Question

Certain electric current for half an hour collects \(11.2\,L\) of hydrogen at NTP. Same current passed for one hour deposits how much silver?

• A. \(216\,g\)
• B. \(108\,g\)
• C. \(47\,g\)
• D. \(60\,g\)

Hint:

1 Faraday deposits 1 gram-equivalent.

Answer 1

The Correct Option is A

To solve this problem, we need to use the principles of electrolysis specifically related to Faraday's laws, which help us calculate the amount of substance deposited during the process.

Step 1: Calculate the charge passed during hydrogen collection

Hydrogen collected at NTP (Normal Temperature and Pressure) implies using molar volume concepts:

The molar volume of any gas at NTP is approximately \(22.4 \, \text{L/mol}\).

Given that \(11.2 \, \text{L}\) of hydrogen is collected:

The amount of hydrogen in moles is:

\[\text{Moles of } H_2 = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{mol}\]

The reaction at the cathode for hydrogen generation is:

\[2 \text{H}^+ + 2e^- \rightarrow \text{H}_2\]

This equation shows that 2 moles of electrons produce 1 mole of hydrogen gas.

For \(0.5 \, \text{mol}\) of \(\text{H}_2\), \(1 \, \text{mol}\) of electrons is required.

Hence, the charge passed is:

\[Q = \text{n} \cdot F = 1 \, \text{mol} \cdot 96500 \, \text{C/mol} = 96500 \, \text{C}\]

Step 2: Calculate the charge passed in 1 hour

Since this charge was for half an hour, the same current passed for one hour would have double the charge:

\[Q_{1 \text{ hour}} = 2 \times 96500 \, \text{C} = 193000 \, \text{C}\]

Step 3: Calculate the amount of silver deposited

Silver is deposited according to the reaction:

\[\text{Ag}^+ + e^- \rightarrow \text{Ag}\]

This indicates that 1 mole of electrons deposits 1 mole of silver.

The amount of silver deposited is given by:

\[\text{Mass of Ag} = \frac{Q \cdot M}{n \cdot F}\]

Where:

• \(Q = 193000 \, \text{C}\) (the total charge)
• \(M = 108 \, \text{g/mol}\) (molar mass of silver)
• \(n = 1\) (number of moles of electrons per mole of silver)
• \(F = 96500 \, \text{C/mol}\) (Faraday's constant)

\[\text{Mass of Ag} = \frac{193000 \cdot 108}{1 \cdot 96500} = 216 \, \text{g}\]

Therefore, the correct answer is 216 g.