redox mixed practice ws l
directions: answer the following questions, using your notes and other worksheets, to the best of your ability.
for numbers 1-6, list the oxidation numbers above each element in the following compounds:
1. li₂so₄
2. no₃⁻
3. na₂s₂o₃
4. co₃²⁻
5. pcl₅
6. h₂
for numbers 7-12, assign oxidation numbers, identify which element is reduced and which element is oxidized, and list spectator ion, if any.
7. 2al + 3fe²⁺ → 2al³⁺ + 3fe
element reduced:
element oxidized:
spectator ion:
8. zn + 2hcl → zncl₂ + h₂
element reduced:
element oxidized:
spectator ion:
9. 2febr₃ + 3cl₂ → 2fecl₃ + 3br₂
element reduced:
element oxidized:
spectator ion:
10. 2hi → h₂ + i₂
element reduced:
element oxidized:
spectator ion:
practice balancing:
11. ___ mg + ___ n₂ → ___ mg₃n₂
element reduced:
element oxidized:
spectator ion:
12. ___ cubr + ___ f₂ → ___ cuf + ___ br₂
element reduced:
element oxidized:
spectator ion:

Question

Accepted Answer

### Problem 1: Assign Oxidation Numbers to $ \text{Li}_2\text{SO}_4 $
# Explanation:
## Step 1: Recall oxidation number rules
- Lithium ($ \text{Li} $) is an alkali metal, so its oxidation number is $ +1 $.
- Oxygen ($ \text{O} $) usually has an oxidation number of $ -2 $ (except in peroxides, superoxides, or with fluorine).
- Let the oxidation number of sulfur ($ \text{S} $) be $ x $.

## Step 2: Set up the equation for charge balance
The compound is neutral, so the sum of oxidation numbers is zero. There are 2 Li atoms, 1 S atom, and 4 O atoms.
$$
2(+1) + x + 4(-2) = 0
$$

## Step 3: Solve for $ x $
$$
2 + x - 8 = 0 \
x - 6 = 0 \
x = +6
$$

# Answer:
- $ \text{Li} $: $ +1 $
- $ \text{S} $: $ +6 $
- $ \text{O} $: $ -2 $

### Problem 2: Assign Oxidation Numbers to $ \text{NO}_3^- $
# Explanation:
## Step 1: Recall oxidation number rules
- Oxygen ($ \text{O} $) has an oxidation number of $ -2 $ (in most cases).
- The ion has a charge of $ -1 $. Let the oxidation number of nitrogen ($ \text{N} $) be $ x $.

## Step 2: Set up the equation for charge balance
There is 1 N atom and 3 O atoms.
$$
x + 3(-2) = -1
$$

## Step 3: Solve for $ x $
$$
x - 6 = -1 \
x = +5
$$

# Answer:
- $ \text{N} $: $ +5 $
- $ \text{O} $: $ -2 $

### Problem 3: Assign Oxidation Numbers to $ \text{Na}_2\text{S}_2\text{O}_3 $
# Explanation:
## Step 1: Recall oxidation number rules
- Sodium ($ \text{Na} $) is an alkali metal, so its oxidation number is $ +1 $.
- Oxygen ($ \text{O} $) has an oxidation number of $ -2 $.
- Let the oxidation number of each sulfur ($ \text{S} $) be $ x $ (there are 2 S atoms).

## Step 2: Set up the equation for charge balance
The compound is neutral, so the sum of oxidation numbers is zero. There are 2 Na atoms, 2 S atoms, and 3 O atoms.
$$
2(+1) + 2x + 3(-2) = 0
$$

## Step 3: Solve for $ x $
$$
2 + 2x - 6 = 0 \
2x - 4 = 0 \
2x = 4 \
x = +2
$$

# Answer:
- $ \text{Na} $: $ +1 $
- $ \text{S} $: $ +2 $
- $ \text{O} $: $ -2 $

### Problem 4: Assign Oxidation Numbers to $ \text{CO}_3^{2-} $
# Explanation:
## Step 1: Recall oxidation number rules
- Oxygen ($ \text{O} $) has an oxidation number of $ -2 $.
- The ion has a charge of $ -2 $. Let the oxidation number of carbon ($ \text{C} $) be $ x $.

