Problem 69 - A Complete Resource Book in Chemistry for JEE Main

Question

The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If \(0.3\) g of nickel was deposited in the first cell, the amount of chromium deposited is (atomic weight of \(\mathrm{Ni}=59\), \(\mathrm{Cr}=52)\) (a) \(0.130 \mathrm{~g}\) (b) \(0.236 \mathrm{~g}\) (c) \(0.176 \mathrm{~g}\) (d) \(1.76 \mathrm{~g}\)

Step-by-Step Solution

Answer

Option (c) \( 0.176 \mathrm{~g} \) is correct.

Step 1: Understand the Problem

We need to determine the mass of chromium deposited from an electrolytic cell when 0.3 g of nickel is deposited in another cell. The atomic weights provided are necessary for calculating the number of moles of metals deposited.

Step 2: Calculate Moles of Nickel Deposited

The formula to convert mass to moles is \( n = \frac{m}{M} \), where \( n \) is moles, \( m \) is mass, and \( M \) is molecular weight. For nickel: \( n_{Ni} = \frac{0.3}{59} \approx 0.00508 \) moles.

Step 3: Use Faraday’s Laws of Electrolysis

According to Faraday's first law, the same quantity of electricity will deposit amounts of substances that are proportional to their equivalent weights. The equivalent weight \( EW = \frac{M}{z} \), where \( M \) is atomic weight and \( z \) is valency.

Step 4: Equivalent Weight Comparison

For nickel, assuming its common valency of 2, \( EW_{Ni} = \frac{59}{2} = 29.5 \). For chromium, assuming its common valency of 3, \( EW_{Cr} = \frac{52}{3} \approx 17.33 \).

Step 5: Calculate Moles of Chromium Deposited

Since the quantity of electricity is the same, the number of electrons transferred (moles \( \times \) valency) must be the same: \( n_{Ni} \times 2 = n_{Cr} \times 3 \). Solving for \( n_{Cr} \), \( n_{Cr} = \frac{0.00508 \times 2}{3} \approx 0.003388 \) moles.

Step 6: Calculate Mass of Chromium Deposited

Use the formula \( m = n \times M \) to find the mass of chromium. For chromium: \( m_{Cr} = 0.003388 \times 52 \approx 0.1761 \) g.

Step 7: Match with Given Options

The calculated mass of chromium, \( 0.176 \) g, matches option (c). Therefore, the correct answer is option (c) \( 0.176 \mathrm{~g} \).

Key Concepts

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis form the foundation of our understanding of how electricity causes chemical changes, particularly in electrolytic cells. These laws were established by Michael Faraday in the early 19th century, and they have two main components. The first law of electrolysis states that the mass of a substance altered at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the cell. This means simply that the more electricity applied, the more material is deposited at the electrodes.

• The formula associated with the first law is \( m = Z imes Q \), where \( m \) is the mass deposited, \( Z \) is the electrochemical equivalent, and \( Q \) is the total electric charge passed during the process.
• The second law of electrolysis posits that for the same amount of electricity, the masses of substances altered are proportional to their respective equivalent weights.

These laws are crucial because they help predict and calculate how much metal can be deposited under a given amount of electricity. In the case of the exercise, the weights of nickel and chromium deposited are dictated by these laws.

Equivalent Weight Calculation

Equivalent weight plays an essential role in the field of chemistry, particularly when discussing electrochemical reactions. It refers to the mass of a given substance that will combine with or displace a fixed quantity of another substance. It can also be defined as the mass of a substance that will donate or accept 1 mole of electrons.

• The formula for calculating equivalent weight is \( EW = \frac{M}{z} \), where \( M \) is the atomic or molecular weight, and \( z \) represents the valency of the substance.
• Understanding this helps in predicting the outcomes in electrolysis. For example, nickel typically has a valency of 2, thus \( EW_{Ni} = \frac{59}{2} \).
• Chromium often has a valency of 3, so \( EW_{Cr} = \frac{52}{3} \).

Equivalent weight calculations are practical for ensuring that you achieve the correct proportions of substances in your electrochemical reactions. This was central in the exercise, guiding us to understand the amount of chromium that would be deposited.

Atomic Weights in Chemistry

Atomic weights are a fundamental concept in chemistry, serving as a standard property of elements. These weights are determined by the sum of an atom's protons and neutrons, making them unique to each element. The atomic weight provides essential data for many calculations in both theoretical and practical chemistry.

• The atomic weight of an element is typically expressed in atomic mass units (amu). For example, nickel has an atomic weight of 59 amu, while chromium has an atomic weight of 52 amu.
• These weights are crucial when calculating the molar mass, which is required for determining how much of an element is involved in a reaction.
• In electrochemistry, knowing atomic weights assists in determining how mass relates to moles and helps execute further chemical calculations with precision.

In the problem we're analyzing, understanding the atomic weights of nickel and chromium was vital in converting mass to moles so we could apply Faraday’s laws effectively. This illustrates the central role atomic weights play in simplifying complex chemical processes.