Consider the reaction `:`
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2Cr^(3+)+7H_(2)O`
What is the quantity of electricity in coulombs needed to reduce `1 mol e` of `Cr_(2)O_(7)^(2-)` ?

Explanation

To determine the quantity of electricity in coulombs needed to reduce 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \), we can follow these steps:

### Step 1: Identify the number of electrons involved in the reduction
From the given reaction:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
we can see that 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) requires 6 moles of electrons (6 \( e^- \)) for reduction.

### Step 2: Calculate the charge required for 6 moles of electrons
The charge of 1 mole of electrons is known as Faraday's constant, which is approximately \( 96500 \) coulombs. Therefore, the total charge required for 6 moles of electrons is calculated as follows:
\[ \text{Charge} = \text{Number of moles of electrons} \times \text{Faraday's constant} \]
\[ \text{Charge} = 6 \, \text{moles} \times 96500 \, \text{C/mole} \]

### Step 3: Perform the multiplication
Now we multiply:
\[ \text{Charge} = 6 \times 96500 = 579000 \, \text{C} \]

### Step 4: Conclusion
Thus, the quantity of electricity in coulombs needed to reduce 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) is \( 579000 \) coulombs.

### Summary
The final answer is:
\[ \text{Charge required} = 579000 \, \text{C} \]

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