## Step 2: Set up the equation for charge balance
There is 1 C atom and 3 O atoms.
$$
x + 3(-2) = -2
$$

## Step 3: Solve for $ x $
$$
x - 6 = -2 \
x = +4
$$

# Answer:
- $ \text{C} $: $ +4 $
- $ \text{O} $: $ -2 $

### Problem 5: Assign Oxidation Numbers to $ \text{PCl}_5 $
# Explanation:
## Step 1: Recall oxidation number rules
- Chlorine ($ \text{Cl} $) has an oxidation number of $ -1 $ (except when combined with oxygen or fluorine, or in interhalogens).
- Let the oxidation number of phosphorus ($ \text{P} $) be $ x $.

## Step 2: Set up the equation for charge balance
The compound is neutral, so the sum of oxidation numbers is zero. There is 1 P atom and 5 Cl atoms.
$$
x + 5(-1) = 0
$$

## Step 3: Solve for $ x $
$$
x - 5 = 0 \
x = +5
$$

# Answer:
- $ \text{P} $: $ +5 $
- $ \text{Cl} $: $ -1 $

### Problem 6: Assign Oxidation Numbers to $ \text{H}_2 $
# Explanation:
- For a diatomic molecule of an element (like $ \text{H}_2 $, $ \text{O}_2 $, $ \text{N}_2 $, etc.), the oxidation number of the element is $ 0 $ because the atoms have equal electronegativity and share electrons equally.

# Answer:
- $ \text{H} $: $ 0 $

### Problem 7: Analyze $ 2\text{Al} + 3\text{Fe}^{2+}
ightarrow 2\text{Al}^{3+} + 3\text{Fe} $
# Explanation:
## Step 1: Identify oxidation states
- $ \text{Al} $ (reactant): $ 0 $; $ \text{Al}^{3+} $ (product): $ +3 $ (oxidation, loss of electrons)
- $ \text{Fe}^{2+} $ (reactant): $ +2 $; $ \text{Fe} $ (product): $ 0 $ (reduction, gain of electrons)

## Step 2: Identify spectator ions (if any)
This is an ionic reaction in solution, but the only ions are $ \text{Fe}^{2+} $ and $ \text{Al}^{3+} $ (and electrons). There are no spectator ions (ions that don't change oxidation state) here.

# Answer:
- Element Reduced: $ \text{Fe} $ (oxidation state decreases from $ +2 $ to $ 0 $)
- Element Oxidized: $ \text{Al} $ (oxidation state increases from $ 0 $ to $ +3 $)
- Spectator ion: None

### Problem 8: Analyze $ \text{Zn} + 2\text{HCl}
ightarrow \text{ZnCl}_2 + \text{H}_2 $
# Explanation:
## Step 1: Identify oxidation states
- $ \text{Zn} $ (reactant): $ 0 $; $ \text{Zn}^{2+} $ (in $ \text{ZnCl}_2 $): $ +2 $ (oxidation, loss of electrons)
- $ \text{H}^+ $ (in $ \text{HCl} $): $ +1 $; $ \text{H}_2 $ (product): $ 0 $ (reduction, gain of electrons)
- $ \text{Cl}^- $ (in $ \text{HCl} $ and $ \text{ZnCl}_2 $): $ -1 $ (no change, spectator ion)

# Answer:
- Element Reduced: $ \text{H} $ (oxidation state decreases from $ +1 $ to $ 0 $)
- Element Oxidized: $ \text{Zn} $ (oxidation state increases from $ 0 $ to $ +2 $)
- Spectator ion: $ \text{Cl}^- $

