Oxidation And Reduction Reactions Practice Problems

Oxidation and reduction reactions, often called redox reactions, are fundamental chemical processes involving the transfer of electrons between chemical species. Mastering the concepts of oxidation and reduction is crucial for understanding various phenomena in chemistry, biology, and industrial processes. To solidify this understanding, working through practice problems is essential. This article will delve into various types of oxidation and reduction reaction practice problems, providing detailed explanations and step-by-step solutions to enhance your grasp of redox reactions.

Understanding Oxidation and Reduction

Before diving into practice problems, let's recap the basic principles of oxidation and reduction.

• Oxidation: This is the loss of electrons by a molecule, atom, or ion. Oxidation results in an increase in the oxidation state.
• Reduction: This is the gain of electrons by a molecule, atom, or ion. Reduction results in a decrease in the oxidation state.

In any redox reaction, oxidation and reduction always occur together. One substance loses electrons (is oxidized), while another substance gains electrons (is reduced). The substance that loses electrons is called the reducing agent (or reductant) because it causes the reduction of another substance. Conversely, the substance that gains electrons is called the oxidizing agent (or oxidant) because it causes the oxidation of another substance.

Key Concepts to Remember

• Oxidation State: A number assigned to an element in a chemical compound that represents the number of electrons lost or gained (or shared) by an atom of that element.
• LEO says GER: A mnemonic to remember the definitions: Lose Electrons Oxidation, Gain Electrons Reduction.
• OIL RIG: Another mnemonic: Oxidation Is Loss, Reduction Is Gain.

Practice Problems: Determining Oxidation States

The first step in identifying redox reactions is determining the oxidation states of the elements involved. Here are some practice problems to help you master this skill.

Problem 1: Determine the oxidation state of each element in potassium permanganate ($KMnO_4$).

Solution:

1. Identify the known oxidation states:
   • Potassium (K) is in Group 1, so its oxidation state is +1.
   • Oxygen (O) typically has an oxidation state of -2.
2. Set up an equation: The sum of the oxidation states in a neutral compound is zero. Therefore: $+1 + Mn + 4(-2) = 0$
3. Solve for Mn: $1 + Mn - 8 = 0$ $Mn = +7$

Thus, the oxidation states are: K (+1), Mn (+7), and O (-2).

Problem 2: Determine the oxidation state of each element in the dichromate ion ($Cr_2O_7^{2-}$).

Solution:

1. Identify the known oxidation state:
   • Oxygen (O) typically has an oxidation state of -2.
2. Set up an equation: The sum of the oxidation states in an ion equals the charge of the ion. Therefore: $2Cr + 7(-2) = -2$
3. Solve for Cr: $2Cr - 14 = -2$ $2Cr = 12$ $Cr = +6$

Thus, the oxidation states are: Cr (+6) and O (-2).

Problem 3: Determine the oxidation state of each element in sulfuric acid ($H_2SO_4$).

Solution:

1. Identify the known oxidation states:
   • Hydrogen (H) typically has an oxidation state of +1.
   • Oxygen (O) typically has an oxidation state of -2.
2. Set up an equation: The sum of the oxidation states in a neutral compound is zero. Therefore: $2(+1) + S + 4(-2) = 0$
3. Solve for S: $2 + S - 8 = 0$ $S = +6$

Thus, the oxidation states are: H (+1), S (+6), and O (-2).

Problem 4: Determine the oxidation state of each element in ammonium ion ($NH_4^+$).

Solution:

1. Identify the known oxidation state:
   • Hydrogen (H) typically has an oxidation state of +1.
2. Set up an equation: The sum of the oxidation states in an ion equals the charge of the ion. Therefore: $N + 4(+1) = +1$
3. Solve for N: $N + 4 = +1$ $N = -3$

Thus, the oxidation states are: N (-3) and H (+1).

Problem 5: Determine the oxidation state of each element in perchloric acid ($HClO_4$).

Solution:

1. Identify the known oxidation states:
   • Hydrogen (H) typically has an oxidation state of +1.
   • Oxygen (O) typically has an oxidation state of -2.
2. Set up an equation: The sum of the oxidation states in a neutral compound is zero. Therefore: $+1 + Cl + 4(-2) = 0$
3. Solve for Cl: $1 + Cl - 8 = 0$ $Cl = +7$

Thus, the oxidation states are: H (+1), Cl (+7), and O (-2).

