`K_(2)Cr_(2)O_(7)+SO_(2) overset(H+)to`products
the oxidation state of chromium changes from

Options

A. `+12` to +6
B. `+3` to +6
C. `+6` to +3
D. `+6` to +4

Answer

To determine the change in the oxidation state of chromium in the reaction between potassium dichromate (K₂Cr₂O₇) and sulfur dioxide (SO₂) in acidic medium, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between potassium dichromate and sulfur dioxide in acidic medium can be represented as:
\[ K_2Cr_2O_7 + 3SO_2 + 4H_2SO_4 \rightarrow 2Cr_2(SO_4)_3 + K_2SO_4 + 4H_2O \]

### Step 2: Determine the oxidation state of chromium in potassium dichromate (K₂Cr₂O₇)
In potassium dichromate:
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of chromium (Cr) be \( x \).

The formula for the oxidation state can be set up as follows:
\[ 2(+1) + 2x + 7(-2) = 0 \]
\[ 2 + 2x - 14 = 0 \]
\[ 2x - 12 = 0 \]
\[ 2x = 12 \]
\[ x = +6 \]

Thus, the oxidation state of chromium in K₂Cr₂O₇ is +6.

### Step 3: Determine the oxidation state of chromium in the product (Cr₂(SO₄)₃)
In the product, chromium is part of chromium(III) sulfate:
- The sulfate ion (SO₄) has an oxidation state of -2.
- Let the oxidation state of chromium in the product be \( y \).

The formula for the oxidation state can be set up as follows:
\[ 2y + 3(-2) = 0 \]
\[ 2y - 6 = 0 \]
\[ 2y = 6 \]
\[ y = +3 \]

Thus, the oxidation state of chromium in Cr₂(SO₄)₃ is +3.

### Step 4: Determine the change in oxidation state
Now we can summarize the change in oxidation state of chromium:
- Initial oxidation state in K₂Cr₂O₇: +6
- Final oxidation state in Cr₂(SO₄)₃: +3

### Conclusion
The oxidation state of chromium changes from +6 to +3 during the reaction.

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