### Problem 9: Analyze $ 2\text{FeBr}_3 + 3\text{Cl}_2
ightarrow 2\text{FeCl}_3 + 3\text{Br}_2 $
# Explanation:
## Step 1: Identify oxidation states
- $ \text{Fe} $: $ +3 $ (reactant and product, no change)
- $ \text{Br}^- $ (in $ \text{FeBr}_3 $): $ -1 $; $ \text{Br}_2 $ (product): $ 0 $ (oxidation, loss of electrons)
- $ \text{Cl}_2 $ (reactant): $ 0 $; $ \text{Cl}^- $ (in $ \text{FeCl}_3 $): $ -1 $ (reduction, gain of electrons)

## Step 2: Identify spectator ions (if any)
$ \text{Fe}^{3+} $ doesn't change oxidation state, but it's part of the compound. Wait, no—$ \text{Fe} $ is $ +3 $ in both $ \text{FeBr}_3 $ and $ \text{FeCl}_3 $, so is it a spectator? Wait, the reaction is about $ \text{Br}^- $ and $ \text{Cl}_2 $. Wait, actually, $ \text{Fe}^{3+} $ is a spectator ion? Wait, no, the ions are $ \text{Fe}^{3+} $, $ \text{Br}^- $, $ \text{Cl}_2 $, $ \text{Cl}^- $, $ \text{Br}_2 $. Wait, $ \text{Fe}^{3+} $ remains $ +3 $, so it's a spectator ion? Wait, no, in the formula, $ \text{FeBr}_3 $ and $ \text{FeCl}_3 $ are ionic, but $ \text{Fe}^{3+} $ doesn't change oxidation state. Wait, but the question is about spectator ions (ions that don't participate in redox). So $ \text{Fe}^{3+} $ is a spectator ion? Wait, no, let's check again.

Wait, the oxidation states:
- $ \text{Br} $: $ -1 $ (reactant) → $ 0 $ (product) (oxidized)
- $ \text{Cl} $: $ 0 $ (reactant) → $ -1 $ (product) (reduced)
- $ \text{Fe} $: $ +3 $ (reactant) → $ +3 $ (product) (no change, so $ \text{Fe}^{3+} $ is a spectator ion)

# Answer:
- Element Reduced: $ \text{Cl} $ (oxidation state decreases from $ 0 $ to $ -1 $)
- Element Oxidized: $ \text{Br} $ (oxidation state increases from $ -1 $ to $ 0 $)
- Spectator ion: $ \text{Fe}^{3+} $

### Problem 10: Analyze $ 2\text{HI}
ightarrow \text{H}_2 + \text{I}_2 $
# Explanation:
## Step 1: Identify oxidation states
- $ \text{H} $ in $ \text{HI} $: $ +1 $; $ \text{H}_2 $: $ 0 $ (reduction, gain of electrons)
- $ \text{I} $ in $ \text{HI} $: $ -1 $; $ \text{I}_2 $: $ 0 $ (oxidation, loss of electrons)

## Step 2: Identify spectator ions (if any)
This is a decomposition of $ \text{HI} $ (a covalent compound in solution, but as ions: $ \text{H}^+ $ and $ \text{I}^- $). Wait, no, $ \text{HI} $ is a strong acid, so in solution, $ \text{H}^+ $ and $ \text{I}^- $. But the reaction is $ 2\text{H}^+ + 2\text{I}^-
ightarrow \text{H}_2 + \text{I}_2 $. There are no spectator ions because both $ \text{H}^+ $ and $ \text{I}^- $ change oxidation states.

# Answer:
- Element Reduced: $ \text{H} $ (oxidation state decreases from $ +1 $ to $ 0 $)
- Element Oxidized: $ \text{I} $ (oxidation state increases from $ -1 $ to $ 0 $)
- Spectator ion: None

### Problem 11: Balance $ \text{Mg} + \text{N}_2
ightarrow \text{Mg}_3\text{N}_2 $
# Explanation:
## Step 1: Count atoms
- Reactants: $ \text{Mg} $: 1; $ \text{N} $: 2
- Products: $ \text{Mg} $: 3; $ \text{N} $: 2

