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

\( \mathrm{AgNO}_{3} \) deposits the maximum mass of silver at the cathode.

Step 1: Calculate the Charge Transfer at the Cathode

Using the formula for total charge \( Q \), which is \( Q = I \times t \), we calculate the total charge transferred. Here, \( I = 10 \) A and \( t = 3600 \) s (1 hour). Thus, \( Q = 10 \times 3600 = 36000 \) C.

Step 2: Determine the Number of Moles of Electrons Transferred

To find the number of moles of electrons, use Faraday's constant (\( F \)), where \( F = 96500 \) C/mol. The number of moles of electrons \( n_e \) is given by \( n_e = \frac{Q}{F} = \frac{36000}{96500} \approx 0.373 \) moles.

Step 3: Analyze Each Salt's Reaction

For each salt, determine the number of moles per metal deposited: - \( \mathrm{ZnSO}_{4} \): \( \mathrm{Zn}^{2+} + 2e^{-} \to \mathrm{Zn} \) (Moles of Zn = \( \frac{0.373}{2} \)) - \( \mathrm{FeCl}_{3} \): \( \mathrm{Fe}^{3+} + 3e^{-} \to \mathrm{Fe} \) (Moles of Fe = \( \frac{0.373}{3} \)) - \( \mathrm{HfCl}_{4} \): \( \mathrm{Hf}^{4+} + 4e^{-} \to \mathrm{Hf} \) (Moles of Hf = \( \frac{0.373}{4} \)) - \( \mathrm{AgNO}_{3} \): \( \mathrm{Ag}^{+} + e^{-} \to \mathrm{Ag} \) (Moles of Ag = \( 0.373 \))

Step 4: Calculate Mass of Metal Deposited for Each Salt

Use the moles of each metal and their respective atomic weights to calculate mass: - Zn: \( \frac{0.373}{2} \times 65 \approx 12.11 \) g - Fe: \( \frac{0.373}{3} \times 56 \approx 6.96 \) g - Hf: \( \frac{0.373}{4} \times 178 \approx 16.58 \) g - Ag: \( 0.373 \times 108 \approx 40.28 \) g

Step 5: Determine Maximum Mass Deposition

Compare the calculated masses: Zn (12.11 g), Fe (6.96 g), Hf (16.58 g), and Ag (40.28 g). The maximum mass deposited is that of silver from \( \mathrm{AgNO}_{3} \) at 40.28 g.

Key Concepts

Faraday's Laws of Electrolysis

Faraday's Laws of Electrolysis are fundamental in understanding the electrochemical processes involving the depositing or dissolving of material at electrodes. The first law states that the amount of substance that undergoes reduction/oxidation at an electrode is directly proportional to the electric charge passed through the electrolyte.

The second law provides that the amounts of chemical substances that react at an electrode are proportional to their equivalent weights when the same quantity of electricity is passed.

• This means that electrolysis is not random but follows a predictable pattern based on charge and material properties.
• Thus, by knowing the charge (Coulombs) and the equivalent weight of the substance, one can calculate the extent of electrolysis.

Through these laws, students can relate directly the conditions of current and duration (as seen with the example of a 10-ampere current for an hour) to predict outcomes accurately.

Charge Transfer

In electrolysis, the notion of charge transfer is foundational since it indicates how much electricity is applied to the electrolyte. The formula to calculate charge transfer is simple:

• \[ Q = I \times t \]

where:

• \(Q\) is the total charge in Coulombs (C),
• \(I\) is the current in amperes (A), and
• \(t\) is the time in seconds (s).

For example, in our actual problem, a 10 A current flows for 3600 seconds (or one hour), giving a charge of 36,000 C.

This measure of electrical charge is crucial to determine how many electrons are available to drive the reduction (or deposition) of metal ions.

Moles of Electrons

The concept of moles of electrons is closely tied to Faraday's constant. This constant, approximately 96,500 C/mol, is the charge of one mole of electrons. To find out how many moles of electrons are passed during electrolysis, one uses the equation:

• \[ n_e = \frac{Q}{F} \]

where:

• \(n_e\) is the moles of electrons,
• \(Q\) is the total charge, and
• \(F\) is Faraday's constant.

In our example, with a charge transfer of 36,000 C, the moles of electrons is calculated as \( \frac{36000}{96500} \) which equals approximately 0.373 moles.

This calculation is pivotal because the moles of electrons relate directly to how much material can be deposited or dissolved during electrolysis.

Metal Deposition

Metal deposition is a fascinating outcome of electrolysis, where metal ions in solution gain electrons and transition to a metallic state at the cathode. Each ion of a metal requires a specific number of electrons to deposit as a neutral metal atom. For example:

• Zn\(^{2+}\) requires 2 electrons,
• Fe\(^{3+}\) requires 3 electrons,
• Hf\(^{4+}\) requires 4 electrons, and
• Ag\(^{+}\) requires just 1 electron.

This means for every set number of electrons, some specific amount of the metal ion can be converted. By knowing the number of electrons available (moles), you can predict how much of each metal will deposit.

In the example, maximum mass deposition was found for silver due to the minimal electron requirement for its deposition, demonstrating how electrolysis conditions directly impact the efficiency and economy of metal recovery processes.