Problem 131 - A Complete Resource Book in Chemistry for JEE Main

Question

Three faraday of electricity is passed through aqueous solutions of \(\mathrm{AgNO}_{3}\), \(\mathrm{NiSO}_{4}\) and \(\mathrm{CrCl}_{3}\) kept in three vessels using inert electrodes. The ratio in moles in which the metals \(\mathrm{Ag}, \mathrm{Ni}\) and \(\mathrm{Cr}\) will be deposited is (a) \(1: 2: 3\) (b) \(2: 3: 6\) (c) \(6: 3: 2\) (d) \(3: 2: 6\)

Step-by-Step Solution

Answer

The ratio of moles in which the metals are deposited is \( 6: 3: 2 \). Option (c) is correct.

Step 1: Determine Faraday's Law of Electrolysis

Faraday's Law states that the amount of substance deposited at an electrode is directly proportional to the quantity of electricity that flows through the solution. The mass deposited is given by \( m = \frac{Q}{F} \times \frac{M}{n} \), where \( Q \) is the total charge, \( F \) is Faraday's constant (approximately 96500 C/mol), \( M \) is the molar mass, and \( n \) is the number of electrons exchanged per mole of substance.

Step 2: Identify Valency for Each Element

Ag, Ni, and Cr undergo reactions where they accept or give away electrons. For \( \text{Ag} \), \( n = 1 \) because \( \text{Ag}^+ \to \text{Ag} \). For \( \text{Ni} \), \( n = 2 \) because \( \text{Ni}^{2+} \to \text{Ni} \). For \( \text{Cr} \), \( n = 3 \) since \( \text{Cr}^{3+} \to \text{Cr} \).

Step 3: Calculate the Moles Deposited for Each Metal

Given 3 Faraday of electricity, calculate the moles deposited using \( \text{moles} = \frac{Q}{nF} \). For Ag, moles are \( \frac{3}{1} = 3 \). For Ni, moles are \( \frac{3}{2} = 1.5 \) or \( 3:2 \). For Cr, moles are \( \frac{3}{3} = 1 \) or \( 3:3 \), translating to \( 1:3 \).

Step 4: Find the Simplified Ratio

The calculated moles are for Ag, Ni, and Cr are \( 3: 1.5: 1 \). Simplifying, the ratio of moles becomes \( 6: 3: 2 \).

Step 5: Compare with Given Options

With the simplified ratio \( 6: 3: 2 \), compare to the options provided. The correct option is \( c) 6: 3: 2 \).

Key Concepts

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis is a fundamental principle in electrochemistry. It helps in determining how much of a certain substance is deposited or dissolved during an electrolytic process. The law states that the mass of substance altered at an electrode during electrolysis is directly proportional to the amount of electric charge passed through the circuit.

The formula to calculate the mass of a substance deposited is given by \( m = \frac{Q}{F} \times \frac{M}{n} \).

• \( Q \) represents the total electric charge in coulombs.
• \( F \) is Faraday's constant, approximately 96,500 C/mol.
• \( M \) is the molar mass of the substance.
• \( n \) indicates the number of electrons exchanged per mole of substance in the reaction.

This means if we know the amount of electricity passed and the properties of the substance involved, we can determine how much is deposited at an electrode. In our exercise, using 3 Faradays helps us calculate the proportional deposits of Ag, Ni, and Cr.

Electrode Reactions

An electrode reaction involves the transfer of electrons between the electrode and the ions in the electrolytic solution. Each metal in the solution will either gain or lose electrons, converting ions into their solid form.

For example, with silver \((\text{Ag})\), the reaction is \( \text{Ag}^+ + e^- \to \text{Ag} \). This indicates one electron is involved per mole of silver ions. In contrast, Nickel \((\text{Ni})\) undergoes \( \text{Ni}^{2+} + 2e^- \to \text{Ni} \), needing two electrons per mole. Finally, Chromium \((\text{Cr})\) requires \( \text{Cr}^{3+} + 3e^- \to \text{Cr} \).

So, each metal has a different valency, which affects the number of moles deposited for each faraday of electricity. Silver requires one electron, Nickel two, and Chromium three, illustrating different valency and molar requirements for deposition.

Electrochemical Cells

Electrochemical cells are devices that allow chemical reactions to produce electrical energy or use electrical energy to drive non-spontaneous reactions. They are composed of two electrodes (anode and cathode) and an electrolyte.

In the context of electrolysis, these cells specifically use electrical energy to cause a chemical change. This process involves the reduction and oxidation of ions at each respective electrode, leading to the deposition of metal on the cathode.

With the given exercise, each vessel with Ag, Ni, and Cr acts as an electrochemical cell, where the passage of 3 Faradays of electricity results in the deposition of metals according to their required electron exchange. This further reinforces understanding of Faraday's Law, where the ratio of deposited metals 6:3:2 corresponds to the specific electrochemical behavior exhibited by each metal in the solution.