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, \(I_{2}\) is formed which required \(40 \mathrm{ml}\) of \(0.1 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) 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

The current efficiency is 8%.

Step 1: Determine the Amount of Iodine Formed Electrochemically

To find the amount of iodine formed at the anode during electrolysis, we use the relation between charge, moles of electrons, and Faraday's constant.Given that the current (I) is 0.5 A and time (t) is 9650 seconds, the charge (Q) passed is:\[ Q = I \times t = 0.5 \text{ A} \times 9650 \text{ s} = 4825 \text{ C}. \]The reaction shows that 2 moles of electrons produce 1 mole of \(I_2\). Thus, for iodine:\[ \text{Moles of } e^- = \frac{Q}{F} = \frac{4825}{96500} = 0.05 \text{ moles}. \]Hence, the theoretical amount of \(I_2\) formed is:\[ \text{Moles of } I_2 = \frac{0.05}{2} = 0.025 \text{ moles.} \]

Step 2: Calculate the Amount of Iodine Required for Titration

The iodine produced reacts with thiosulfate in a titration process. The volume of \(Na_2S_2O_3\) used was 40 mL with a molarity of 0.1 M.Calculate the moles of thiosulfate used:\[ \text{Moles of } S_2O_3^{2-} = 0.1 \text{ M} \times 0.04 \text{ L} = 0.004 \text{ moles}. \]According to the titration reaction equation, 1 mole of \(I_2\) reacts with 2 moles of \(S_2O_3^{2-}\), therefore the moles of \(I_2\) calculated are:\[ \text{Moles of } I_2 = \frac{0.004}{2} = 0.002 \text{ moles}. \]

Step 3: Calculate Current Efficiency

Current efficiency is the ratio of the actual amount of iodine formed to the theoretical amount, expressed as a percentage.Using the results of steps 1 and 2:\[ \text{Current Efficiency} = \left( \frac{\text{Actual moles of } I_2}{\text{Theoretical moles of } I_2} \right) \times 100\%= \left( \frac{0.002}{0.025} \right) \times 100\% = 8\%. \]

Key Concepts

Faraday's Law of Electrolysis

Faraday's Law is crucial in understanding electrolysis. It states that the mass of a substance altered at an electrode during electrolysis is directly proportional to the quantity of electricity that passes through the electrolyte. It fundamentally connects electrical charge with chemical change. In the context of electrolysis, Faraday's constant (F) represents the charge of one mole of electrons, approximately 96500 Coulombs.
The equation used to calculate the amount of substance is:
\[ Q = n \times F \]Where

• \( Q \) is the total electric charge in Coulombs,
• \( n \) is the amount of substance in moles, and
• \( F \) is Faraday’s constant.

Therefore, to find out how much of a substance is produced or consumed in an electrochemical cell, you need the total charge and Faraday's constant. In our exercise, we calculated the moles of electrons to understand the iodine formation at the anode.
This fundamental law helps determine the theoretical yield in electrolysis, as seen in our initial calculation of iodine.

Current Efficiency in Electrolysis

Current efficiency measures how effectively an electrochemical cell converts electrical charge into a desired chemical product. It is the ratio of the actual yield of product to the theoretical yield predicted by Faraday's law, expressed as a percentage.
In practical terms, this tells us how much of the current applied during electrolysis goes toward producing the intended chemical reaction versus being lost to side reactions or inefficiencies.
In the electrolysis of potassium iodide (KI), we determined the current efficiency by comparing the actual amount of iodine formed with the predicted amount:

• **Calculated moles of \( I_2 \)** from Faraday's calculation provided a theoretical amount.
• **Measured moles of \( I_2 \)** came from a titration process with thiosulfate.

The current efficiency of 8% reflects significant losses, suggesting not all the electrical energy was used to produce iodine. Understanding the efficiency helps in optimizing industrial electrochemical processes, aiming for higher yield and better resource use.

Electrochemical Reactions Explained

Electrochemical reactions involve the transfer of electrons, bringing together concepts from both chemistry and electricity. During such reactions, oxidation and reduction occur at different electrodes:

• **Oxidation**, the loss of electrons, happens at the anode.
• **Reduction**, the gain of electrons, takes place at the cathode.

In the context of the exercise, iodine is oxidized at the anode, forming iodine gas \(I_2\) by losing electrons. Meanwhile, the iodide ions are "sacrificed" to release these electrons needed for the electrochemical process.
These reactions are central to the function of electrochemical cells, batteries, and even industrial processes that use electrolysis to produce materials. Understanding how these reactions function is key in predicting the outcomes of electrical energy application in chemical processes.