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 t 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

The solution of \( \mathrm{AgNO}_3 \) deposits the maximum mass of metal (Ag) at the cathode.

Step 1: Determine the Relationship between Current and Mass Deposit

The mass of a metal deposited during electrolysis is given by Faraday's law of electrolysis, which states that \( m = \frac{Q \times M}{n \times F} \), where \( m \) is the mass of the substance deposited, \( Q \) is the total electric charge passed through the substance, \( M \) is the molar mass of the element, \( n \) is the number of electrons required to reduce one ion of the metal, and \( F \) is Faraday's constant \((96500 \text{ C mol}^{-1})\).

Step 2: Calculate Total Charge (Q)

The total charge \( Q \) in Coulombs can be calculated by the formula \( Q = I \times t \), where \( I \) is the current in amperes and \( t \) is the time in seconds. Here, \( I = 10 \) A and \( t = 1 \) hour \((3600 \text{ seconds})\), so \( Q = 10 \times 3600 = 36000 \text{ C}\).

Step 3: Calculate Mass Deposit for Each Salt

Using Faraday's law, calculate the mass deposited for each salt:(a) For \( \text{ZnSO}_4 \), \( n=2 \), so \( m = \frac{36000 \times 65}{2 \times 96500} = 12.08 \text{ grams} \).(b) For \( \text{FeCl}_3 \), \( n=3 \), so \( m = \frac{36000 \times 56}{3 \times 96500} = 6.96 \text{ grams} \).(c) For \( \text{HfCl}_4 \), \( n=4 \), so \( m = \frac{36000 \times 178}{4 \times 96500} = 16.65 \text{ grams} \).(d) For \( \text{AgNO}_3 \), \( n=1 \), so \( m = \frac{36000 \times 108}{1 \times 96500} = 40.36 \text{ grams} \).

Step 4: Compare the Mass Deposits

Compare the calculated masses for each metal: - \( \text{Zn} = 12.08 \text{ grams} \)- \( \text{Fe} = 6.96 \text{ grams} \)- \( \text{Hf} = 16.65 \text{ grams} \)- \( \text{Ag} = 40.36 \text{ grams} \)Silver \((\text{Ag})\) yields the highest mass deposit at \( 40.36 \text{ grams} \).

Key Concepts

Faraday's Law

Faraday's Law of Electrolysis forms the backbone for understanding how mass is deposited on an electrode during electrolysis. Michael Faraday established this law, which relates the amount of substance that transitions during electrolysis to the total charge passed through the circuit. The formula derived from Faraday's Law is \( m = \frac{Q \times M}{n \times F} \). Here, \( m \) is the mass of the substance deposited in grams, \( Q \) represents the total electric charge in coulombs, \( M \) is the molar mass of the metal, \( n \) is the count of electrons needed to transport one ion of the metal, and \( F \) stands for Faraday's constant, approximately \( 96500 \, \text{C/mol} \). Faraday's Law implies that knowing the total electrical charge and the specifics of the element's molar mass helps us accurately determine the mass deposited during electrolysis. This principle is crucial for processes such as electroplating and electrorefining.

Electrochemical Deposition

Electrochemical deposition occurs when a metal ion in a solution is transformed into its metallic form and settles onto an electrode in response to an electric current. This process is core to electrolysis and finds applications in various industries, from creating semiconductor components to coating materials for corrosion resistance. In the exercise, the solutions of different metal salts undergo this process when subjected to a constant current. The specifics of each metal — its molar mass and electron requirement for reduction — influence the process's efficiency and outcome. For instance, in the example, while silver nitrate (\( \text{AgNO}_3 \)\ () produces the highest mass deposit, it's partly due to silver ions needing fewer electrons (\( n=1 \)) compared to, say, zirconium from hafnium chloride (\( \text{HfCl}_4 \)\ () that requires four electrons (\( n=4 \)). The effectiveness of electrochemical deposition depends on parameters such as current duration and magnitude, solution concentration, and electrode surface area, all aligning with Faraday's laws for precise prediction.

Molar Mass

Molar mass represents the mass of a given substance (element or compound) divided by the amount of substance. It's a fundamental concept in chemistry, expressed in grams per mole (\( \text{g/mol} \)\ () and crucial when calculating the mass of a substance formed or consumed. It's essentially the sum of the atomic masses of all atoms present in a molecule. For analysis or electrolysis calculations, knowing the molar mass allows us to link the amount in moles to the mass in grams. This is vital while applying Faraday's Law, as evident where each salt has a specific molar mass which, when coupled with the total charge and the charge number, determines the deposited metal's mass. For example, the molar mass of silver is 108 \( \text{g/mol} \)\ (, thereby influencing the final mass deposited according to its quantity times the moles involved.

Chemical Stoichiometry

Chemical stoichiometry involves the calculation of reactants and products in chemical reactions. It provides the quantitative relationship between reactants consumed and products formed. This concept is key in deciphering the outcome of electrolysis reactions within solutions, defining how much of each reactant gets converted into a product. The stoichiometric calculations consider coefficients from balanced chemical equations to relate quantities of different substances. During electrolysis, the number of moles of electrons needed to neutralize one mole of cations is pivotal. For instance, in our given exercise, calculating the necessary electrons for reductions:

• Zn from \( \text{ZnSO}_4 \) needs 2 electrons.
• Fe from \( \text{FeCl}_3 \) involves 3.
• Hf from \( \text{HfCl}_4 \) requires 4.
• Ag from \( \text{AgNO}_3 \) takes just 1.

Understanding these stoichiometric fundamentals aids in predicting the precise mass outcome from an electrolysis reaction, confirming whether the expected yield is achieved or aligned to theoretical predictions.