What is the oxidizing agent in the following equation ?
`HAsO_(2)(aq)+Sn^(2+)(aq)+H^(+)(aq)rarrAs(s)+Sn^(4+)(aq)+H_(2)Ol`

Options

A. `HAsO_(2)(aq)`
B. `Sn^(2+)(sq)`
C. `H^(+)`(aq)
D. `Sn^(4+)(aq)`

Explanation

To determine the oxidizing agent in the given reaction, we need to analyze the oxidation states of the elements involved and identify which species is being reduced. Here’s a step-by-step solution:

### Step 1: Write down the reaction
The reaction provided is:
\[ \text{HAsO}_2(aq) + \text{Sn}^{2+}(aq) + \text{H}^+(aq) \rightarrow \text{As}(s) + \text{Sn}^{4+}(aq) + \text{H}_2O(l) \]

### Step 2: Assign oxidation states
- In \(\text{HAsO}_2\):
- Hydrogen (H) = +1
- Oxygen (O) = -2 (there are 2 O, contributing -4)
- Let the oxidation state of Arsenic (As) be \(x\).
- The overall charge is neutral, so:
\[ +1 + x - 4 = 0 \implies x = +3 \]
Thus, the oxidation state of As in \(\text{HAsO}_2\) is +3.

- In \(\text{Sn}^{2+}\), the oxidation state of Sn is +2.

- In \(\text{As}(s)\), the oxidation state of As is 0 (elemental state).

- In \(\text{Sn}^{4+}\), the oxidation state of Sn is +4.

### Step 3: Identify changes in oxidation states
- Arsenic (As) changes from +3 in \(\text{HAsO}_2\) to 0 in \(\text{As}(s)\). This is a reduction (gain of electrons).
- Tin (Sn) changes from +2 in \(\text{Sn}^{2+}\) to +4 in \(\text{Sn}^{4+}\). This is an oxidation (loss of electrons).

### Step 4: Determine the oxidizing agent
The oxidizing agent is the species that gets reduced (gains electrons) in the reaction. In this case, \(\text{HAsO}_2\) is reduced as the oxidation state of As decreases from +3 to 0. Therefore, \(\text{HAsO}_2\) is the oxidizing agent.

### Conclusion
The oxidizing agent in the reaction is \(\text{HAsO}_2\).

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