JEE Chemistry 2025 Question with Solution

Given: Volume of NaCl solution = $0.6 \, \text{L}$, time of electrolysis = $300 \, \text{s}$, final $pH = 12$.

Find: The current used during electrolysis.

During electrolysis of aqueous NaCl, hydroxide ions are produced, making the solution basic.

At the cathode:

$$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$$

Thus, $1$ mole of electrons produces $1$ mole of $OH^-$.

Using

$$pH + pOH = 14$$

we get

$$pOH = 14 - 12 = 2$$

So,

$$[OH^-] = 10^{-2} = 0.01 \, \text{M}$$

Moles of $OH^-$ in $0.6 \, \text{L}$ solution:

$$n(OH^-) = 0.01 \times 0.6 = 0.006 \, \text{mol}$$

Hence,

$$n(e^-) = 0.006 \, \text{mol}$$

Using Faraday's law:

$$Q = n(e^-) \times F = 0.006 \times 96500 = 579 \, \text{C}$$

Now,

$$I = \frac{Q}{t} = \frac{579}{300} = 1.93 \, \text{A}$$

Therefore, the current is $1.93 \, \text{A}$. Since the question asks for the nearest integer, the nearest integer value is $2$.

Given: Electrolysis of $600 \, \text{mL}$ aqueous NaCl solution for $5 \, \text{min}$ gives final $pH = 12$.

Find: Current used.

Step 1: Convert the given data.

$$600 \, \text{mL} = 0.6 \, \text{L}, \qquad 5 \, \text{min} = 300 \, \text{s}$$

Step 2: Use the relation between $pH$ and $pOH$.

$$pOH = 14 - 12 = 2$$

Therefore,

$$[OH^-] = 10^{-2} \, \text{mol L}^{-1}$$

Step 3: Find total moles of $OH^-$ produced.

$$n(OH^-) = 10^{-2} \times 0.6 = 6 \times 10^{-3} \, \text{mol}$$

Step 4: Relate hydroxide formed to electrons transferred. From

$$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$$

we get

$$n(e^-) = n(OH^-) = 6 \times 10^{-3} \, \text{mol}$$

Step 5: Calculate the charge passed.

$$Q = nF = 6 \times 10^{-3} \times 96500 = 579 \, \text{C}$$

Step 6: Calculate the current.

$$I = \frac{Q}{t} = \frac{579}{300} = 1.93 \, \text{A}$$

the solution gives $1.93 \, \text{A}$. Because the question explicitly asks for the nearest integer, the final numerical value answer should be $2$.