When an electric current is passed through acidified water, 112 ml of hydrogen gas at N.T.P. is collected at the cathode in 965 seconds. The current passed, in amperes is:

Options

A. `1.0` A
B. `0.5` A
C. `0.1` A
D. `2.0` A

Answer

To find the current passed in amperes when an electric current is passed through acidified water, we can follow these steps:

### Step 1: Understand the Reaction
When electric current is passed through acidified water, hydrogen ions (H⁺) are reduced at the cathode to form hydrogen gas (H₂). The half-reaction can be represented as:
\[ 2H^+ + 2e^- \rightarrow H_2 \]

### Step 2: Calculate Moles of Hydrogen Gas
We know that at Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters (or 22400 mL). Given that 112 mL of hydrogen gas is collected, we can calculate the number of moles of hydrogen gas produced:
\[ \text{Number of moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar volume at NTP}} = \frac{112 \, \text{mL}}{22400 \, \text{mL/mol}} = 0.005 \, \text{mol} \]

### Step 3: Use Faraday's Law
According to Faraday's law of electrolysis, the amount of substance produced at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The relationship can be expressed as:
\[ Q = n \cdot F \]
Where:
- \( Q \) is the total electric charge (in coulombs),
- \( n \) is the number of moles of electrons transferred,
- \( F \) is Faraday's constant (approximately 96500 C/mol).

From the half-reaction, we see that 2 moles of electrons are required to produce 1 mole of hydrogen gas. Therefore, for 0.005 moles of \( H_2 \):
\[ n = 2 \times 0.005 = 0.01 \, \text{mol of electrons} \]

### Step 4: Calculate Total Charge
Now we can calculate the total charge \( Q \) using Faraday's constant:
\[ Q = n \cdot F = 0.01 \, \text{mol} \times 96500 \, \text{C/mol} = 965 \, \text{C} \]

### Step 5: Calculate Current
Current \( I \) can be calculated using the formula:
\[ I = \frac{Q}{t} \]
Where \( t \) is the time in seconds. Given \( t = 965 \, \text{s} \):
\[ I = \frac{965 \, \text{C}}{965 \, \text{s}} = 1 \, \text{A} \]

### Final Answer
The current passed is:
\[ \boxed{1 \, \text{A}} \]