons}\)

To deposit one mole of the substance, the total charge required is \( n \times F \):

\(\text{Charge} = n \times 96485 \text{ coulombs}\)

Examples:

Silver (Ag): The reaction \(Ag^+ + e^- → Ag\) involves one electron (\( n = 1 \)). Hence, one Faraday deposits one mole of silver.

Aluminum (Al): The reaction \(Al^{3+} + 3e^- → Al\) involves three electrons (\( n = 3 \)). Hence, one Faraday deposits only one-third of a mole of aluminum.

Therefore, the statement that one Faraday deposits one mole of the substance is false because it is only true if the substance involves a one-electron transfer per ion. It does not hold for substances involving multi-electron transfers.Summary:

Statement I is correct because it correctly describes the relationship between one coulomb of electric charge and the electrochemical equivalent of the substance.

Statement II is false because one Faraday deposits one mole of electrons, not necessarily one mole of the substance. The number of moles deposited depends on the specific electrochemical reaction and the number of electrons involved.

Thus, the correct answer is: Statement I is correct but Statement II is false.