Problem 124 - A Complete Resource Book in Chemistry for JEE Main

Question

One litre of \(1 \mathrm{M} \mathrm{CuSO}_{4}\) solution is electrolysed. After passing 2F of electricity, molarity of \(\mathrm{CuSO}_{4}\) solution will be (a) \(\mathrm{M} / 2\) (b) \(\mathrm{M} / 4\) (c) \(\mathrm{M}\) (d) 0

Step-by-Step Solution

Answer

The molarity of the solution after electrolysis is 0M (d).

Step 1: Understand Electrolysis of CuSO4

Copper sulfate (\(\mathrm{CuSO}_{4}\)) in aqueous solution dissociates into \(\mathrm{Cu}^{2+}\) and \(\mathrm{SO}_4^{2-}\) ions. During electrolysis, \(\mathrm{Cu}^{2+}\) ions will gain electrons at the cathode to deposit copper metal. The reaction is \(\mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu}\). Each mole of \(\mathrm{Cu}^{2+}\) ions requires 2 Faradays (F) of electricity to be deposited as copper metal.

Step 2: Calculate Moles of Cu Deposited

Since 2F of electricity are passed, this is enough to deposit 1 mole of copper from the solution, as each mole of \(\mathrm{Cu}^{2+}\) requires 2F to be reduced to metallic Cu. Therefore, 1 mole of \(\mathrm{Cu}^{2+}\) is removed from the solution.

Step 3: Find Initial Moles of CuSO4 in Solution

Given that the original solution is 1 liter of 1M \(\mathrm{CuSO}_{4}\), there is initially 1 mole of \(\mathrm{Cu}^{2+}\) (and thus CuSO4) in the solution.

Step 4: Determine Remaining Moles of CuSO4

After electrolysis, 1 mole of \(\mathrm{Cu}^{2+}\) is removed, deposited as metal on the cathode. Therefore, no moles of \(\mathrm{CuSO}_{4}\) remain in the solution.

Step 5: Calculate the Final Molarity of CuSO4 Solution

The final number of moles of \(\mathrm{CuSO}_{4}\) in 1 liter of solution is 0 moles. Thus the molarity of the \(\mathrm{CuSO}_{4}\) solution after electrolysis is 0M.

Key Concepts

Molarity

Molarity is a fundamental concept in chemistry, referring to the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. The formula to calculate molarity is given by: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]Understanding molarity is crucial because it helps describe how concentrated a solution is. For instance, a 1M solution of copper sulfate (\(\mathrm{CuSO}_{4}\)) means there is 1 mole of copper sulfate dissolved in 1 liter of water. This concentration provides a clear picture of how much copper sulfate is present in the solution.Moreover, molarity plays a major role in calculating reactions in solutions, as seen in electrolysis. When any part of the solute is consumed, like copper ions during the cathodic reaction in electrolysis, the molarity of the remaining solution changes. This change is essential for predicting the reaction's progress and understanding the material balance in the solution.

Copper sulfate electrolyte

Copper sulfate (\(\mathrm{CuSO}_{4}\)) acts as an electrolyte in electrolysis. An electrolyte is a substance that conducts electricity when dissolved in water because it dissociates into ions. In an aqueous solution, copper sulfate dissociates into copper ions (\(\mathrm{Cu}^{2+}\)) and sulfate ions (\(\mathrm{SO}_4^{2-}\)).During the electrolytic process, the copper ions move towards the cathode where they gain electrons and are reduced to metallic copper. This reaction can be represented as:- Cathodic reaction: \(\mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu}\)The sulfate ions, on the other hand, do not take part in the electron transfer and remain in the solution. This characteristic of copper sulfate makes it an effective electrolyte for various electroplating processes, where a metal layer is deposited onto a surface through electrical currents. This dissociation and movement of ions are what allow the electrolyte solution to conduct electricity efficiently.

Faraday's law of electrolysis

Faraday's law of electrolysis is a crucial principle explaining the quantitative relationship between the amount of electricity used during electrolysis and the amount of substance that is deposited or dissolved at the electrodes. There are two main laws:1. **Faraday's First Law:** - The amount of substance deposited or dissolved at an electrode is directly proportional to the quantity of electricity (i.e., charge) that passes through the electrolyte.2. **Faraday's Second Law:** - For exactly the same quantity of electricity passed, the amount of substance deposited or dissolved is inversely proportional to the equivalent weight of the substance.In practical terms, this means that in the electrolysis of a \(\mathrm{CuSO}_{4}\) solution, the deposition of copper depends on the electric charge passed through the solution. As per the solution, passing 2 Faradays (F) through a 1M solution deposits 1 mole of copper. This is because each mole of \(\mathrm{Cu}^{2+}\) requires 2 Faradays of electricity for complete reduction to metallic copper. Understanding Faraday's laws is essential for efficient industrial processes that depend on electrolysis, such as electroplating and purification of metals.