The charge required for the reduction of `1 mol` `Cr_(2)O_(7)^(2-)` ions to `Cr^(3+)` is

Options

A. `96500C`
B. `2xx96500C`
C. `3xx96500C`
D. `6xx96500C`

Explanation

To solve the problem of determining the charge required for the reduction of 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) ions to \( \text{Cr}^{3+} \), we will follow these steps:

### Step 1: Write the half-reaction for the reduction
The reduction of dichromate ions \( \text{Cr}_2\text{O}_7^{2-} \) to chromium ions \( \text{Cr}^{3+} \) can be represented as:
\[ \text{Cr}_2\text{O}_7^{2-} + \text{H}^+ + \text{e}^- \rightarrow \text{Cr}^{3+} + \text{H}_2\text{O} \]

### Step 2: Balance the chromium atoms
In \( \text{Cr}_2\text{O}_7^{2-} \), there are 2 chromium atoms. Therefore, we need to produce 2 \( \text{Cr}^{3+} \) ions:
\[ \text{Cr}_2\text{O}_7^{2-} + \text{H}^+ + \text{e}^- \rightarrow 2 \text{Cr}^{3+} + \text{H}_2\text{O} \]

### Step 3: Balance the oxygen atoms
The dichromate ion has 7 oxygen atoms. To balance the oxygen, we add 7 water molecules on the product side:
\[ \text{Cr}_2\text{O}_7^{2-} + \text{H}^+ + \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 4: Balance the hydrogen atoms
Now, we have 14 hydrogen atoms from the 7 water molecules. To balance the hydrogen, we add 14 \( \text{H}^+ \) ions on the reactant side:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 5: Balance the charges
Now we balance the charges. The left side has a total charge of:
\[ -2 + 14 = +12 \]
The right side has a total charge of:
\[ 2 \times +3 = +6 \]
To balance the charges, we need to add 6 electrons to the left side:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 6: Calculate the total charge
From the balanced equation, we see that 6 moles of electrons are required for the reduction. The charge associated with 1 mole of electrons (1 Faraday) is approximately 96500 Coulombs. Therefore, for 6 moles of electrons:
\[ \text{Total charge} = 6 \times 96500 \, \text{C} = 579000 \, \text{C} \]

### Final Answer
The charge required for the reduction of 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) ions to \( \text{Cr}^{3+} \) is:
\[ \text{579000 Coulombs} \quad \text{(or } 6 \times 96500 \text{ Coulombs)} \]