For reaction :
`Cr_(2)O_(7)^(-2)+14H^(+)rarr2Cr^(+3)+7H_(2)O`, How many `e^(-)s` are required

Options

A. 3
B. 4
C. 5
D. 6

Answer

To determine how many electrons are required for the reaction:

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

we will follow these steps:

### Step 1: Determine the oxidation states
First, we need to determine the oxidation states of chromium in the dichromate ion (\( \text{Cr}_2\text{O}_7^{2-} \)) and in the chromium ion (\( \text{Cr}^{3+} \)).

- In \( \text{Cr}_2\text{O}_7^{2-} \):
- Let the oxidation state of chromium be \( x \).
- The oxidation state of oxygen is \(-2\).
- The equation for the overall charge is:

\[ 2x + 7(-2) = -2 \]

Simplifying this gives:

\[ 2x - 14 = -2 \implies 2x = 12 \implies x = +6 \]

- In \( \text{Cr}^{3+} \):
- The oxidation state of chromium is \( +3 \).

### Step 2: Calculate the change in oxidation state
Now we calculate the change in oxidation state for chromium:

- From \( +6 \) in \( \text{Cr}_2\text{O}_7^{2-} \) to \( +3 \) in \( \text{Cr}^{3+} \), the change is:

\[ 6 - 3 = 3 \]

### Step 3: Determine the total change for the reaction
Since there are 2 chromium atoms in \( \text{Cr}_2\text{O}_7^{2-} \), the total change in oxidation state for both chromium atoms is:

\[ 2 \times 3 = 6 \]

### Step 4: Conclude the number of electrons
The total decrease in oxidation state (which corresponds to the number of electrons gained) is 6. Therefore, the reaction requires 6 electrons.

### Final Answer
Thus, the number of electrons required for the reaction is:

\[ \boxed{6} \]

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