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) passed through the solution.
The 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:
\[ Q = 10 \text{ A} \times 6540 \text{ seconds} = 65400 \text{ C} \]

### Step 2: Determine the number of moles of electrons transferred.
Using Faraday's constant, which is \( 1 \text{ Faraday} = 96500 \text{ C} \), we can find the number of moles of electrons (n) transferred:
\[ n = \frac{Q}{F} \]
where:
- \( F \) = Faraday's constant (96500 C)

Substituting the values:
\[ n = \frac{65400 \text{ C}}{96500 \text{ C/mol}} \approx 0.678 \text{ mol} \]

### Step 3: Calculate the equivalent weight of cobalt.
The equivalent weight (E) of cobalt can be calculated using the formula:
\[ E = \frac{\text{Molar mass}}{\text{n-factor}} \]
where:
- Molar mass of Co = 59 g/mol
- n-factor for Co in CoCl₂ = 2 (since it is Co²⁺)

Substituting the values:
\[ E = \frac{59 \text{ g/mol}}{2} = 29.5 \text{ g/equiv} \]

### Step 4: Calculate the mass of cobalt deposited.
The mass (m) of cobalt deposited can be calculated using the formula:
\[ m = n \times E \]
Substituting the values:
\[ m = 0.678 \text{ mol} \times 29.5 \text{ g/equiv} \approx 20.0 \text{ g} \]

### Final Answer:
Approximately 20 grams of cobalt metal will be deposited.

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