Acidified potassium dichromate reacts with potassium iodide and oxidiese it to `I_2`. What is the oxidation state of chromium in the products of the reaction?

Options

A. `+4`
B. `+6`
C. `+3`
D. `+2`

Explanation

To determine the oxidation state of chromium in the products of the reaction between acidified potassium dichromate (K2Cr2O7) and potassium iodide (KI), we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
When potassium dichromate reacts with potassium iodide in an acidic medium (like sulfuric acid), the reaction can be written as follows:

\[ K_2Cr_2O_7 + 6KI + 7H_2SO_4 \rightarrow 2Cr_2(SO_4)_3 + 3I_2 + 6K_2SO_4 + 7H_2O \]

### Step 2: Identify the oxidation states of the elements involved.
- In potassium dichromate (K2Cr2O7), the oxidation state of chromium (Cr) is +6.
- In potassium iodide (KI), the oxidation state of iodine (I) is -1.
- In the product iodine (I2), the oxidation state of iodine is 0.

### Step 3: Determine the change in oxidation states.
- Iodine is oxidized from -1 (in KI) to 0 (in I2), indicating that it loses electrons.
- Chromium is reduced from +6 (in K2Cr2O7) to +3 (in Cr2(SO4)3), indicating that it gains electrons.

### Step 4: Identify the oxidation state of chromium in the products.
In the product chromium sulfate (Cr2(SO4)3), the oxidation state of chromium is +3.

### Conclusion:
The oxidation state of chromium in the products of the reaction is +3.

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