If `0.224 L` of `H_(2)(g)` is formed at the cathode of one cell at `STP,` how much of `Mg` is formed at the cathode of the other electrolytic cell arranged in series ?

Options

A. `0.24g`
B. `2.4g`
C. `0.48g`
D. `4.8g`

Explanation

To solve the problem step by step, we need to analyze the reactions occurring at the cathodes of both electrolytic cells and relate the amount of hydrogen gas produced to the amount of magnesium formed.

### Step 1: Determine the moles of hydrogen gas produced.
Given that `0.224 L` of `H₂(g)` is formed at STP, we can calculate the number of moles of hydrogen gas produced using the molar volume of a gas at STP, which is `22.4 L/mol`.

\[ \text{Moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar Volume at STP}} = \frac{0.224 \, L}{22.4 \, L/mol} = 0.01 \, mol \]

### Step 2: Write the half-reaction for hydrogen production.
At the cathode, hydrogen ions are reduced to form hydrogen gas. The half-reaction can be written as:

\[ 2H^+ + 2e^- \rightarrow H_2 \]

This shows that 2 moles of electrons are required to produce 1 mole of hydrogen gas.

### Step 3: Relate the moles of electrons to magnesium production.
Since the cells are arranged in series, the same amount of electric charge (or current) flows through both cells. The reduction of magnesium ions at the cathode of the magnesium cell can be represented by the half-reaction:

\[ Mg^{2+} + 2e^- \rightarrow Mg \]

This also requires 2 moles of electrons to produce 1 mole of magnesium.

### Step 4: Calculate the moles of magnesium produced.
From the previous steps, we know that `0.01 mol` of hydrogen gas was produced, which means `0.01 mol` of electrons were used. Since the reaction for magnesium also requires `2 moles of electrons` to produce `1 mole of magnesium`, the moles of magnesium produced will be the same as the moles of hydrogen produced:

\[ \text{Moles of } Mg = 0.01 \, mol \]

### Step 5: Convert moles of magnesium to grams.
The molar mass of magnesium (Mg) is approximately `24 g/mol`. Therefore, we can calculate the mass of magnesium produced:

\[ \text{Mass of } Mg = \text{Moles of } Mg \times \text{Molar Mass of } Mg = 0.01 \, mol \times 24 \, g/mol = 0.24 \, g \]

### Final Answer:
The amount of magnesium formed at the cathode of the other electrolytic cell is `0.24 grams`.

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