Problem 172 - A Complete Resource Book in Chemistry for JEE Main

Question

In the electrolysis of \(\mathrm{KI}, \mathrm{I}_{2}\) is formed at the anode by the reaction; \(2 \mathrm{I} \longrightarrow \mathrm{I}_{2}+2 \mathrm{e}^{-}\) After the passage of current of \(0.5\) ampere for 9650 seconds, \(\mathrm{I}_{2}\) is formed which required \(40 \mathrm{ml}\) of \(0.1 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} .5 \mathrm{H}_{2} \mathrm{O}\) solution in the following reaction; \(\mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}+2 \mathrm{I}^{-}\) What is the current efficiency?

Step-by-Step Solution

Answer

Current efficiency is 8%.

Step 1: Determine the amount of Iodine formed

First, calculate the moles of Na2S2O3 used in the titration:\[\text{Moles of } \text{Na}_2\text{S}_2\text{O}_3 = \text{Molarity} \times \text{Volume in Liters} = 0.1 \times \frac{40}{1000} = 0.004 \text{ moles}\]Since 1 mole of \(\text{I}_2\) reacts with 2 moles of \(\text{S}_2\text{O}_3^{2-}\), the moles of \(\text{I}_2\) formed are half of the moles of \(\text{S}_2\text{O}_3^{2-}\):\[\text{Moles of } \text{I}_2 = \frac{0.004}{2} = 0.002 \text{ moles}\]

Step 2: Calculate the theoretical amount of iodine from Faraday's law

Using Faraday's law, calculate the theoretical moles of iodine produced. The charge passed \(Q\) is given by:\[Q = I \times t = 0.5 \times 9650 = 4825 \text{ C}\]Since 1 mole of electrons corresponds to a charge of \(96500 \text{ C}\) (Faraday constant), the moles of electrons are:\[\text{Moles of electrons} = \frac{4825}{96500} = 0.05 \text{ moles}\]From the reaction \(2 \text{I}^- \rightarrow \text{I}_2 + 2 \text{e}^-\), it takes 2 moles of electrons to form 1 mole of \(\text{I}_2\), thus the theoretical moles of \(\text{I}_2\) should be:\[\text{Moles of } \text{I}_2 = \frac{0.05}{2} = 0.025 \text{ moles}\]

Step 3: Calculate current efficiency

The current efficiency is given by the ratio of the actual moles of \(\text{I}_2\) formed to the theoretical moles of \(\text{I}_2\) that could be formed:\[\text{Current Efficiency} = \frac{\text{Actual moles of } \text{I}_2}{\text{Theoretical moles of } \text{I}_2} \times 100 = \frac{0.002}{0.025} \times 100 = 8\%\]

Key Concepts

Understanding Electrolysis

Electrolysis is an essential concept in chemistry where electrical energy is used to drive a non-spontaneous chemical reaction. It involves the movement of ions in an electrolyte solution between two electrodes. During electrolysis, oxidation occurs at the anode and reduction at the cathode. For the electrolysis of potassium iodide (KI), iodine (\(\text{I}_2\)) is formed at the anode. This process requires a direct current to decompose the ionic compound into its elements. When considering electrolysis, it is important to determine factors such as voltage, current, and the concentration of the electrolyte, as they affect the efficiency of the process. By controlling these conditions, one can facilitate the desired chemical transformation.

Exploring Faraday's Law

Faraday's law of electrolysis quantifies the relationship between the charge passed through an electrolyte and the amount of substance transformed at the electrodes. It states that the mass of a substance transformed during electrolysis is directly proportional to the total charge passed through the electrolyte. This principle translates into a formula: \[ Q = nF \] where \( Q \) is the total electric charge, \( n \) is the number of moles of electrons involved, and \( F \) is Faraday's constant (\(96500 \text{ C/mol} \)).This law allows us to calculate the theoretical yield of a product in an electrochemical reaction by using the measured current and time. Understanding Faraday's law is crucial for determining the efficiency of electrolysis and predicting the amount of product formed in electrochemical processes.

Grasping the Mole Concept

The mole concept is a fundamental chemical measurement that helps quantify the amount of substance. A mole represents \(6.022 \times 10^{23}\) particles, which could be atoms, molecules, ions, or electrons. In the context of the electrolysis of potassium iodide, the mole concept helps convert the measured volume and concentration of substances into a number that corresponds to the actual amounts involved in the reactions.Using the mole concept with stoichiometry, we can calculate the necessary reactants and predict the products of a chemical reaction. For example, the determination of iodine (\(\text{I}_2\)) formed via electrolysis depends significantly on converting molarity and volume into moles, which is key for calculating current efficiency.

The Intricacies of Redox Reactions

Redox reactions, or reduction-oxidation reactions, involve the transfer of electrons between species. In these reactions, one substance is oxidized by losing electrons, while another is reduced by gaining electrons. Electrolysis of a compound like potassium iodide is a classic example of a redox reaction.In the given exercise, the iodide ion (\(\text{I}^-\)) loses electrons to become iodine (\(\text{I}_2\)) at the anode, exemplifying oxidation. Understanding redox processes is vital for manipulating and harnessing chemical reactions in electrochemical cells. By analyzing how electrons are transferred, one can comprehend the oxidation states and balance chemical equations necessary for accurate computations in electrolysis.