The conversion of `K_(2)Cr_(2)O_(7)` into `Cr_(2)(SO_(4))_(3)` is

Options

A. Oxidation
B. Reduction
C. Decomposition
D. Substitution

Answer

To determine the nature of the conversion of \( K_2Cr_2O_7 \) into \( Cr_2(SO_4)_3 \), we will analyze the oxidation states of chromium in both compounds.

### Step 1: Determine the oxidation state of chromium in \( K_2Cr_2O_7 \)

1. **Identify the oxidation states of the other elements**:
- Potassium (\( K \)) has an oxidation state of \( +1 \).
- Oxygen (\( O \)) has an oxidation state of \( -2 \).

2. **Set up the equation**:
- The formula for potassium dichromate is \( K_2Cr_2O_7 \).
- The total charge of the compound is 0.
- Therefore, we can write the equation:
\[ 2(+1) + 2x + 7(-2) = 0 \]
- Simplifying this gives:
\[ 2 + 2x - 14 = 0 \]
\[ 2x - 12 = 0 \]
\[ 2x = 12 \quad \Rightarrow \quad x = 6 \]

3. **Conclusion**:
- The oxidation state of chromium in \( K_2Cr_2O_7 \) is \( +6 \).

### Step 2: Determine the oxidation state of chromium in \( Cr_2(SO_4)_3 \)

1. **Identify the oxidation states of the other elements**:
- Sulfate (\( SO_4^{2-} \)) has a charge of \( -2 \).

2. **Set up the equation**:
- The formula for chromium(III) sulfate is \( Cr_2(SO_4)_3 \).
- The total charge of the compound is 0.
- Therefore, we can write the equation:
\[ 2x + 3(-2) = 0 \]
- Simplifying this gives:
\[ 2x - 6 = 0 \]
\[ 2x = 6 \quad \Rightarrow \quad x = 3 \]

3. **Conclusion**:
- The oxidation state of chromium in \( Cr_2(SO_4)_3 \) is \( +3 \).

### Step 3: Analyze the change in oxidation state

- The oxidation state of chromium changes from \( +6 \) in \( K_2Cr_2O_7 \) to \( +3 \) in \( Cr_2(SO_4)_3 \).
- This change indicates that chromium is being reduced (since the oxidation state decreases).

### Final Conclusion

The conversion of \( K_2Cr_2O_7 \) into \( Cr_2(SO_4)_3 \) is a **reduction reaction**.

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