Practice Problems: Identifying Redox Reactions

Now that you're familiar with determining oxidation states, let's identify redox reactions. Remember, a redox reaction involves a change in the oxidation state of at least two elements.

Problem 1: Is the following reaction a redox reaction? $NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$

Solution:

1. Determine the oxidation states of all elements:
   • NaOH: Na (+1), O (-2), H (+1)
   • HCl: H (+1), Cl (-1)
   • NaCl: Na (+1), Cl (-1)
   • $H_2O$: H (+1), O (-2)
2. Check for changes in oxidation states:
   • Na remains +1
   • O remains -2
   • H remains +1
   • Cl remains -1

Since there are no changes in oxidation states, this is not a redox reaction. This is an acid-base neutralization reaction.

Problem 2: Is the following reaction a redox reaction? $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$

Solution:

1. Determine the oxidation states of all elements:
   • Mg (s): 0 (elemental form)
   • $O_2$ (g): 0 (elemental form)
   • MgO: Mg (+2), O (-2)
2. Check for changes in oxidation states:
   • Mg changes from 0 to +2 (oxidation)
   • O changes from 0 to -2 (reduction)

Since there are changes in oxidation states, this is a redox reaction. Magnesium is oxidized, and oxygen is reduced.

Problem 3: Is the following reaction a redox reaction? $Cu(s) + 2AgNO_3(aq) \rightarrow 2Ag(s) + Cu(NO_3)_2(aq)$

Solution:

1. Determine the oxidation states of all elements:
   • Cu (s): 0 (elemental form)
   • $AgNO_3$: Ag (+1), N (+5), O (-2)
   • Ag (s): 0 (elemental form)
   • $Cu(NO_3)_2$: Cu (+2), N (+5), O (-2)
2. Check for changes in oxidation states:
   • Cu changes from 0 to +2 (oxidation)
   • Ag changes from +1 to 0 (reduction)
   • N remains +5
   • O remains -2

Since there are changes in oxidation states, this is a redox reaction. Copper is oxidized, and silver is reduced.

Problem 4: Is the following reaction a redox reaction? $Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)$

Solution:

1. Determine the oxidation states of all elements:
   • $Fe_2O_3$: Fe (+3), O (-2)
   • CO: C (+2), O (-2)
   • Fe (s): 0 (elemental form)
   • $CO_2$: C (+4), O (-2)
2. Check for changes in oxidation states:
   • Fe changes from +3 to 0 (reduction)
   • C changes from +2 to +4 (oxidation)
   • O remains -2

Since there are changes in oxidation states, this is a redox reaction. Iron is reduced, and carbon is oxidized.

Problem 5: Is the following reaction a redox reaction? $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$

Solution:

1. Determine the oxidation states of all elements:
   • $H_2$ (g): 0 (elemental form)
   • $Cl_2$ (g): 0 (elemental form)
   • HCl: H (+1), Cl (-1)
2. Check for changes in oxidation states:
   • H changes from 0 to +1 (oxidation)
   • Cl changes from 0 to -1 (reduction)

Since there are changes in oxidation states, this is a redox reaction. Hydrogen is oxidized, and chlorine is reduced.

Practice Problems: Identifying Oxidizing and Reducing Agents

Once you've identified a redox reaction, the next step is to determine the oxidizing and reducing agents.

Problem 1: Identify the oxidizing and reducing agents in the reaction: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

Solution:

1. Determine the oxidation states of all elements:
   • Zn (s): 0
   • $Cu^{2+}$ (aq): +2
   • $Zn^{2+}$ (aq): +2
   • Cu (s): 0
2. Identify oxidation and reduction:
   • Zn changes from 0 to +2 (oxidation)
   • Cu changes from +2 to 0 (reduction)
3. Identify oxidizing and reducing agents:
   • Zn is oxidized, so it is the reducing agent.
   • $Cu^{2+}$ is reduced, so it is the oxidizing agent.

