Problem 31 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A current of \(1.70 \mathrm{~A}\) is passed through \(300.0 \mathrm{~mL}\) of \(0.160 \mathrm{M}\) solution of a \(\mathrm{ZnSO}_{4}\) for \(230 \mathrm{sec}\). with a current efficiency of \(90 \%\). Find out the molarity of \(\mathrm{Zn}^{2+}\) after the deposition of \(\mathrm{Zn}\). Assume the volume of the solution to remain constant during the electrolysis.

Step-by-Step Solution

Answer

The final molarity of \( \text{Zn}^{2+} \) is 0.154 M.

Step 1: Calculate the Charge Passed

First, find out the total charge (in Coulombs) that passed through the solution. Use the formula: \[ Q = I imes t \] where \( I = 1.70 \text{ A} \) and \( t = 230 \text{ seconds} \). Thus, \[ Q = 1.70 \times 230 = 391 \text{ Coulombs} \].

Step 2: Correct for Current Efficiency

Apply the current efficiency to the total charge to find the actual charge used for zinc deposition. Current efficiency is given as 90%, so the effective charge is:\[ Q_{ ext{actual}} = 0.90 \times 391 = 351.9 \text{ Coulombs} \].

Step 3: Determine Moles of Electrons

Calculate the moles of electrons involved in the deposition using Faraday's constant \( F = 96500 \text{ C/mol} \):\[ n_e = \frac{Q_{ ext{actual}}}{F} = \frac{351.9}{96500} = 0.00365 \text{ moles of } e^- \].

Step 4: Relate to Moles of Zinc

Since the reduction equation for \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn(s)} \) shows that 2 moles of electrons are needed for 1 mole of zinc, calculate the moles of zinc deposited:\[ n_{\text{Zn}} = \frac{0.00365}{2} = 0.001825 \text{ moles of Zn deposited} \].

Step 5: Initial Moles of Zn in Solution

Calculate the initial moles of \( \text{Zn}^{2+} \) in the solution:\[ \text{Initial moles of } \text{Zn}^{2+} = \text{Concentration} \times \text{Volume} = 0.160 \times 0.300 = 0.048 \text{ moles} \].

Step 6: Calculate Remaining Moles of Zn

Find the remaining moles of \( \text{Zn}^{2+} \) in the solution after electrolysis:\[ \text{Moles of } \text{Zn}^{2+} \text{ remaining} = 0.048 - 0.001825 = 0.046175 \text{ moles} \].

Step 7: Determine Final Molarity

Finally, calculate the molarity of \( \text{Zn}^{2+} \) ions after deposition. The volume remains 300.0 mL, which is 0.300 L:\[ \text{Molarity} = \frac{\text{Moles of } \text{Zn}^{2+}}{\text{Volume}} = \frac{0.046175}{0.300} = 0.154 \text{ M} \].

Key Concepts

Faraday's Laws of Electrolysis

Faraday's Laws of Electrolysis are fundamental concepts in electrochemistry. They describe the principles governing the amount of substance deposited at an electrode during electrolysis. Faraday's First Law states that the mass of an element deposited during electrolysis is directly proportional to the charge passed through the system. This can be expressed in the equation:

• \[ m = \frac{Q}{F} \times \frac{M}{n} \]

where \( m \) is the mass of the substance, \( Q \) is the total electric charge, \( F \) is Faraday's constant (approximately \( 96500 \) Coulombs per mole), \( M \) is the molar mass of the substance, and \( n \) is the number of moles of electrons exchanged.
This means that to deposit more of a substance, you need more charge or time.

Faraday's Second Law of Electrolysis states that, if the same quantity of electricity is passed through different solutions, the mass deposited or dissolved is proportional to the equivalent weight of the substances involved. These laws allow us to calculate exactly how much material will be transformed in an electrochemical reaction.

Current Efficiency

In the context of electrochemistry, current efficiency refers to the percentage of the electric current that is effectively used for the desired chemical reaction during electrolysis. This is not always 100% due to various side reactions that may occur or energy consumed in non-targeted processes. It's calculated as actual charge used divided by the total charge passed, often expressed as a percentage:

• \[ \text{Current Efficiency} = \left( \frac{Q_{\text{actual}}}{Q_{\text{total}}} \right) \times 100\%\]

A current efficiency of 90% means that only 90% of the supplied current contributed to the desired reaction, while the remaining 10% was lost to other processes.
This concept is crucial for optimizing industrial electrochemical processes as it impacts the cost-efficiency and effectiveness of the reaction. Optimizing current efficiency can help in minimizing energy waste and cost.

Molarity Calculation

Molarity is a measure of the concentration of a solute in a solution. It is expressed as the number of moles of solute per liter of solution. Calculating molarity involves knowing the amount of solute present and the total volume of the solution:

• \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume in liters}} \]

In electrolysis problems, such as calculating remaining \( \text{Zn}^{2+} \) concentration, it's important to know how much of the solute was initially present and how much was deposited or reacted during the electrolysis.
With continuous reactions like the deposition of \( \text{Zn} \), calculating the molarity post-reaction involves subtracting the moles of product deposited from the initial moles of the solute, then dividing by the unchanged volume.
This careful adjustment ensures that the post-reaction molarity reflects the true concentration of the solute after electrolysis has occurred.