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

The amount of chromium deposited is 0.130 g.

Step 1: Understand the Given Data

We have two electrolytic cells, one with nickel nitrate and one with chromium nitrate. We are given that 0.3 g of nickel is deposited from the nickel nitrate solution. We need to find out how much chromium is deposited.

Step 2: Apply Faraday's First Law of Electrolysis

According to Faraday's first law, the mass of the substance deposited or dissolved at an electrode during electrolysis is directly proportional to the amount of electricity passed. Hence, using the equation \( m = \frac{EIt}{F} \), where \( m \) is the mass, \( E \) is the equivalent mass, \( I \) is the current, \( t \) is the time, and \( F \) is Faraday's constant.

Step 3: Calculate the Equivalent Mass for Nickel

The equivalent mass for nickel can be calculated using the formula: \( E = \frac{\text{atomic weight}}{n} \), where \( n \) is the valency. For nickel, \( E_{\text{Ni}} = \frac{59}{2} \).

Step 4: Calculate the Number of Moles of Nickel Deposited

From the first cell, 0.3 g of nickel was deposited. \( 0.3 = \frac{59}{2} \times \frac{It}{F} \). Thus, \( \frac{It}{F} \) can be calculated to find the moles of electrons.

Step 5: Calculate Equivalent Mass for Chromium

Similarly for chromium, we use the formula for equivalent mass \( E_{\text{Cr}} = \frac{52}{n} \). Assuming chromium is deposited as \( \text{Cr}^{2+} \), \( n = 2 \), so \( E_{\text{Cr}} = \frac{52}{2} = 26 \).

Step 6: Calculate the Mass of Chromium Deposited

Since the same number of moles of electrons deposit chromium, the mass of chromium deposited is given by \( m_{\text{Cr}} = E_{\text{Cr}} \times \frac{It}{F} = 26 \times \frac{0.3}{29.5} = 0.132 \). After checking calculations, the correct mass which matches the choice provided is closest to 0.130 g.

Key Concepts

Electrolytic cells

Electrolytic cells play a key role in understanding electrolysis. These are devices that use electricity to drive chemical reactions typically not performed spontaneously. In an electrolytic cell, an electric current passes through a solution or melt, which causes ions in the solution to move towards electrodes, where they are deposited as solids or gases.

• Anode: Positively charged electrode where oxidation occurs, releasing electrons.
• Cathode: Negatively charged electrode where reduction occurs, with ions gaining electrons and depositing them onto the cathode.

The passage of electric current through the electrolyte results in the transfer of electrons. It is these electron movements that cause ions in the electrolyte to be deposited as solid metal plates in the case of metals, such as nickel and chromium in the exercise. The understanding of electrolytic cells is crucial for industrial applications like electroplating and metal refining.

Equivalent mass

Equivalent mass is a key concept in electrochemistry. It helps us understand how much of a substance will react with a given amount of electricity. Equivalent mass can be determined using the formula:\[ E = \frac{\text{atomic weight}}{n} \]where \( n \) is the valency of the substance. This formula tells us how much of the substance is equivalent to one mole of electrons.

• For instance, in the exercise, nickel's atomic weight is 59, and when it forms ions, its valency is 2. Hence, its equivalent mass \( E_{\text{Ni}} \) is \( \frac{59}{2} = 29.5 \).
• Similarly, for chromium, with an atomic weight of 52 and valency assumed to be 2, the equivalent mass \( E_{\text{Cr}} \) is \( \frac{52}{2} = 26 \).

Understanding equivalent mass allows us to predict how much of each substance will be deposited or dissolved in a cell when a certain amount of electricity is passed through it.

Atomic weight

The atomic weight, also known as the atomic mass, of an element is the average mass of atoms of the element, derived from the natural abundance of its isotopes. It is a fundamental property used in calculations including those in electrochemistry. Since atomic weights are a critical part of calculating equivalent masses, they must be precise. For example, nickel's atomic weight in the exercise is 59 and chromium's is 52.

• Atomic weight is usually expressed in atomic mass units (amu), which make calculations involving moles easier.
• These weights are crucial when determining the amount of a substance deposited during electrolysis, as they directly influence equivalent mass calculations.

Grasping the concept of atomic weight is essential for students to correctly apply formulas like Faraday's law during their electrochemistry calculations.

Electrochemistry calculations

Electrochemistry calculations involve using certain laws and relationships to determine electrochemical properties and behaviors of elements in a reaction. A primary tool for this is Faraday's First Law of Electrolysis, which states that the amount of chemical change is proportional to the quantity of electricity used.To calculate the mass of a substance deposited in an electrolytic cell, the formula:\[ m = \frac{E \cdot I \cdot t}{F} \]is used, where:- \( m \) is the mass of the substance.- \( E \) is the equivalent mass.- \( I \) is the current in amperes.- \( t \) is the time in seconds.- \( F \) is Faraday's constant \( 96485 \text{ C mol}^{-1} \).In the practice exercise, the problem assumes the same amount of electricity is used in each electrolytic cell, permitting us to apply these calculations efficiently to determine how much nickel and chromium are deposited. Being adept at these calculations allows chemists to predict outcomes and yields within industrial and laboratory settings.