JEE Chemistry 2024 Question with Solution

Given: Volume of $O_2$ displaced at S.T.P. is $5600 \, \text{mL}$ and molar mass of $Ag$ is $108 \, \text{g/mol}$.

Find: The mass of silver displaced by the same quantity of electricity.

At S.T.P., $1$ mole of any gas occupies $22400 \, \text{mL}$. Therefore, moles of $O_2$ are

$$\frac{5600}{22400} = 0.25$$

For electrolysis of water:

$$2H_2O \rightarrow 4H^+ + O_2 + 4e^-$$

So, $1$ mole of $O_2$ is produced by $4$ faradays of electricity. Hence, $0.25$ mole of $O_2$ is produced by

$$0.25 \times 4 = 1$$

faraday of electricity.

For silver deposition:

$$Ag^+ + e^- \rightarrow Ag$$

Thus, $1$ faraday deposits $1$ mole of $Ag$. Therefore, mass of silver deposited is

$$1 \times 108 = 108 \, \text{g}$$

Therefore, the mass of silver displaced is $108 \, \text{g}$.

Given: The same quantity of electricity displaces $5600 \, \text{mL}$ of $O_2$ at S.T.P.

Find: Mass of $Ag$ displaced.

Using the gas volume at S.T.P.:

$$\text{moles of } O_2 = \frac{5.6}{22.4} = 0.25$$

Since formation of $1$ mole of $O_2$ requires $4$ moles of electrons, the charge passed corresponds to

$$0.25 \times 4 = 1$$

faraday.

Now for silver:

$$Ag^+ + e^- \rightarrow Ag$$

So, $1$ faraday deposits $1$ mole of $Ag$.

Hence,

$$\text{mass of Ag} = 1 \times 108 = 108 \, \text{g}$$

Therefore, the numerical value of the answer is 108.

The answer key shown as $7.2 \, \text{g}$ does not agree with the solution working. The extracted solution consistently gives $108 \, \text{g}$.