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 \(2 \mathrm{~F}\) 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 \( \mathrm{CuSO}_{4} \) becomes 0 after electrolysis; the answer is (d) 0.

Step 1: Understanding the Problem

We have a 1L solution of 1 M \( \mathrm{CuSO}_{4} \). We need to find the change in molarity of the solution after passing \( 2 \mathrm{~F} \) (Faradays) of electricity through electrolysis.

Step 2: Determine the Electrochemical Equation

The electrolysis of \( \mathrm{CuSO}_{4} \) leads to the reduction of \( \mathrm{Cu}^{2+} \) ions to \( \mathrm{Cu} \) metal. The reaction at the cathode is: \[ \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu} \]This indicates that 1 mole of \( \mathrm{Cu}^{2+} \) ions requires 2 moles of electrons (or 2 Faradays) to be completely deposited as copper metal.

Step 3: Calculate Moles of Copper deposited

Since 2 Faradays of electricity are passed, the equation \( \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu} \) means that 1 mole of \( \mathrm{Cu}^{2+} \) ions will be deposited. The initial solution has 1 mole of \( \mathrm{Cu}^{2+} \) ions (since it's 1M in 1L solution).

Step 4: Determine the Remaining Molarity

Before electrolysis, there were 1 mole of \( \mathrm{Cu}^{2+} \) in the solution, and after depositing through electrolysis, all \( 1 \) mole will be deposited. Therefore, the molarity of \( \mathrm{CuSO}_{4} \) becomes 0M.

Step 5: Solution Verification

Verify the calculations by ensuring that the amount of electricity passed could indeed deposit the entire initial amount of \( \mathrm{Cu}^{2+} \), which it does. Therefore, the molarity of \( \mathrm{CuSO}_{4} \) post electrolysis is 0M, confirming option (d).

Key Concepts

Electrochemical Reactions

Electrochemical reactions are fascinating processes where electricity is used to cause a chemical change. They are the heart of electrolysis, where electrical energy transforms into chemical energy. To understand how these reactions work, picture a setup with two electrodes submerged in a liquid solution. When an electric current is passed through the system, ions in the solution begin to move. - At the positively charged electrode, called the anode, oxidation occurs. This means electrons are lost by ions or molecules.- At the negatively charged electrode, the cathode, reduction takes place, and ions gain electrons. In the context of electrolysis discussed here, when copper sulfate (\( \mathrm{CuSO}_4 \)) is electrolysed, the \( \mathrm{Cu}^{2+} \) ions in the solution are attracted to the cathode and gain electrons to form copper metal. This change is an example of how electrical energy can directly alter chemical species, showcasing the dynamic nature of electrochemical reactions.

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis are critical in predicting the outcomes of electrolysis processes. Established by Michael Faraday in the 19th century, these laws connect the amount of electrical charge passed through the electrolyte to the mass of substance transformed at the electrodes.The first law states that the mass of a substance altered at an electrode during electrolysis is directly proportional to the amount of electric charge (electric current multiplied by time) passed through the electrolyte.The second law clarifies that when the same amount of electricity passes through different electrolytes, the mass of substances transformed is proportional to their equivalent weights. In this exercise, when 2 Faradays are passed through a \( \mathrm{CuSO}_4 \) solution, it means that enough electrons have flowed to balance 1 mole of \( \mathrm{Cu}^{2+} \) ions, making them neutral copper atoms. By directly applying Faraday's laws, we can predict how much copper will be deposited and how the molarity of the solution changes.

Copper Sulfate Solution

Copper sulfate solution, a staple in various chemical experiments and industrial processes, is a blue, crystalline chemical compound dissolved in water. Known for its distinctive color, it is represented chemically as \( \mathrm{CuSO}_4 \) and contains copper ions (\( \mathrm{Cu}^{2+} \)) and sulfate ions (\( \mathrm{SO}_4^{2-} \)).When it comes to electrolysis, its importance lies in its ionic components:- The \( \mathrm{Cu}^{2+} \) ions can be reduced at the cathode to deposit pure copper metal.- The stable sulfate ions (\( \mathrm{SO}_4^{2-} \)) largely remain unaltered.For this exercise, starting with 1 liter of a 1 M \( \mathrm{CuSO}_4 \) solution means there is 1 mole of these copper ions ready to transform chemically through the process of electrolysis. As the current passes, the \( \mathrm{Cu}^{2+} \) ions are consumed, decreasing the molarity of the solution accordingly.

Molarity Calculation

Molarity is an essential concept in chemistry, describing the concentration of a solution. It's defined as the number of moles of solute per liter of solution. This measure is essential for understanding how reactions proceed, including electrolysis.In this specific problem, you're starting with 1 mole of copper ions in a 1-liter solution, meaning the molarity is 1 M. After passing 2 Faradays of electricity through the solution, which corresponds to removing 1 mole of \( \mathrm{Cu}^{2+} \) ions (as per Faraday's laws), no copper ions remain in the solution.Thus, the molarity calculation after electrolysis is straightforward. Subtracting 1 mole from the initial concentration results in 0 moles of copper ions per liter, bringing the final molarity of \( \mathrm{CuSO}_4 \) to 0 M. Here, molarity helps quantify the extent of chemical change the electrolysis process brings about.