Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

If the aqueous solutions of the following salts are electrolysed for 1 hour with 10 ampere current, which solution will deposit the maximum mass of the metal at the cathode? The atomic weights are \(\mathrm{Fe}=56, \mathrm{Zn}=\) \(65, \mathrm{Ag}=108, \mathrm{Hf}=178\) and \(\mathrm{W}=184\) (a) \(\mathrm{ZnSO}_{4}\) (b) \(\mathrm{FeCl}_{3}\) (c) \(\mathrm{HfCl}_{4}\) (d) \(\mathrm{AgNO}_{3}\)

Step-by-Step Solution

Answer

Silver from AgNO₃ will deposit the maximum mass at the cathode, about 40.33 grams.

Step 1: Understand Faraday's Laws of Electrolysis

Faraday's first law states that the mass of a substance deposited at an electrode during electrolysis is proportional to the total electric charge passed through the substance. It's given by \[\text{mass (m)} = \frac{Q \times M}{n \times F}\] where \(Q\) is the total electric charge, \(M\) is the molar mass of the element, \(n\) is the number of electrons in the reduction half-equation of the metal ion, and \(F\) is Faraday's constant (approximately 96500 C/mol).

Step 2: Calculate the Total Charge Q

The charge \(Q\) passed through each solution is calculated using \(Q = I \times t\), where \(I\) is the current and \(t\) is the time. For a 10 A current over 1 hour, \[Q = 10 \times 3600 = 36000 \text{ C}\] because 1 hour is 3600 seconds.

Step 3: Determine Molar Mass and Electrons per Metal

Note the molar masses (\(M\)) and the number of electrons (\(n\)) needed to deposit each metal: (a) Zn: \(\mathrm{ZnSO}_{4}\), \(M = 65\), \(n = 2\) (b) Fe: \(\mathrm{FeCl}_{3}\), \(M = 56\), \(n = 3\) (c) Hf: \(\mathrm{HfCl}_{4}\), \(M = 178\), \(n = 4\) (d) Ag: \(\mathrm{AgNO}_{3}\), \(M = 108\), \(n = 1\).

Step 4: Compute Deposited Mass for Each Metal

Use the formula from Step 1 to compute the mass for each option: (a) \(\text{mass}_{\text{Zn}} = \frac{36000 \times 65}{2 \times 96500} \approx 12.10 \text{ g}\) (b) \(\text{mass}_{\text{Fe}} = \frac{36000 \times 56}{3 \times 96500} \approx 6.96 \text{ g}\) (c) \(\text{mass}_{\text{Hf}} = \frac{36000 \times 178}{4 \times 96500} \approx 16.61 \text{ g}\) (d) \(\text{mass}_{\text{Ag}} = \frac{36000 \times 108}{1 \times 96500} \approx 40.33 \text{ g}\).

Step 5: Determine Maximum Deposited Mass

After calculating the masses, we see that silver (Ag) has the highest deposited mass at the cathode: approximately 40.33 grams.

Key Concepts

Electrolysis

Electrolysis is a fascinating chemical process where electrical energy is used to drive a non-spontaneous chemical reaction. In simpler terms, it helps in breaking down compounds into their components using an electric current. This process takes place in an electrolytic cell, where the electrolyte solution contains ions that move towards electrodes to undergo chemical reactions. These ions are crucial because, without them, the current cannot flow through the solution.
When electricity passes through the solution, it causes different reactions at each electrode. The positive ions move to the cathode (negative electrode) to gain electrons and become neutral, while negative ions move to the anode (positive electrode), where they lose electrons. This movement and exchange of electrons is how we end up depositing metal at one of the electrodes, typically the cathode.

Molar Mass Calculation

Molar mass is an important concept in chemistry because it allows us to convert between the mass of a substance and the number of moles, thereby making it possible to analyze reactions at a molecular level. The molar mass of an element is simply the mass of one mole of atoms of a given element, expressed in grams per mole (g/mol). Molar masses are usually found on the periodic table and are equivalent to the atomic weight of elements.
In the context of electrolysis, understanding molar mass is crucial for calculating how much of a substance will be deposited on the electrode. The greater the molar mass, the fewer moles are needed to produce a desired weight of the element during the reaction. This concept is deeply intertwined with the calculation of deposited mass during electrolysis, as seen in the textbook example where different metals with varying molar masses were examined.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical reactions involving the transfer of electrons between two substances. During these reactions, one substance loses electrons, known as oxidation, while another gains electrons, a process called reduction.
In electrolysis, redox reactions are fundamental. At the cathode, reduction takes place as metal ions gain electrons and deposit as a layer of metal. For instance, in the case of AgNO₃, the silver ions (Ag⁺) gain an electron to form neutral silver atoms (Ag) at the cathode. The opposite reaction, oxidation, occurs at the anode, where other ions give up electrons.
Understanding these reactions helps one predict the products of electrolysis and manipulate the conditions for optimal metal recovery.

Metal Deposition

Metal deposition is the central outcome in the process of electrolysis when a metal is being extracted from its compound and deposited as a solid layer on the cathode. The amount of metal deposited is determined by various factors, including the charge passed through the solution, the molar mass of the metal, and the number of electrons required for each atom of metal to form.
During electrolysis, as explained in the example, metals like silver, zinc, iron, and hafnium were assessed for which would deposit the most mass on the cathode. Silver emerged as the metal with the maximum deposition due to its higher position in terms of electron transfer efficiency in the redox process and its lower number of electrons required per ion for deposition.

Faraday's Constant

Faraday's Constant is a key element in calculating various phenomena in electrochemistry, particularly in the electrolysis process. It is a constant that represents the charge of one mole of electrons, approximately 96500 Coulombs/mol. This figure allows chemists to bridge the gap between the macroscopic world of grams and the microscopic world of electrons.
In our textbook example, Faraday’s constant was employed to find the mass of metal deposited at the cathode. By using the formula: \[\text{mass} = \frac{Q \times M}{n \times F}\]where \(Q\) is the total electric charge, \(M\) is the molar mass, \(n\) is the number of electrons per metal ion, and \(F\) is Faraday's constant, we can calculate how much of a specific metal is deposited. Faraday's constant thus offers a practical tool for translating electrical energy into chemical amounts.