https://file.aiquickdraw.com/mathbot/file/17287019101382yomqmho.jpg

Let's solve the oxidation-reduction problems step by step.

### Part A:
**Identify the changes in oxidation number and how many electrons are lost or gained:**

1. **\( I_2O_5(s) + 5CO(g) \rightarrow I_2(s) + 5CO_2(g) \)**
   - **Oxidation states:**
     - In \( I_2O_5 \), iodine (I) has an oxidation state of +5 (since oxygen is -2, and the molecule must sum to zero).
     - In \( I_2(s) \), iodine (I) is 0 (as it is in elemental form).
     - In \( CO \), carbon (C) is +2.
     - In \( CO_2 \), carbon (C) is +4.
   - **Oxidation-reduction:**
     - Iodine goes from +5 to 0 (reduction, gains 5 electrons).
     - Carbon goes from +2 to +4 (oxidation, loses 2 electrons).
   
2. **\( 2Hg^{2+}(aq) + N_2H_4(aq) \rightarrow 2Hg(l) + N_2(g) + 4H^+(aq) \)**
   - **Oxidation states:**
     - \( Hg^{2+} \) starts at +2.
     - In \( Hg(l) \), mercury is 0 (elemental form).
     - In \( N_2H_4 \), nitrogen is -2 (since hydrogen is +1).
     - In \( N_2(g) \), nitrogen is 0 (elemental form).
   - **Oxidation-reduction:**
     - Mercury goes from +2 to 0 (reduction, gains 2 electrons).
     - Nitrogen goes from -2 to 0 (oxidation, loses 2 electrons).

3. **\( 3H_2S(aq) + 2H^+(aq) + 2NO_3^-(aq) \rightarrow 3S(s) + 2NO(g) + 4H_2O(l) \)**
   - **Oxidation states:**
     - In \( H_2S \), sulfur (S) is -2.
     - In \( S(s) \), sulfur (S) is 0 (elemental form).
     - In \( NO_3^- \), nitrogen is +5.
     - In \( NO \), nitrogen is +2.
   - **Oxidation-reduction:**
     - Sulfur goes from -2 to 0 (oxidation, loses 2 electrons).
     - Nitrogen goes from +5 to +2 (reduction, gains 3 electrons).

### Part B:
**Determine if these are redox reactions and identify oxidation number changes:**

4. **\( PBr_3(l) + 3H_2O(l) \rightarrow H_3PO_3(aq) + 3HBr(aq) \)**
   - Phosphorus (P) remains +3 in both \( PBr_3 \) and \( H_3PO_3 \), and bromine (Br) stays at -1. There is no change in oxidation numbers, so this is **not a redox reaction**.

5. **\( NaI(aq) + 3HOCl(aq) \rightarrow NaIO_3(aq) + 3HCl(aq) \)**
   - **Oxidation states:**
     - In \( NaI \), iodine (I) is -1.
     - In \( NaIO_3 \), iodine (I) is +5.
     - In \( HOCl \), chlorine (Cl) is +1.
     - In \( HCl \), chlorine (Cl) is -1.
   - Iodine goes from -1 to +5 (oxidation), and chlorine goes from +1 to -1 (reduction). This is a **redox reaction**.

6. **\( 3SO_2(g) + 2HNO_3(aq) \rightarrow 3H_2SO_4(aq) + 2NO(g) \)**
   - **Oxidation states:**
     - In \( SO_2 \), sulfur (S) is +4.
     - In \( H_2SO_4 \), sulfur (S) is +6.
     - In \( HNO_3 \), nitrogen is +5.
     - In \( NO \), nitrogen is +2.
   - Sulfur goes from +4 to +6 (oxidation), and nitrogen goes from +5 to +2 (reduction). This is a **redox reaction**.

Do you need further details or explanations?

### Follow-up questions:
1. What is the significance of oxidation states in redox reactions?
2. How do you determine if a reaction is a redox reaction?
3. Why do some reactions not involve changes in oxidation number?
4. What are some common oxidizing and reducing agents?
5. How can the number of electrons lost and gained in a reaction be balanced?

**Tip:** Always remember that in a redox reaction, one element is oxidized (loses electrons), and another is reduced (gains electrons).

In each of the following balanced oxidation-reduction equations, identify the elements that undergo changes in oxidation number and indicate how many electrons are lost or gained. Also, indicate whether certain reactions involve oxidation-reduction and identify changes in oxidation numbers.

Explore detailed step-by-step solutions for oxidation-reduction reactions, including identifying elements that undergo oxidation number changes, electron transfers, and redox analysis. Includes examples such as I2O5 and CO reactions, NaI with HOCl, and more. Suitable for advanced chemistry students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation