Identify the oxidant and reductant in the following reactions:
a. `10H^(o+)(aq)+4Zn(s)+NO_(3)^(ө)(aq)rarr4Zn^(2+)(aq)+NH_(4)^(o+)(aq)+3H_(2)O(l)`
b. `I_(2)(g)+H_(2)S(g)rarr2Hl(g)+S(s)`

Explanation

To identify the oxidant and reductant in the given reactions, we need to analyze the changes in oxidation states of the elements involved in each reaction. Let's go through each reaction step by step.

### Reaction a:
**Given Reaction:**
\[ 10H^+(aq) + 4Zn(s) + NO_3^-(aq) \rightarrow 4Zn^{2+}(aq) + NH_4^+(aq) + 3H_2O(l) \]

**Step 1: Assign Oxidation States**
- For \( H^+ \): +1
- For \( Zn \): 0 (elemental state)
- For \( N \) in \( NO_3^- \): +5
- For \( Zn^{2+} \): +2
- For \( N \) in \( NH_4^+ \): -3
- For \( O \) in \( H_2O \): -2

**Step 2: Determine Changes in Oxidation States**
- Zinc changes from 0 to +2 (oxidation).
- Nitrogen changes from +5 in \( NO_3^- \) to -3 in \( NH_4^+ \) (reduction).

**Step 3: Identify the Oxidant and Reductant**
- **Oxidant**: The species that is reduced (gains electrons). Here, \( NO_3^- \) is the oxidant because nitrogen is reduced from +5 to -3.
- **Reductant**: The species that is oxidized (loses electrons). Here, \( Zn \) is the reductant because it is oxidized from 0 to +2.

### Conclusion for Reaction a:
- **Oxidant**: \( NO_3^- \)
- **Reductant**: \( Zn \)

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### Reaction b:
**Given Reaction:**
\[ I_2(g) + H_2S(g) \rightarrow 2HI(g) + S(s) \]

**Step 1: Assign Oxidation States**
- For \( I_2 \): 0 (elemental state)
- For \( H_2S \): H: +1, S: -2
- For \( HI \): H: +1, I: -1
- For \( S \): 0 (elemental state)

**Step 2: Determine Changes in Oxidation States**
- Iodine changes from 0 in \( I_2 \) to -1 in \( HI \) (reduction).
- Sulfur changes from -2 in \( H_2S \) to 0 in elemental sulfur (oxidation).

**Step 3: Identify the Oxidant and Reductant**
- **Oxidant**: The species that is reduced. Here, \( I_2 \) is the oxidant because iodine is reduced from 0 to -1.
- **Reductant**: The species that is oxidized. Here, \( H_2S \) is the reductant because sulfur is oxidized from -2 to 0.

### Conclusion for Reaction b:
- **Oxidant**: \( I_2 \)
- **Reductant**: \( H_2S \)

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### Summary of Answers:
- **For Reaction a**:
- Oxidant: \( NO_3^- \)
- Reductant: \( Zn \)

- **For Reaction b**:
- Oxidant: \( I_2 \)
- Reductant: \( H_2S \)

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