Problem 2: Identify the oxidizing and reducing agents in the reaction: $2Al(s) + 3Sn^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Sn(s)$

Solution:

1. Determine the oxidation states of all elements:
   • Al (s): 0
   • $Sn^{2+}$ (aq): +2
   • $Al^{3+}$ (aq): +3
   • Sn (s): 0
2. Identify oxidation and reduction:
   • Al changes from 0 to +3 (oxidation)
   • Sn changes from +2 to 0 (reduction)
3. Identify oxidizing and reducing agents:
   • Al is oxidized, so it is the reducing agent.
   • $Sn^{2+}$ is reduced, so it is the oxidizing agent.

Problem 3: Identify the oxidizing and reducing agents in the reaction: $5Fe^{2+}(aq) + MnO_4^-(aq) + 8H^+(aq) \rightarrow 5Fe^{3+}(aq) + Mn^{2+}(aq) + 4H_2O(l)$

Solution:

1. Determine the oxidation states of all elements:
   • $Fe^{2+}$ (aq): +2
   • $MnO_4^-$: Mn (+7), O (-2)
   • $H^+$ (aq): +1
   • $Fe^{3+}$ (aq): +3
   • $Mn^{2+}$ (aq): +2
   • $H_2O$: H (+1), O (-2)
2. Identify oxidation and reduction:
   • Fe changes from +2 to +3 (oxidation)
   • Mn changes from +7 to +2 (reduction)
3. Identify oxidizing and reducing agents:
   • $Fe^{2+}$ is oxidized, so it is the reducing agent.
   • $MnO_4^-$ is reduced, so it is the oxidizing agent.

Problem 4: Identify the oxidizing and reducing agents in the reaction: $Cr_2O_7^{2-}(aq) + 6Fe^{2+}(aq) + 14H^+(aq) \rightarrow 2Cr^{3+}(aq) + 6Fe^{3+}(aq) + 7H_2O(l)$

Solution:

1. Determine the oxidation states of all elements:
   • $Cr_2O_7^{2-}$: Cr (+6), O (-2)
   • $Fe^{2+}$ (aq): +2
   • $H^+$ (aq): +1
   • $Cr^{3+}$ (aq): +3
   • $Fe^{3+}$ (aq): +3
   • $H_2O$: H (+1), O (-2)
2. Identify oxidation and reduction:
   • Cr changes from +6 to +3 (reduction)
   • Fe changes from +2 to +3 (oxidation)
3. Identify oxidizing and reducing agents:
   • $Fe^{2+}$ is oxidized, so it is the reducing agent.
   • $Cr_2O_7^{2-}$ is reduced, so it is the oxidizing agent.

Problem 5: Identify the oxidizing and reducing agents in the reaction: $2H_2S(g) + O_2(g) \rightarrow 2S(s) + 2H_2O(l)$

Solution:

1. Determine the oxidation states of all elements:
   • $H_2S$: H (+1), S (-2)
   • $O_2$ (g): 0
   • S (s): 0
   • $H_2O$: H (+1), O (-2)
2. Identify oxidation and reduction:
   • S changes from -2 to 0 (oxidation)
   • O changes from 0 to -2 (reduction)
3. Identify oxidizing and reducing agents:
   • $H_2S$ is oxidized, so it is the reducing agent.
   • $O_2$ is reduced, so it is the oxidizing agent.

Practice Problems: Balancing Redox Reactions

Balancing redox reactions can be more complex than balancing non-redox reactions. Common methods include the half-reaction method and the oxidation number method. Here, we'll focus on the half-reaction method.

Problem 1: Balance the following redox reaction in acidic solution: $MnO_4^-(aq) + Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + Fe^{3+}(aq)$

Solution:

