In the reaction
` 2K MnO_4 + 16 HCl rarr 5Cl_2 + 2MnCl_2 + 2 KCl+ 8H_2O` the reduced product is .

Options

A. `Cl_2`
B. ` KCl`
C. ` H_2O`
D. `MnCl_2`

Explanation

To determine the reduced product in the given reaction:

**Step 1: Identify the reaction and its components.**
The reaction is:
\[ 2 \text{KMnO}_4 + 16 \text{HCl} \rightarrow 5 \text{Cl}_2 + 2 \text{MnCl}_2 + 2 \text{KCl} + 8 \text{H}_2\text{O} \]

**Step 2: Assign oxidation states to the elements in the reactants and products.**
- In KMnO4:
- K: +1
- Mn: +7
- O: -2 (4 O atoms contribute -8, so Mn must be +7 to balance)

- In HCl:
- H: +1
- Cl: -1

- In Cl2:
- Cl: 0 (as it is in elemental form)

- In MnCl2:
- Mn: +2
- Cl: -1 (2 Cl atoms contribute -2, so Mn must be +2 to balance)

- In KCl:
- K: +1
- Cl: -1

- In H2O:
- H: +1
- O: -2

**Step 3: Determine changes in oxidation states.**
- Mn changes from +7 (in KMnO4) to +2 (in MnCl2). This indicates a reduction (decrease in oxidation state).
- Cl changes from -1 (in HCl) to 0 (in Cl2). This indicates oxidation (increase in oxidation state).

**Step 4: Identify the reduced product.**
The species that is reduced is the one whose oxidation state decreases. Here, Mn is reduced from +7 to +2 in the formation of MnCl2.

**Conclusion:**
The reduced product in the reaction is **MnCl2**.

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