When electric current is passed through acidified water for 1930 s, 1120 mL of `H_(2)` gas is collected (at STP) at the cathode. What is the current passed in amperes?

Options

A. `0.05`
B. `0.50`
C. `5.0`
D. 50

Explanation

To find the current passed in amperes when electric current is passed through acidified water for 1930 seconds and 1120 mL of H₂ gas is collected at STP, we can follow these steps:

### Step 1: Calculate the number of moles of H₂ gas collected
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters (or 22400 mL).

Using the formula for the number of moles:
\[ \text{Number of moles (n)} = \frac{\text{Volume (V)}}{22.4 \text{ L}} = \frac{1120 \text{ mL}}{22400 \text{ mL}} = 0.05 \text{ moles} \]

### Step 2: Calculate the weight of H₂ gas
The molecular weight of hydrogen (H₂) is 2 g/mol. Therefore, the weight (W) of H₂ gas can be calculated as:
\[ W = n \times \text{Molecular weight} = 0.05 \text{ moles} \times 2 \text{ g/mol} = 0.1 \text{ g} \]

### Step 3: Calculate the electrochemical equivalent (Z)
The electrochemical equivalent (Z) is given by the formula:
\[ Z = \frac{\text{Molecular weight}}{n \times F} \]
where:
- \( n \) is the number of electrons exchanged (for H₂, n = 2),
- \( F \) is Faraday's constant, approximately 96500 C/mol.

Substituting the values:
\[ Z = \frac{2 \text{ g/mol}}{2 \times 96500 \text{ C/mol}} = \frac{2}{193000} \text{ g/C} \]

### Step 4: Use Faraday's first law of electrolysis
According to Faraday's first law:
\[ W = Z \times I \times t \]
where:
- \( W \) is the weight of the substance deposited (0.1 g),
- \( I \) is the current in amperes,
- \( t \) is the time in seconds (1930 s).

Rearranging the formula to find the current (I):
\[ I = \frac{W}{Z \times t} \]

### Step 5: Substitute the values to find the current
Substituting the values we calculated:
\[ I = \frac{0.1 \text{ g}}{\left(\frac{2}{193000} \text{ g/C}\right) \times 1930 \text{ s}} \]
Calculating \( Z \times t \):
\[ Z \times t = \frac{2}{193000} \times 1930 = \frac{3860}{193000} \approx 0.0200 \text{ g} \]
Now substituting back:
\[ I = \frac{0.1}{0.0200} = 5 \text{ A} \]

### Final Answer
The current passed is **5 amperes**.

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