How many grams of cobalt metal will be deposited when a solution of cobat (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes? (1 Faraday = 96,500C, Atomic mass of Co= 59u)

Options

A. 0.66
B. `4.0`
C. `20.0`
D. `40.0`

Answer

To solve the problem of how many grams of cobalt metal will be deposited when a solution of cobalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes, we can follow these steps:

### Step 1: Calculate the Total Charge (Q)
The total charge (Q) can be calculated using the formula:
\[ Q = I \times T \]
where:
- \( I \) = current in amperes (10 A)
- \( T \) = time in seconds

First, we need to convert the time from minutes to seconds:
\[ T = 109 \text{ minutes} \times 60 \text{ seconds/minute} = 6540 \text{ seconds} \]

Now, substituting the values:
\[ Q = 10 \text{ A} \times 6540 \text{ s} = 65400 \text{ C} \]

### Step 2: Calculate the Equivalent Weight (E)
The equivalent weight (E) can be calculated using the formula:
\[ E = \frac{\text{Molar Mass}}{n} \times F \]
where:
- Molar Mass of Co = 59 g/mol
- \( n \) = number of electrons transferred (for Co²⁺ to Co, \( n = 2 \))
- \( F \) = Faraday's constant = 96500 C/mol

Now, substituting the values:
\[ E = \frac{59 \text{ g/mol}}{2} \times 96500 \text{ C/mol} \]
\[ E = 29.5 \text{ g} \times 96500 \text{ C/mol} = 2846750 \text{ g/C} \]

### Step 3: Calculate the Weight of Cobalt Deposited
The weight of cobalt deposited can be calculated using the formula:
\[ \text{Weight} = Q \times \frac{E}{F} \]

Substituting the values:
\[ \text{Weight} = 65400 \text{ C} \times \frac{29.5 \text{ g}}{96500 \text{ C}} \]
\[ \text{Weight} = 65400 \times 0.000305 \text{ g/C} \]
\[ \text{Weight} \approx 20 \text{ g} \]

### Conclusion
The weight of cobalt metal deposited is approximately **20 grams**.

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