r1 – redox half reactions
1. write the redox half reaction for the following reactions and indicates if the original substance was oxidized or reduced.
a) formation of sodium ion
b) formation of chloride ion
c) formation of bromide ion
d) formation of copper (ii) ion
e) formation of oxide ion
f) formation of silver ion

2. for the following reaction:
3 k₍ₛ₎ + al(clo₃)₃₍ₐq₎ → al₍ₛ₎ + 3 kclo₃₍ₐq₎
a) write the total and net ionic equations and build the half-reactions.
b) what is the reducing agent?
c) what is the oxidizing agent?
d) what is being reduced?
e) what is being oxidized?

3. write the half reaction showing ni²⁺₍ₐq₎ as an oxidizing agent.

4. write the half reactions showing sn²⁺₍ₐq₎ as both an oxidizing agent and a reducing agent.

Question

r1 – redox half reactions
1. write the redox half reaction for the following reactions and indicates if the original substance was oxidized or reduced.
   a) formation of sodium ion
   b) formation of chloride ion
   c) formation of bromide ion
   d) formation of copper (ii) ion
   e) formation of oxide ion
   f) formation of silver ion

2. for the following reaction:
   3 k₍ₛ₎ + al(clo₃)₃₍ₐq₎ → al₍ₛ₎ + 3 kclo₃₍ₐq₎
   a) write the total and net ionic equations and build the half-reactions.
   b) what is the reducing agent?
   c) what is the oxidizing agent?
   d) what is being reduced?
   e) what is being oxidized?

3. write the half reaction showing ni²⁺₍ₐq₎ as an oxidizing agent.

4. write the half reactions showing sn²⁺₍ₐq₎ as both an oxidizing agent and a reducing agent.

Accepted Answer

### Problem 1a: Formation of sodium ion
# Explanation:
## Step 1: Identify the reaction
Sodium metal (Na) forms sodium ion ($ \text{Na}^+ $). The reaction is $ \text{Na}
ightarrow \text{Na}^+ + e^- $.
## Step 2: Determine oxidation/reduction
Loss of electrons means oxidation. Sodium loses an electron to form $ \text{Na}^+ $, so it is oxidized.

# Answer:
Half - reaction: $ \text{Na}
ightarrow \text{Na}^+ + e^- $; Original substance (Na) is oxidized.

### Problem 1b: Formation of chloride ion
# Explanation:
## Step 1: Identify the reaction
Chlorine gas ($ \text{Cl}_2 $) or chlorine atom (Cl) forms chloride ion ($ \text{Cl}^- $). The reaction is $ \text{Cl} + e^-
ightarrow \text{Cl}^- $ (or $ \text{Cl}_2 + 2e^-
ightarrow 2\text{Cl}^- $).
## Step 2: Determine oxidation/reduction
Gain of electrons means reduction. Chlorine gains an electron to form $ \text{Cl}^- $, so it is reduced.

# Answer:
Half - reaction: $ \text{Cl} + e^-
ightarrow \text{Cl}^- $ (or $ \text{Cl}_2 + 2e^-
ightarrow 2\text{Cl}^- $); Original substance (Cl or $ \text{Cl}_2 $) is reduced.

### Problem 1c: Formation of bromide ion
# Explanation:
## Step 1: Identify the reaction
Bromine (Br) forms bromide ion ($ \text{Br}^- $). The reaction is $ \text{Br} + e^-
ightarrow \text{Br}^- $ (or $ \text{Br}_2 + 2e^-
ightarrow 2\text{Br}^- $).
## Step 2: Determine oxidation/reduction
Gain of electrons means reduction. Bromine gains an electron to form $ \text{Br}^- $, so it is reduced.

# Answer:
Half - reaction: $ \text{Br} + e^-
ightarrow \text{Br}^- $ (or $ \text{Br}_2 + 2e^-
ightarrow 2\text{Br}^- $); Original substance (Br or $ \text{Br}_2 $) is reduced.

### Problem 1d: Formation of copper (II) ion
# Explanation:
## Step 1: Identify the reaction
Copper metal (Cu) forms copper (II) ion ($ \text{Cu}^{2+} $). The reaction is $ \text{Cu}
ightarrow \text{Cu}^{2+} + 2e^- $.
## Step 2: Determine oxidation/reduction
Loss of electrons means oxidation. Copper loses 2 electrons to form $ \text{Cu}^{2+} $, so it is oxidized.

# Answer:
Half - reaction: $ \text{Cu}
ightarrow \text{Cu}^{2+} + 2e^- $; Original substance (Cu) is oxidized.

### Problem 1e: Formation of oxide ion
# Explanation:
## Step 1: Identify the reaction
Oxygen (O) forms oxide ion ($ \text{O}^{2-} $). The reaction is $ \text{O} + 2e^-
ightarrow \text{O}^{2-} $ (or $ \text{O}_2 + 4e^-
ightarrow 2\text{O}^{2-} $).
## Step 2: Determine oxidation/reduction
Gain of electrons means reduction. Oxygen gains 2 electrons to form $ \text{O}^{2-} $, so it is reduced.

# Answer:
Half - reaction: $ \text{O} + 2e^-
ightarrow \text{O}^{2-} $ (or $ \text{O}_2 + 4e^-
ightarrow 2\text{O}^{2-} $); Original substance (O or $ \text{O}_2 $) is reduced.

### Problem 1f: Formation of silver ion
# Explanation:
## Step 1: Identify the reaction
Silver metal (Ag) forms silver ion ($ \text{Ag}^+ $). The reaction is $ \text{Ag}
ightarrow \text{Ag}^+ + e^- $.
## Step 2: Determine oxidation/reduction
Loss of electrons means oxidation. Silver loses an electron to form $ \text{Ag}^+ $, so it is oxidized.

# Answer:
Half - reaction: $ \text{Ag}
ightarrow \text{Ag}^+ + e^- $; Original substance (Ag) is oxidized.

### Problem 2a: Total and net ionic equations, half - reactions
# Explanation:
## Step 1: Write total ionic equation
The reaction is $ 3\text{K}(s) + \text{Al}(\text{ClO}_3)_3(aq)
ightarrow \text{Al}(s) + 3\text{KClO}_3(aq) $.
Dissociate $ \text{Al}(\text{ClO}_3)_3 $ and $ \text{KClO}_3 $:
$ 3\text{K}(s) + \text{Al}^{3+}(aq) + 3\text{ClO}_3^-(aq)
ightarrow \text{Al}(s) + 3\text{K}^+(aq) + 3\text{ClO}_3^-(aq) $.

## Step 2: Write net ionic equation
Cancel spectator ions ($ \text{ClO}_3^- $):
$ 3\text{K}(s) + \text{Al}^{3+}(aq)
ightarrow \text{Al}(s) + 3\text{K}^+(aq) $.

## Step 3: Write half - reactions
Oxidation (K loses electrons): $ \text{K}
ightarrow \text{K}^+ + e^- $ (multiply by 3: $ 3\text{K}
ightarrow 3\text{K}^+ + 3e^- $).
Reduction (Al³⁺ gains electrons): $ \text{Al}^{3+} + 3e^-
ightarrow \text{Al} $.

# Answer:
Total ionic equation: $ 3\text{K}(s) + \text{Al}^{3+}(aq) + 3\text{ClO}_3^-(aq)
ightarrow \text{Al}(s) + 3\text{K}^+(aq) + 3\text{ClO}_3^-(aq) $
Net ionic equation: $ 3\text{K}(s) + \text{Al}^{3+}(aq)
ightarrow \text{Al}(s) + 3\text{K}^+(aq) $
Oxidation half - reaction: $ 3\text{K}
ightarrow 3\text{K}^+ + 3e^- $
Reduction half - reaction: $ \text{Al}^{3+} + 3e^-
ightarrow \text{Al} $

### Problem 2b: Reducing agent
# Explanation:
The reducing agent is the substance oxidized (loses electrons). In the reaction, K is oxidized (loses electrons to form $ \text{K}^+ $), so K is the reducing agent.

# Answer:
Reducing agent: $ \text{K}(s) $

### Problem 2c: Oxidizing agent
# Explanation:
The oxidizing agent is the substance reduced (gains electrons). In the reaction, $ \text{Al}^{3+} $ (from $ \text{Al}(\text{ClO}_3)_3 $) is reduced (gains electrons to form Al), so $ \text{Al}(\text{ClO}_3)_3 $ (or $ \text{Al}^{3+} $) is the oxidizing agent.

# Answer:
Oxidizing agent: $ \text{Al}(\text{ClO}_3)_3(aq) $ (or $ \text{Al}^{3+}(aq) $)

### Problem 2d: Substance being reduced
# Explanation:
The substance being reduced gains electrons. $ \text{Al}^{3+} $ (from $ \text{Al}(\text{ClO}_3)_3 $) gains electrons to form Al, so $ \text{Al}^{3+} $ (or $ \text{Al}(\text{ClO}_3)_3 $) is being reduced.

# Answer:
Substance being reduced: $ \text{Al}^{3+}(aq) $ (or $ \text{Al}(\text{ClO}_3)_3(aq) $)

### Problem 2e: Substance being oxidized
# Explanation:
The substance being oxidized loses electrons. K loses electrons to form $ \text{K}^+ $, so K is being oxidized.

# Answer:
Substance being oxidized: $ \text{K}(s) $

### Problem 3: $ \text{Ni}^{2+}(aq) $ as oxidizing agent
# Explanation:
An oxidizing agent is reduced (gains electrons). $ \text{Ni}^{2+} $ will gain electrons to form a lower - oxidation - state Ni (e.g., Ni metal). The half - reaction is $ \text{Ni}^{2+} + 2e^-
ightarrow \text{Ni} $.

# Answer:
Half - reaction: $ \text{Ni}^{2+}(aq) + 2e^-
ightarrow \text{Ni}(s) $

### Problem 4: $ \text{Sn}^{2+}(aq) $ as oxidizing and reducing agent
# Explanation:
- As oxidizing agent (reduced, gains electrons): $ \text{Sn}^{2+} $ gains electrons to form Sn (lower oxidation state). Reaction: $ \text{Sn}^{2+} + 2e^-
ightarrow \text{Sn} $.
- As reducing agent (oxidized, loses electrons): $ \text{Sn}^{2+} $ loses electrons to form $ \text{Sn}^{4+} $ (higher oxidation state). Reaction: $ \text{Sn}^{2+}
ightarrow \text{Sn}^{4+} + 2e^- $.

# Answer:
As oxidizing agent: $ \text{Sn}^{2+}(aq) + 2e^-
ightarrow \text{Sn}(s) $
As reducing agent: $ \text{Sn}^{2+}(aq)
ightarrow \text{Sn}^{4+}(aq) + 2e^- $

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QUESTION IMAGE

Question

Explanation:

Problem 1a: Formation of sodium ion

Step 1: Identify the reaction

Step 2: Determine oxidation/reduction

Step 1: Identify the reaction

Step 2: Determine oxidation/reduction

Step 1: Identify the reaction

Step 2: Determine oxidation/reduction

Answer:

Problem 1b: Formation of chloride ion