How many electrons are involved in the following redox reaction?
`Cr_(2)O_(7)^(2-)+Fe^(2+)+C_(2)O_(4)^(2-)toCr^(3+)+Fe^(3+)+CO_(2)`

Options

A. 3
B. 4
C. 6
D. 7

Explanation

To determine how many electrons are involved in the given redox reaction:

**Reaction:**
\[ \text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} + \text{C}_2\text{O}_4^{2-} \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+} + \text{CO}_2 \]

### Step 1: Identify the oxidation and reduction half-reactions.

1. **Oxidation Half-Reaction:**
The oxidation half-reaction involves the conversion of \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \).
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + 1e^- \]
(1 electron is lost in this process)

2. **Reduction Half-Reaction:**
The reduction half-reaction involves the conversion of \( \text{Cr}_2\text{O}_7^{2-} \) to \( \text{Cr}^{3+} \).
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
(6 electrons are gained in this process)

3. **Another Oxidation Half-Reaction:**
The conversion of \( \text{C}_2\text{O}_4^{2-} \) to \( \text{CO}_2 \):
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^- \]
(2 electrons are lost in this process)

### Step 2: Balance the half-reactions.

Now we need to balance the electrons in the half-reactions so that they can be combined.

- The oxidation of \( \text{Fe}^{2+} \) produces 1 electron.
- The oxidation of \( \text{C}_2\text{O}_4^{2-} \) produces 2 electrons.
- The reduction of \( \text{Cr}_2\text{O}_7^{2-} \) consumes 6 electrons.

### Step 3: Equalize the number of electrons.

To equalize the electrons, we can multiply the oxidation reactions by appropriate coefficients:

- Multiply the \( \text{Fe}^{2+} \) oxidation half-reaction by 6:
\[ 6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^- \]

- Multiply the \( \text{C}_2\text{O}_4^{2-} \) oxidation half-reaction by 3:
\[ 3\text{C}_2\text{O}_4^{2-} \rightarrow 6\text{CO}_2 + 6e^- \]

### Step 4: Combine the half-reactions.

Now we can combine all the half-reactions:

1. From \( \text{Cr}_2\text{O}_7^{2-} \):
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

2. From \( 6\text{Fe}^{2+} \):
\[ 6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^- \]

3. From \( 3\text{C}_2\text{O}_4^{2-} \):
\[ 3\text{C}_2\text{O}_4^{2-} \rightarrow 6\text{CO}_2 + 6e^- \]

### Step 5: Count the total electrons involved.

In the overall reaction, we see that:

- 6 electrons are lost from \( 6\text{Fe}^{2+} \) and \( 3\text{C}_2\text{O}_4^{2-} \).
- 6 electrons are gained by \( \text{Cr}_2\text{O}_7^{2-} \).

Thus, the total number of electrons involved in the overall redox reaction is **6 electrons**.

### Final Answer:
**Total electrons involved: 6 electrons.**

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