1. Write the unbalanced half-reactions:
   • Oxidation: $Fe^{2+}(aq) \rightarrow Fe^{3+}(aq)$
   • Reduction: $MnO_4^-(aq) \rightarrow Mn^{2+}(aq)$
2. Balance the atoms (except O and H):
   • Oxidation: $Fe^{2+}(aq) \rightarrow Fe^{3+}(aq)$ (already balanced)
   • Reduction: $MnO_4^-(aq) \rightarrow Mn^{2+}(aq)$ (already balanced for Mn)
3. Balance oxygen by adding $H_2O$:
   • Oxidation: $Fe^{2+}(aq) \rightarrow Fe^{3+}(aq)$ (no oxygen)
   • Reduction: $MnO_4^-(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l)$
4. Balance hydrogen by adding $H^+$:
   • Oxidation: $Fe^{2+}(aq) \rightarrow Fe^{3+}(aq)$ (no hydrogen)
   • Reduction: $8H^+(aq) + MnO_4^-(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l)$
5. Balance the charge by adding electrons:
   • Oxidation: $Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-$
   • Reduction: $5e^- + 8H^+(aq) + MnO_4^-(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l)$
6. Make the number of electrons equal in both half-reactions:
   • Multiply the oxidation half-reaction by 5: $5Fe^{2+}(aq) \rightarrow 5Fe^{3+}(aq) + 5e^-$
   • Reduction half-reaction remains the same: $5e^- + 8H^+(aq) + MnO_4^-(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l)$
7. Add the half-reactions and cancel out electrons: $5Fe^{2+}(aq) + 5e^- + 8H^+(aq) + MnO_4^-(aq) \rightarrow 5Fe^{3+}(aq) + 5e^- + Mn^{2+}(aq) + 4H_2O(l)$
8. Final balanced equation: $5Fe^{2+}(aq) + MnO_4^-(aq) + 8H^+(aq) \rightarrow 5Fe^{3+}(aq) + Mn^{2+}(aq) + 4H_2O(l)$

Problem 2: Balance the following redox reaction in basic solution: $Cr_2O_7^{2-}(aq) + I^-(aq) \rightarrow Cr^{3+}(aq) + I_2(s)$

Solution:

1. Write the unbalanced half-reactions:
   • Oxidation: $I^-(aq) \rightarrow I_2(s)$
   • Reduction: $Cr_2O_7^{2-}(aq) \rightarrow Cr^{3+}(aq)$
2. Balance the atoms (except O and H):
   • Oxidation: $2I^-(aq) \rightarrow I_2(s)$
   • Reduction: $Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq)$
3. Balance oxygen by adding $H_2O$:
   • Oxidation: $2I^-(aq) \rightarrow I_2(s)$ (no oxygen)
   • Reduction: $Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$
4. Balance hydrogen by adding $H^+$:
   • Oxidation: $2I^-(aq) \rightarrow I_2(s)$ (no hydrogen)
   • Reduction: $14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$
5. Balance the charge by adding electrons:
   • Oxidation: $2I^-(aq) \rightarrow I_2(s) + 2e^-$
   • Reduction: $6e^- + 14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$
6. Make the number of electrons equal in both half-reactions:
   • Multiply the oxidation half-reaction by 3: $6I^-(aq) \rightarrow 3I_2(s) + 6e^-$
   • Reduction half-reaction remains the same: $6e^- + 14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$
7. Add the half-reactions and cancel out electrons: $6I^-(aq) + 6e^- + 14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 3I_2(s) + 6e^- + 2Cr^{3+}(aq) + 7H_2O(l)$
8. Simplify the equation: $6I^-(aq) + 14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 3I_2(s) + 2Cr^{3+}(aq) + 7H_2O(l)$
9. Since the solution is basic, neutralize $H^+$ by adding $OH^-$ to both sides: $6I^-(aq) + 14H^+(aq) + 14OH^-(aq) + Cr_2O_7^{2-}(aq) \rightarrow 3I_2(s) + 2Cr^{3+}(aq) + 7H_2O(l) + 14OH^-(aq)$
10. Combine $H^+$ and $OH^-$ to form $H_2O$: $6I^-(aq) + 14H_2O(l) + Cr_2O_7^{2-}(aq) \rightarrow 3I_2(s) + 2Cr^{3+}(aq) + 7H_2O(l) + 14OH^-(aq)$
11. Cancel out water molecules: $6I^-(aq) + 7H_2O(l) + Cr_2O_7^{2-}(aq) \rightarrow 3I_2(s) + 2Cr^{3+}(aq) + 14OH^-(aq)$
12. Final balanced equation: $Cr_2O_7^{2-}(aq) + 6I^-(aq) + 7H_2O(l) \rightarrow 2Cr^{3+}(aq) + 3I_2(s) + 14OH^-(aq)$

Problem 3: Balance the following redox reaction in acidic solution: $H_2O_2(aq) + MnO_4^-(aq) \rightarrow O_2(g) + Mn^{2+}(aq)$

Solution:

1. Write the unbalanced half-reactions:
   • Oxidation: $H_2O_2(aq) \rightarrow O_2(g)$
   • Reduction: $MnO_4^-(aq) \rightarrow Mn^{2+}(aq)$
2. Balance the atoms (except O and H):
   • Oxidation: $H_2O_2(aq) \rightarrow O_2(g)$ (already balanced for O)
   • Reduction: $MnO_4^-(aq) \rightarrow Mn^{2+}(aq)$ (already balanced for Mn)
3. Balance oxygen by adding $H_2O$:
   • Oxidation: $H_2O_2(aq) \rightarrow O_2(g)$ (already balanced for O)
   • Reduction: $MnO_4^-(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l)$
4. Balance hydrogen by adding $H^+$:
   • Oxidation: $H_2O_2(aq) \rightarrow O_2(g) + 2H^+(aq)$
   • Reduction: $8H^+(aq) + MnO_4^-(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l)$
5. Balance the charge by adding electrons:
   • Oxidation: $H_2O_2(aq) \rightarrow O_2(g) + 2H^+(aq) + 2e^-$
   • Reduction: $5e^- + 8H^+(aq) + MnO_4^-(aq) \rightarrow Mn^{2+}(aq) + 4H_2O(l)$
6. Make the number of electrons equal in both half-reactions:
   • Multiply the oxidation half-reaction by 5: $5H_2O_2(aq) \rightarrow 5O_2(g) + 10H^+(aq) + 10e^-$
   • Multiply the reduction half-reaction by 2: $10e^- + 16H^+(aq) + 2MnO_4^-(aq) \rightarrow 2Mn^{2+}(aq) + 8H_2O(l)$
7. Add the half-reactions and cancel out electrons: $5H_2O_2(aq) + 10e^- + 16H^+(aq) + 2MnO_4^-(aq) \rightarrow 5O_2(g) + 10H^+(aq) + 10e^- + 2Mn^{2+}(aq) + 8H_2O(l)$
8. Simplify the equation: $5H_2O_2(aq) + 16H^+(aq) + 2MnO_4^-(aq) \rightarrow 5O_2(g) + 10H^+(aq) + 2Mn^{2+}(aq) + 8H_2O(l)$
9. Cancel out $H^+$ ions: $5H_2O_2(aq) + 6H^+(aq) + 2MnO_4^-(aq) \rightarrow 5O_2(g) + 2Mn^{2+}(aq) + 8H_2O(l)$
10. Final balanced equation: $2MnO_4^-(aq) + 5H_2O_2(aq) + 6H^+(aq) \rightarrow 2Mn^{2+}(aq) + 5O_2(g) + 8H_2O(l)$

Problem 4: Balance the following redox reaction in basic solution: $Cl_2(g) \rightarrow ClO_3^-(aq) + Cl^-(aq)$

Solution:

