Calculate the quantity of electricity needed in coulombs to reduce one centimole of dichromate in acid medium, to chromic state

Options

A. 5790
B. 5780
C. 5770
D. 5760

Explanation

To calculate the quantity of electricity needed in coulombs to reduce one centimole of dichromate (\( \text{Cr}_2\text{O}_7^{2-} \)) in acidic medium to chromic state (\( \text{Cr}^{3+} \)), we can follow these steps:

### Step 1: Write the half-reaction
The reduction half-reaction for dichromate in acidic medium is:

\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 2: Identify the number of electrons involved
From the half-reaction, we can see that 6 electrons (\(6 \text{e}^-\)) are required to reduce one mole of dichromate to chromic ions.

### Step 3: Determine the amount of substance
Since we are dealing with one centimole of dichromate, we need to convert this to moles:

\[ 1 \text{ centimole} = 0.01 \text{ moles} \]

### Step 4: Calculate the total number of electrons needed
To find the total number of electrons needed for 0.01 moles of dichromate, we multiply the number of moles by the number of electrons per mole:

\[ \text{Total electrons} = 0.01 \text{ moles} \times 6 \text{ electrons/mole} = 0.06 \text{ moles of electrons} \]

### Step 5: Convert moles of electrons to coulombs
Using Faraday's constant, which is approximately \(96500 \, \text{C/mol}\), we can convert moles of electrons to coulombs:

\[ \text{Charge (C)} = \text{moles of electrons} \times \text{Faraday's constant} \]

\[ \text{Charge (C)} = 0.06 \text{ moles} \times 96500 \, \text{C/mol} = 5790 \, \text{C} \]

### Final Answer
The quantity of electricity needed to reduce one centimole of dichromate in acidic medium to chromic state is **5790 coulombs**.

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