2. can oxidation occur without reduction? explain.
3. why would a nail corrode more quickly in saltwater than in distilled water?
4. a. what is the oxidation number of any atom in the elemental state?
b. what is the oxidation number of any monatomic ion?
c. what is the sum of the oxidation numbers in a neutral compound equal?
d. what is the sum of the oxidation numbers in a polyatomic ion equal?
5. determine the oxidation number of phosphorous in each substance.
a. p₄o₈
b. p
c. p₄o₆
d. h₃po₄
e. po₃³⁻
f. po₄³⁻
6. determine the oxidation numbers of each element in each of the following compounds.
a. s₂o₃
b. cl₂
c. co₃²⁻
d. no₃
e. kcl
f. he
g. no₂
h. h₃n
7. if a substance is
educed\ does it gain or lose electrons? ______ if a substance is \oxidized\ does it gain or lose electrons? ______
8. distinguish between an oxidizing agent and a reducing agent.
9. use the changes in oxidation numbers to identify which atom is oxidized and which is reduced. then, determine which reactant is the oxidizing agent and which is the reducing agent. (show your work.)
a. h₂ + cl₂ → 2 hcl
oxidized ______
reduced ______
oxidizing agent ______
reducing agent ______

Question

2. can oxidation occur without reduction? explain.
3. why would a nail corrode more quickly in saltwater than in distilled water?
4. a. what is the oxidation number of any atom in the elemental state?
   b. what is the oxidation number of any monatomic ion?
   c. what is the sum of the oxidation numbers in a neutral compound equal?
   d. what is the sum of the oxidation numbers in a polyatomic ion equal?
5. determine the oxidation number of phosphorous in each substance.
   a. p₄o₈
   b. p
   c. p₄o₆
   d. h₃po₄
   e. po₃³⁻
   f. po₄³⁻
6. determine the oxidation numbers of each element in each of the following compounds.
   a. s₂o₃
   b. cl₂
   c. co₃²⁻
   d. no₃
   e. kcl
   f. he
   g. no₂
   h. h₃n
7. if a substance is
educed\ does it gain or lose electrons? ______ if a substance is \oxidized\ does it gain or lose electrons? ______
8. distinguish between an oxidizing agent and a reducing agent.
9. use the changes in oxidation numbers to identify which atom is oxidized and which is reduced. then, determine which reactant is the oxidizing agent and which is the reducing agent. (show your work.)
   a. h₂ + cl₂ → 2 hcl
      oxidized ______
      reduced ______
      oxidizing agent ______
      reducing agent ______

Accepted Answer

### Question 2: Can oxidation occur without reduction? Explain.
# Brief Explanations:
Oxidation - reduction (redox) reactions involve the transfer of electrons. Oxidation is the loss of electrons, and reduction is the gain of electrons. In any redox reaction, when one species loses electrons (oxidized), another must gain those electrons (reduced). So, oxidation and reduction are complementary processes and cannot occur independently.
# Answer:
No, oxidation cannot occur without reduction. Oxidation is the loss of electrons, and reduction is the gain of electrons. In a redox reaction, electrons lost by one species (during oxidation) must be gained by another species (during reduction), so they occur simultaneously.

### Question 3: Why would a nail corrode more quickly in saltwater than in distilled water?
# Brief Explanations:
Corrosion of a nail (usually iron) is an electrochemical process (rusting, a redox reaction). Saltwater contains dissolved ions (like $ \text{Na}^+ $ and $ \text{Cl}^- $), which increase the electrical conductivity of the water. This allows the electrochemical reactions (oxidation of $ \text{Fe} $ and reduction of $ \text{O}_2 $) involved in corrosion to proceed more rapidly compared to distilled water, which has very low conductivity due to the absence of significant ions.
# Answer:
A nail corrodes faster in saltwater because saltwater contains dissolved ions (e.g., $ \text{Na}^+ $, $ \text{Cl}^- $) that increase the electrical conductivity of the water. This enhanced conductivity accelerates the electrochemical (redox) reactions involved in the corrosion (rusting) process, unlike distilled water, which has low conductivity due to a lack of significant ions.

### Question 4:
#### a. What is the oxidation number of any atom in the elemental state?
# Brief Explanations:
By definition, the oxidation number of an atom in its elemental state (e.g., $ \text{O}_2 $, $ \text{Fe} $, $ \text{Ne} $) is 0. This is because in the elemental form, the atom has no net loss or gain of electrons relative to its neutral state.
# Answer:
0

#### b. What is the oxidation number of any monatomic ion?
# Brief Explanations:
For a monatomic ion (e.g., $ \text{Na}^+ $, $ \text{Cl}^- $, $ \text{Mg}^{2+} $), the oxidation number is equal to the charge of the ion. This is because the ion's charge is determined by the number of electrons gained or lost, which defines its oxidation state.
# Answer:
Equal to the charge of the ion (e.g., $ \text{Na}^+ $ has an oxidation number of +1, $ \text{Cl}^- $ has -1).

