For the reduction half reaction such as `Cr_2O_7rarr Cr^+3`by which number we should multiply the `Cr^+3` to balance the Cr atom?

Options

A. 3
B. 4
C. 2
D. (3/2)

Answer

To balance the chromium atoms in the reduction half-reaction \( \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+} \), follow these steps:

### Step-by-Step Solution:

1. **Identify the number of chromium atoms on each side**:
- In the reactant \( \text{Cr}_2\text{O}_7^{2-} \), there are **2 chromium (Cr)** atoms.
- In the product \( \text{Cr}^{3+} \), there is **1 chromium (Cr)** atom.

2. **Determine how many \( \text{Cr}^{3+} \) ions are needed to balance the chromium atoms**:
- Since there are 2 chromium atoms in the reactant, we need to have 2 chromium atoms in the product as well.

3. **Multiply the \( \text{Cr}^{3+} \) by the necessary factor**:
- To balance the 2 chromium atoms from the reactant side, we need to multiply \( \text{Cr}^{3+} \) by **2**.
- This gives us \( 2 \text{Cr}^{3+} \) on the product side.

4. **Write the balanced half-reaction**:
- The balanced reduction half-reaction is:
\[ \text{Cr}_2\text{O}_7^{2-} \rightarrow 2 \text{Cr}^{3+} \]

### Final Answer:
To balance the chromium atoms, you should multiply \( \text{Cr}^{3+} \) by **2**.

---