Problem 69 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An aqueous solution of \(\mathrm{NaCl}\) on electrolysis gives \(\mathrm{H}_{2}(\mathrm{~g}), \mathrm{Cl}_{2}(\mathrm{~g})\) and \(\mathrm{NaOH}\) according to the reaction : \(2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}=2 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) A direct current of 25 amperes with a current efficiency of \(62 \%\) is passed through 20 litres of \(\mathrm{NaCl}\) solution \((20 \%\) by weight). Write down the reactions taking place at the anode and the cathode. How long will it take to produce \(1 \mathrm{~kg}\) of \(\mathrm{Cl}_{2}\) ? What will be the molarity of the solution with respect to hydroxide ion? (Assume no loss due to evaporation.)

Step-by-Step Solution

Answer

It takes approximately 48.7 hours to produce 1 kg of Cl2, and the OH⁻ molarity is 1.408 M.

Step 1: Electrolysis Reaction Revision

We are given the overall balanced equation for the electrolysis of an aqueous NaCl solution: \[ 2 \text{Cl}^- (\text{aq}) + 2 \text{H}_2\text{O} \rightarrow 2 \text{OH}^- (\text{aq}) + \text{H}_2 (\text{g}) + \text{Cl}_2 (\text{g}) \] Our task is to examine individual electrode reactions.

Step 2: Determine Electrochemical Reactions

The reactions at the electrodes are: - At the anode (oxidation): \( 2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \)- At the cathode (reduction): \( 2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \) These reactions occur when passing current through an NaCl solution.

Step 3: Using Faraday's Laws of Electrolysis

To find how long it takes to produce 1 kg of \( \text{Cl}_2 \): 1. First calculate the moles of \( \text{Cl}_2 \): \[ 1\text{ kg } = \frac{1000}{71} \approx 14.08 \text{ moles of } \text{Cl}_2 \]2. Using Faraday's law, calculate the charge required: \[ \text{Charge} = \text{Moles } \times \text{n} \times F \] Here, "n" is the number of electrons exchanged (2 for Cl), and \( F \approx 96500 \text{ C/mol} \). Hence, \( \text{Charge} = 14.08 \times 2 \times 96500 = 2717440 \text{ C} \). 3. Adjust for current efficiency: Actual charge needed = \( \frac{2717440}{0.62} \approx 4382993.55 \text{ C} \).

Step 4: Calculate Time Duration

Using the relationship \( Q = I \times t \), where \( I = 25 \text{ A} \): Solve for \( t \): \[ t = \frac{Q}{I} = \frac{4382993.55}{25} \approx 175319.74 \text{ seconds} \]Converting seconds to hours gives: \[ t \approx \frac{175319.74}{3600} \approx 48.7 \text{ hours} \]

Step 5: Find Molarity of Hydroxide Ion

Since each \( \text{Cl}_2 \) produced generates 2 \( \text{OH}^- \), the moles of \( \text{OH}^- \) produced are: \[ 2 \times 14.08 = 28.16 \text{ moles of } \text{OH}^- \] For 20 liters solution, molarity \( \text{M} \) is: \[ \text{M} = \frac{\text{moles}}{\text{liters}} = \frac{28.16}{20} = 1.408 \text{ M} \]

Key Concepts

Electrode Reactions

In electrolysis, the reactions at each electrode are crucial to understanding how compounds are broken down into different elements and ions. During the electrolysis of aqueous sodium chloride (NaCl), different reactions occur at the anode and cathode.

At the anode, oxidation takes place. The chloride ions (Cl⁻) donate their electrons to form chlorine gas (Cl₂). The reaction can be represented as follows:
- Anode (oxidation): \(2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \)

On the other hand, at the cathode, reduction occurs. Water molecules gain electrons to form hydroxide ions (OH⁻) and hydrogen gas (H₂), represented by:
- Cathode (reduction): \(2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-\)

This exchange of electrons between the electrodes leads to the overall electrolysis process. Each electrode reaction contributes to producing the final products—chlorine gas, hydrogen gas, and sodium hydroxide.

Faraday's Laws of Electrolysis

Faraday's Laws of Electrolysis play an essential role in calculating the quantities of substances produced during electrolysis. These laws state the relationship between the amount of substance produced at an electrode and the quantity of electricity used.

The first law of electrolysis states that the amount of chemical change is directly proportional to the quantity of electricity (charge) passed through the electrolyte. This can be expressed with the formula:
- \(\text{moles of product} \propto Q\)
where \(Q\) is the charge in coulombs.

The second law explains that when the same amount of electricity is passed through different substances, the amounts of different substances liberated or deposited at electrodes are proportional to their equivalent weights.

In the exercise, to produce 1 kg of \(\text{Cl}_2\), first, we must calculate the moles of chlorine. By applying Faraday’s laws, the charge required to produce these moles is found, accounting for electron exchange—which for Cl is 2 electrons per mole of \(\text{Cl}_2\). Therefore, knowing the charge and current, we can calculate the time required for electrolysis.

Molarity Calculations

Molarity is a way of measuring the concentration of a solute in a solution, expressed in moles per liter. This calculation is crucial in understanding the concentration of products formed in an electrolysis reaction.

In the given electrolysis of NaCl, the production of \(\text{OH}^-\) ions directly relates to the number of \(\text{Cl}_2\) molecules formed. Since each \(\text{Cl}_2\) molecule yields 2 hydroxide ions, you can find the moles of \(\text{OH}^-\) by multiplying the moles of \(\text{Cl}_2\) produced by 2.

To calculate molarity, use the formula:
- \(M = \frac{\text{moles of solute}}{\text{liters of solution}}\)

For instance, if the total moles of \(\text{OH}^-\) produced is 28.16 in a 20-liter solution, the molarity would be:
- \(\text{M} = \frac{28.16}{20} = 1.408 \text{ M}\)

This means the final hydroxide ion concentration is 1.408 molar, providing insight into the solution's strength after electrolysis.