#### c. What is the sum of the oxidation numbers in a neutral compound equal?
# Brief Explanations:
In a neutral compound (e.g., $ \text{H}_2\text{O} $, $ \text{NaCl} $), the sum of the oxidation numbers of all atoms is 0. This is because the compound has no net charge, so the total positive oxidation numbers must balance the total negative oxidation numbers.
# Answer:
0

#### d. What is the sum of the oxidation numbers in a polyatomic ion equal?
# Brief Explanations:
For a polyatomic ion (e.g., $ \text{SO}_4^{2-} $, $ \text{NH}_4^+ $), the sum of the oxidation numbers of all atoms is equal to the charge of the ion. This is because the ion has a net charge, and the sum of the oxidation states of its constituent atoms must equal that charge.
# Answer:
Equal to the charge of the polyatomic ion (e.g., $ \text{SO}_4^{2-} $ has a sum of -2, $ \text{NH}_4^+ $ has a sum of +1).

### Question 5: Determine the oxidation number of phosphorous in each substance.
#### a. $ \text{P}_4\text{O}_8 $
# Explanation:
## Step 1: Recall oxygen's oxidation number.
Oxygen in compounds (except peroxides, superoxides) has an oxidation number of -2. Let the oxidation number of P be $ x $.
## Step 2: Set up the equation for the sum of oxidation numbers.
In $ \text{P}_4\text{O}_8 $, there are 4 P atoms and 8 O atoms. The sum of oxidation numbers in a neutral compound is 0. So:
$ 4x + 8(-2) = 0 $
## Step 3: Solve for $ x $.
$ 4x - 16 = 0 $
$ 4x = 16 $
$ x = 4 $
# Answer:
+4

#### b. $ \text{P} $ (elemental state)
# Brief Explanations:
As per the rule for elemental state, the oxidation number of an atom in its elemental form is 0.
# Answer:
0

#### c. $ \text{P}_4\text{O}_6 $
# Explanation:
## Step 1: Oxygen's oxidation number is -2. Let P's oxidation number be $ x $.
## Step 2: Sum of oxidation numbers in neutral compound is 0.
$ 4x + 6(-2) = 0 $
## Step 3: Solve.
$ 4x - 12 = 0 $
$ 4x = 12 $
$ x = 3 $
# Answer:
+3

#### d. $ \text{H}_3\text{PO}_4 $
# Explanation:
## Step 1: Hydrogen in compounds (except metal hydrides) has +1, oxygen has -2. Let P's oxidation number be $ x $.
## Step 2: Sum of oxidation numbers in neutral compound is 0.
$ 3(+1) + x + 4(-2) = 0 $
## Step 3: Solve.
$ 3 + x - 8 = 0 $
$ x - 5 = 0 $
$ x = 5 $
# Answer:
+5

#### e. $ \text{PO}_3^{3-} $ (polyatomic ion)
# Explanation:
## Step 1: Oxygen has -2. Let P's oxidation number be $ x $. The sum of oxidation numbers equals the ion's charge (-3).
$ x + 3(-2) = -3 $
## Step 2: Solve.
$ x - 6 = -3 $
$ x = 3 $
# Answer:
+3

#### f. $ \text{PO}_4^{3-} $ (polyatomic ion)
# Explanation:
## Step 1: Oxygen has -2. Let P's oxidation number be $ x $. Sum equals ion's charge (-3).
$ x + 4(-2) = -3 $
## Step 2: Solve.
$ x - 8 = -3 $
$ x = 5 $
# Answer:
+5

### Question 6: Determine the oxidation numbers of each element in each of the following compounds.
#### a. $ \text{S}_2\text{O}_3 $ (neutral compound)
# Explanation:
## Step 1: Oxygen has -2. Let S's oxidation number be $ x $. Sum of oxidation numbers is 0.
$ 2x + 3(-2) = 0 $
## Step 2: Solve.
$ 2x - 6 = 0 $
$ 2x = 6 $
$ x = 3 $
So, S: +3, O: -2.
# Answer:
S: +3, O: -2

#### b. $ \text{Cl}_2 $ (elemental state)
# Brief Explanations:
In elemental form, the oxidation number of Cl is 0.
# Answer:
Cl: 0

#### c. $ \text{CO}_3^{2-} $ (polyatomic ion)
# Explanation:
## Step 1: Oxygen has -2. Let C's oxidation number be $ x $. Sum equals ion's charge (-2).
$ x + 3(-2) = -2 $
## Step 2: Solve.
$ x - 6 = -2 $
$ x = 4 $
So, C: +4, O: -2.
# Answer:
C: +4, O: -2