1. Write the unbalanced half-reactions:
   • Oxidation: $Cl_2(g) \rightarrow ClO_3^-(aq)$
   • Reduction: $Cl_2(g) \rightarrow Cl^-(aq)$
2. Balance the atoms (except O and H):
   • Oxidation: $Cl_2(g) \rightarrow 2ClO_3^-(aq)$
   • Reduction: $Cl_2(g) \rightarrow 2Cl^-(aq)$
3. Balance oxygen by adding $H_2O$:
   • Oxidation: $Cl_2(g) + 6H_2O(l) \rightarrow 2ClO_3^-(aq)$
   • Reduction: $Cl_2(g) \rightarrow 2Cl^-(aq)$ (no oxygen)
4. Balance hydrogen by adding $H^+$:
   • Oxidation: $Cl_2(g) + 6H_2O(l) \rightarrow 2ClO_3^-(aq) + 12H^+(aq)$
   • Reduction: $Cl_2(g) \rightarrow 2Cl^-(aq)$ (no hydrogen)
5. Balance the charge by adding electrons:
   • Oxidation: $Cl_2(g) + 6H_2O(l) \rightarrow 2ClO_3^-(aq) + 12H^+(aq) + 10e^-$
   • Reduction: $Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)$
6. Make the number of electrons equal in both half-reactions:
   • Oxidation: $Cl_2(g) + 6H_2O(l) \rightarrow 2ClO_3^-(aq) + 12H^+(aq) + 10e^-$
   • Multiply the reduction half-reaction by 5: $5Cl_2(g) + 10e^- \rightarrow 10Cl^-(aq)$
7. Add the half-reactions and cancel out electrons: $Cl_2(g) + 6H_2O(l) + 5Cl_2(g) + 10e^- \rightarrow 2ClO_3^-(aq) + 12H^+(aq) + 10e^- + 10Cl^-(aq)$
8. Simplify the equation: $6Cl_2(g) + 6H_2O(l) \rightarrow 2ClO_3^-(aq) + 12H^+(aq) + 10Cl^-(aq)$
9. Since the solution is basic, neutralize $H^+$ by adding $OH^-$ to both sides: $6Cl_2(g) + 6H_2O(l) + 12OH^-(aq) \rightarrow 2ClO_3^-(aq) + 12H^+(aq) + 12OH^-(aq) + 10Cl^-(aq)$
10. Combine $H^+$ and $OH^-$ to form $H_2O$: $6Cl_2(g) + 6H_2O(l) + 12OH^-(aq) \rightarrow 2ClO_3^-(aq) + 12H_2O(l) + 10Cl^-(aq)$
11. Cancel out water molecules: $6Cl_2(g) + 12OH^-(aq) \rightarrow 2ClO_3^-(aq) + 6H_2O(l) + 10Cl^-(aq)$
12. Simplify the coefficients: $3Cl_2(g) + 6OH^-(aq) \rightarrow ClO_3^-(aq) + 3H_2O(l) + 5Cl^-(aq)$
13. Final balanced equation: $3Cl_2(g) + 6OH^-(aq) \rightarrow ClO_3^-(aq) + 5Cl^-(aq) + 3H_2O(l)$

Problem 5: Balance the following redox reaction in acidic solution: $Cu(s) + HNO_3(aq) \rightarrow Cu^{2+}(aq) + NO_2(g)$

Solution:

1. Write the unbalanced half-reactions:
   • Oxidation: $Cu(s) \rightarrow Cu^{2+}(aq)$
   • Reduction: $HNO_3(aq) \rightarrow NO_2(g)$
2. Balance the atoms (except O and H):
   • Oxidation: $Cu(s) \rightarrow Cu^{2+}(aq)$ (already balanced)
   • Reduction: $HNO_3(aq) \rightarrow NO_2(g)$ (already balanced for N)
3. Balance oxygen by adding $H_2O$:
   • Oxidation: $Cu(s) \rightarrow Cu^{2+}(aq)$ (no oxygen)
   • Reduction: $HNO_3(aq) \rightarrow NO_2(g) + H_2O(l)$
4. Balance hydrogen by adding $H^+$:
   • Oxidation: $Cu(s) \rightarrow Cu^{2+}(aq)$ (no hydrogen)
   • Reduction: $H^+(aq) + HNO_3(aq) \rightarrow NO_2(g) + H_2O(l)$
5. Balance the charge by adding electrons:
   • Oxidation: $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$
   • Reduction: $e^- + H^+(aq) + HNO_3(aq) \rightarrow NO_2(g) + H_2O(l)$
6. Make the number of electrons equal in both half-reactions:
   • Oxidation: $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$
   • Multiply the reduction half-reaction by 2: $2e^- + 2H^+(aq) + 2HNO_3(aq) \rightarrow 2NO_2(g) + 2H_2O(l)$
7. Add the half-reactions and cancel out electrons: $Cu(s) + 2e^- + 2H^+(aq) + 2HNO_3(aq) \rightarrow Cu^{2+}(aq) + 2e^- + 2NO_2(g) + 2H_2O(l)$
8. Final balanced equation: $Cu(s) + 2HNO_3(aq) + 2H^+(aq) \rightarrow Cu^{2+}(aq) + 2NO_2(g) + 2H_2O(l)$

Conclusion

Mastering oxidation and reduction reactions requires a solid understanding of oxidation states, the ability to identify redox reactions, and proficiency in balancing these reactions. By working through these practice problems and understanding the solutions, you can significantly improve your grasp of redox reactions and their applications in various fields. Remember to practice regularly and reinforce your knowledge of the fundamental principles.