## Step 2: Balance Mg
Multiply $ \text{Mg} $ by 3:
$$
3\text{Mg} + \text{N}_2
ightarrow \text{Mg}_3\text{N}_2
$$

## Step 3: Check N
Nitrogen is already balanced (2 on each side).

# Answer:
Balanced equation: $ 3\text{Mg} + \text{N}_2
ightarrow \text{Mg}_3\text{N}_2 $
- Element Reduced: $ \text{N} $ (oxidation state decreases from $ 0 $ to $ -3 $ in $ \text{Mg}_3\text{N}_2 $)
- Element Oxidized: $ \text{Mg} $ (oxidation state increases from $ 0 $ to $ +2 $ in $ \text{Mg}_3\text{N}_2 $)
- Spectator ion: None (it's a combination reaction of elements, no ions)

### Problem 12: Balance $ \text{CuBr} + \text{F}_2
ightarrow \text{CuF} + \text{Br}_2 $
# Explanation:
## Step 1: Count atoms
- Reactants: $ \text{Cu} $: 1; $ \text{Br} $: 1; $ \text{F} $: 2
- Products: $ \text{Cu} $: 1; $ \text{F} $: 1; $ \text{Br} $: 2

## Step 2: Balance Br
Multiply $ \text{CuBr} $ by 2:
$$
2\text{CuBr} + \text{F}_2
ightarrow \text{CuF} + \text{Br}_2
$$

## Step 3: Balance Cu
Multiply $ \text{CuF} $ by 2:
$$
2\text{CuBr} + \text{F}_2
ightarrow 2\text{CuF} + \text{Br}_2
$$

## Step 4: Check F
Fluorine is balanced (2 on each side).

# Answer:
Balanced equation: $ 2\text{CuBr} + \text{F}_2
ightarrow 2\text{CuF} + \text{Br}_2 $
- Element Reduced: $ \text{F} $ (oxidation state decreases from $ 0 $ to $ -1 $ in $ \text{CuF} $)
- Element Oxidized: $ \text{Br} $ (oxidation state increases from $ -1 $ to $ 0 $ in $ \text{Br}_2 $)
- Spectator ion: None (it's a redox reaction with ions? Wait, $ \text{CuBr} $ and $ \text{CuF} $ are ionic, but $ \text{Cu}^+ $ remains $ +1 $? Wait, no:
  - $ \text{Cu} $ in $ \text{CuBr} $: $ +1 $; $ \text{Cu} $ in $ \text{CuF} $: $ +1 $ (so $ \text{Cu}^+ $ is a spectator ion? Wait, no, the reaction is $ 2\text{Cu}^+ + 2\text{Br}^- + \text{F}_2
ightarrow 2\text{Cu}^+ + 2\text{F}^- + \text{Br}_2 $. So $ \text{Cu}^+ $ is a spectator ion!

Wait, let's re-analyze oxidation states:
- $ \text{Cu} $: $ +1 $ (reactant and product) → spectator ion
- $ \text{Br} $: $ -1 $ (reactant) → $ 0 $ (product) (oxidized)
- $ \text{F} $: $ 0 $ (reactant) → $ -1 $ (product) (reduced)

So:
- Element Reduced: $ \text{F} $
- Element Oxidized: $ \text{Br} $
- Spectator ion: $ \text{Cu}^+ $ (or $ \text{Cu} $ in $ \text{CuBr} $ and $ \text{CuF} $)

# Final Answer for Problem 12:
Balanced equation: $ 2\text{CuBr} + \text{F}_2
ightarrow 2\text{CuF} + \text{Br}_2 $
- Element Reduced: $ \text{F} $
- Element Oxidized: $ \text{Br} $
- Spect

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QUESTION IMAGE

Question

Explanation:

Problem 1: Assign Oxidation Numbers to \( \text{Li}_2\text{SO}_4 \)

Step 1: Recall oxidation number rules

Step 2: Set up the equation for charge balance

Step 3: Solve for \( x \)

Step 1: Recall oxidation number rules

Step 2: Set up the equation for charge balance

Step 3: Solve for \( x \)

Step 1: Recall oxidation number rules

Step 2: Set up the equation for charge balance

Step 3: Solve for \( x \)

Answer:

Problem 2: Assign Oxidation Numbers to \( \text{NO}_3^- \)