#### d. $ \text{NO}_3 $ (assuming $ \text{NO}_3^- $ ion, likely a typo; if neutral, but $ \text{NO}_3 $ is unstable, so assuming $ \text{NO}_3^- $)
# Explanation (for $ \text{NO}_3^- $):
## Step 1: Oxygen has -2. Let N's oxidation number be $ x $. Sum equals ion's charge (-1).
$ x + 3(-2) = -1 $
## Step 2: Solve.
$ x - 6 = -1 $
$ x = 5 $
So, N: +5, O: -2.
# Answer (for $ \text{NO}_3^- $):
N: +5, O: -2 (Note: If $ \text{NO}_3 $ is intended as neutral, it's unstable, but likely a typo for $ \text{NO}_3^- $)

#### e. $ \text{KCl} $ (neutral compound)
# Explanation:
Potassium (group 1 metal) has +1. Let Cl's oxidation number be $ x $. Sum is 0.
$ +1 + x = 0 $
$ x = -1 $
So, K: +1, Cl: -1.
# Answer:
K: +1, Cl: -1

#### f. $ \text{He} $ (elemental state)
# Brief Explanations:
In elemental form, the oxidation number of He is 0.
# Answer:
He: 0

#### g. $ \text{NO}_2 $ (neutral compound)
# Explanation:
## Step 1: Oxygen has -2. Let N's oxidation number be $ x $. Sum is 0.
$ x + 2(-2) = 0 $
## Step 2: Solve.
$ x - 4 = 0 $
$ x = 4 $
So, N: +4, O: -2.
# Answer:
N: +4, O: -2

#### h. $ \text{H}_3\text{N} $ (assuming $ \text{NH}_3 $, ammonia)
# Explanation:
Hydrogen has +1. Let N's oxidation number be $ x $. Sum is 0 (neutral compound).
$ 3(+1) + x = 0 $
$ 3 + x = 0 $
$ x = -3 $
So, H: +1, N: -3.
# Answer:
H: +1, N: -3 (Note: $ \text{H}_3\text{N} $ is $ \text{NH}_3 $)

### Question 7: If a substance is “reduced” does it gain or lose electrons? If a substance is “oxidized” does it gain or lose electrons?
# Brief Explanations:
- Reduction: In redox chemistry, reduction is defined as the gain of electrons. When a substance gains electrons, its oxidation number decreases (becomes more negative or less positive).
- Oxidation: Oxidation is defined as the loss of electrons. When a substance loses electrons, its oxidation number increases (becomes more positive or less negative).
# Answer:
- A reduced substance **gains** electrons.
- An oxidized substance **loses** electrons.

### Question 8: Distinguish between an oxidizing agent and a reducing agent.
# Brief Explanations:
- Oxidizing Agent: A substance that causes another substance to be oxidized (by accepting electrons from it). In the process, the oxidizing agent is reduced (gains electrons, oxidation number decreases).
- Reducing Agent: A substance that causes another substance to be reduced (by donating electrons to it). In the process, the reducing agent is oxidized (loses electrons, oxidation number increases).
# Answer:
- **Oxidizing Agent**: Causes oxidation in another substance (accepts electrons) and is itself reduced (gains electrons, oxidation number decreases).
- **Reducing Agent**: Causes reduction in another substance (donates electrons) and is itself oxidized (loses electrons, oxidation number increases).

### Question 9a: $ \text{H}_2 + \text{Cl}_2
ightarrow 2\text{HCl} $
# Explanation:
1. **Oxidation Numbers Initial**:
   - $ \text{H}_2 $: H is 0 (elemental state).
   - $ \text{Cl}_2 $: Cl is 0 (elemental state).
2. **Oxidation Numbers Final (in $ \text{HCl} $)**:
   - H: +1 (in compound), Cl: -1 (in compound).
3. **Identify Oxidized/Reduced**:
   - H in $ \text{H}_2 $ goes from 0 to +1: loses electrons (oxidized).
   - Cl in $ \text{Cl}_2 $ goes from 0 to -1: gains electrons (reduced).
4. **Identify Agents**:
   - Oxidizing Agent: The substance reduced ( $ \text{Cl}_2 $ ), as it causes H to be oxidized.
   - Reducing Agent: The substance oxidized ( $ \text{H}_2 $ ), as it causes Cl to be reduced.

# Answer:
- Oxidized: $ \text{H} $ (in $ \text{H}_2 $)
- Reduced: $ \text{Cl} $ (in $ \text{Cl}_2 $)
- Oxidizing Agent: $ \text{Cl}_2 $
- Reducing Agent: $ \text{H}_2 $

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QUESTION IMAGE

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Explanation:

Question 2: Can oxidation occur without reduction? Explain.

Answer:

Question 3: Why would a nail corrode more quickly in saltwater than in distilled water?