Beryllium occurs naturally in the form of beryl. The metal is produced from its ore by electrolysis after the ore has been converted to the oxide and then to the chloride.How many grams of Be(s) is deposited form a`BeCl_2` solution by a current of 5.0 A that flows for 1.0 h? (Atomic weight:Be=9)

Options

A. (a) 0.84
B. (b) 1.68
C. (c) 1.42
D. (d) 1.08

Answer

To solve the problem of how many grams of beryllium (Be) are deposited from a BeCl₂ solution by a current of 5.0 A flowing for 1.0 hour, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current (I) = 5.0 A
- Time (t) = 1.0 hour = 1.0 × 60 × 60 seconds = 3600 seconds
- Atomic weight of Beryllium (Be) = 9 g/mol
- Beryllium in BeCl₂ has an oxidation state of +2, meaning it requires 2 electrons (N = 2) to be deposited.

2. **Calculate the Total Charge (Q):**
- The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
- Substituting the values:
\[ Q = 5.0 \, \text{A} \times 3600 \, \text{s} = 18000 \, \text{C} \]

3. **Determine the Electrochemical Equivalent (Z):**
- The electrochemical equivalent (Z) is given by:
\[ Z = \frac{\text{Equivalent weight}}{96500} \]
- The equivalent weight is calculated as:
\[ \text{Equivalent weight} = \frac{\text{Atomic weight}}{N} = \frac{9}{2} = 4.5 \, \text{g/equiv} \]
- Therefore, Z becomes:
\[ Z = \frac{4.5}{96500} \, \text{g/C} \]

4. **Calculate the Weight (W) of Beryllium Deposited:**
- Using Faraday's first law of electrolysis:
\[ W = Z \times Q \]
- Substituting the values of Z and Q:
\[ W = \left(\frac{4.5}{96500}\right) \times 18000 \]
- Calculating this gives:
\[ W = \frac{81000}{96500} \approx 0.839 \, \text{g} \]

5. **Final Result:**
- The weight of beryllium deposited is approximately 0.84 grams.

### Summary:
The weight of beryllium deposited from a BeCl₂ solution by a current of 5.0 A flowing for 1.0 hour is approximately **0.84 grams**.

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