Chromium metal can be plated out from an acidic solution containing `CrO_(3)` according to the following the reaction:
`CrO_(3) + 6H^(+) + 6e^(-) rarr Cr + 3H_(2)O`
How long will it take to plate out 1.5 gm of Cr using 12.5 ampere current ? (Atomic weight of Cr =52, 1F =96500 C)

Explanation

To solve the problem of how long it will take to plate out 1.5 grams of chromium using a current of 12.5 amperes, we will follow these steps:

### Step 1: Calculate the number of moles of chromium (Cr)
To find the number of moles of chromium, we use the formula:

\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

Given:
- Mass of Cr = 1.5 g
- Molar mass of Cr = 52 g/mol

Calculating the number of moles:

\[ \text{Number of moles of Cr} = \frac{1.5 \, \text{g}}{52 \, \text{g/mol}} = 0.02885 \, \text{mol} \]

### Step 2: Determine the number of moles of electrons required
From the balanced chemical equation:

\[ \text{CrO}_3 + 6\text{H}^+ + 6e^- \rightarrow \text{Cr} + 3\text{H}_2\text{O} \]

We see that 6 moles of electrons are required to reduce 1 mole of chromium. Therefore, for 0.02885 moles of chromium:

\[ \text{Moles of electrons} = 6 \times 0.02885 = 0.1731 \, \text{mol} \]

### Step 3: Calculate the total charge required
Using Faraday's law, we know that 1 mole of electrons corresponds to 96500 coulombs. Thus, the total charge (Q) required is:

\[ Q = \text{Moles of electrons} \times F \]

Where \( F = 96500 \, \text{C/mol} \):

\[ Q = 0.1731 \, \text{mol} \times 96500 \, \text{C/mol} = 16666.15 \, \text{C} \]

### Step 4: Calculate the time using the current
Using the formula for current:

\[ I = \frac{Q}{t} \quad \Rightarrow \quad t = \frac{Q}{I} \]

Where:
- \( I = 12.5 \, \text{A} \)

Substituting the values:

\[ t = \frac{16666.15 \, \text{C}}{12.5 \, \text{A}} = 1333.29 \, \text{s} \]

### Final Answer
The time required to plate out 1.5 grams of chromium using a current of 12.5 amperes is approximately **1333.29 seconds**.

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