Which of the following samples of reducing agents is /are chemically equivalent to 25mL of 0.2 N `KMnO_(4)` to be reduced to `Mn^(2+)` and water?

Options

A. 25 ml of 0.2 M `FeSO_4` to be oxidized to `Fe^(3+)`
B. 50 ml of 0.1 M `H_3AsO_3` to be oxidized to `H_3AsO_4`
C. 25 ml of 0.1 M `H_2O_2` to be oxidized to `H^+ and O_2`
D. 25 ml of 0.1 M `SnCl_2` to be oxidized to `Sn^(4+)`

Explanation

To solve the problem of determining which reducing agents are chemically equivalent to 25 mL of 0.2 N KMnO4, we will follow these steps:

### Step 1: Calculate the milliequivalents of KMnO4
To find the milliequivalents (mEq) of KMnO4, we use the formula:

\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (in L)} \]

Given:
- Normality (N) = 0.2 N
- Volume (V) = 25 mL = 0.025 L

Calculating:

\[ \text{Milliequivalents} = 0.2 \, \text{N} \times 0.025 \, \text{L} = 0.005 \, \text{equivalents} = 5 \, \text{mEq} \]

### Step 2: Check each option for equivalence
Now we will analyze each option to see if it provides 5 mEq of reducing agent.

#### Option A: 25 mL of 0.2 M FeSO4
- Molarity (M) = 0.2 M
- Volume (V) = 25 mL = 0.025 L
- n-factor for FeSO4 (oxidation state change from Fe^2+ to Fe^3+) = 1

Calculating:

\[ \text{Milliequivalents} = \text{Molarity} \times \text{Volume} \times \text{n-factor} = 0.2 \times 0.025 \times 1 = 0.005 \, \text{equivalents} = 5 \, \text{mEq} \]

**Conclusion**: Option A is correct.

#### Option B: 50 mL of 0.1 N H3AsO3
- Normality (N) = 0.1 N
- Volume (V) = 50 mL = 0.050 L
- n-factor for H3AsO3 (oxidation state change from +3 to +5) = 2

Calculating:

\[ \text{Milliequivalents} = 0.1 \times 0.050 \times 2 = 0.01 \, \text{equivalents} = 10 \, \text{mEq} \]

**Conclusion**: Option B is incorrect.

#### Option C: 25 mL of 0.2 M H2O2
- Molarity (M) = 0.2 M
- Volume (V) = 25 mL = 0.025 L
- n-factor for H2O2 (oxidation state change from -1 to 0) = 2

Calculating:

\[ \text{Milliequivalents} = 0.2 \times 0.025 \times 2 = 0.01 \, \text{equivalents} = 5 \, \text{mEq} \]

**Conclusion**: Option C is correct.

#### Option D: 25 mL of 0.2 M SnCl2
- Molarity (M) = 0.2 M
- Volume (V) = 25 mL = 0.025 L
- n-factor for SnCl2 (oxidation state change from +2 to +4) = 2

Calculating:

\[ \text{Milliequivalents} = 0.2 \times 0.025 \times 2 = 0.01 \, \text{equivalents} = 5 \, \text{mEq} \]

**Conclusion**: Option D is correct.

### Final Answer:
The reducing agents that are chemically equivalent to 25 mL of 0.2 N KMnO4 are:
- Option A (FeSO4)
- Option C (H2O2)
- Option D (SnCl2)