Oxidation-Reduction (Redox) Reactions

⚡ Oxidation States

Part 1 of 7 — Rules for Assigning Oxidation Numbers

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

• Understanding the core concepts covered in Part 1
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📏 Rules for Assigning Oxidation States

Apply these rules in order of priority (Rule 1 overrides Rule 2, etc.):

💡 Tip: Always apply the rules in order — higher-numbered rules yield to lower-numbered ones when there's a conflict.

🔑 Key Concept: Rule 6 is your primary calculation tool — if you know all oxidation states except one, solve for the unknown!

Rule 6 Is Your Calculation Tool

For any compound or polyatomic ion:

$\boxed{\sum \text{(oxidation states)} = \text{overall charge}}$

🧪 Worked Examples

Example 1: $H_{2}SO_{4}$

Problem: Find the oxidation state of sulfur in $H_{2}SO_{4}$.

Solution:

• H = +1 (Rule 5), O = −2 (Rule 4)
• $2(+1) + S + 4(-2) = 0$
• $+2 + S - 8 = 0$
• $\boxed{S = +6}$

Example 2: $MnO_{4}^{-}$ (permanganate ion)

Problem: Find the oxidation state of manganese in $MnO_{4}^{-}$.

Solution:

• O = −2 (Rule 4)
• $\text{Mn} + 4(-2) = -1$ (charge of ion)
• $\text{Mn} - 8 = -1$
• $\boxed{\text{Mn} = +7}$

Example 3: $Cr_{2}O_{7}^{2-}$ (dichromate ion)

Problem: Find the oxidation state of chromium in $Cr_{2}O_{7}^{2-}$.

Solution:

• O = −2 (Rule 4)
• $2(\text{Cr}) + 7(-2) = -2$
• $2\text{Cr} - 14 = -2$
• $2\text{Cr} = +12$
• $\boxed{\text{Cr} = +6}$

Example 4: $Na_{2}O_{2}$ (sodium peroxide)

⚠️ Warning: Peroxides are a common exception — oxygen is −1, not −2!

• Na = +1 (Rule 2, Group 1 metal)
• $2(+1) + 2(\text{O}) = 0$
• $\text{O} = -1$ (peroxide exception!)

Oxidation States Concept Quiz 🎯

Calculate Oxidation States 🧮

Find the oxidation state of the underlined element. Give your answer as a number with sign (e.g., +5 or -2).

1) Sulfur in $SO_{4}^{2-}$

2) Phosphorus in $H_{3}PO_{4}$

3) Manganese in $MnO_{2}$

Oxidation State Rules 🔽

Exit Quiz — Oxidation States ✅

⚡ Identifying Redox Reactions

Part 2 of 7 — OIL RIG and Oxidizing/Reducing Agents

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

• Understanding the core concepts covered in Part 2
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 OIL RIG — The Key Mnemonic

🔑 Key Concept: OIL RIG — Oxidation Is Loss, Reduction Is Gain of electrons.

$\boxed{\text{OIL RIG}}$

How to Spot a Redox Reaction

1. Assign oxidation states to every atom in reactants and products
2. If any oxidation state changes, it's a redox reaction
3. If NO oxidation states change, it's NOT redox (e.g., double replacement)

Example

$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

⚡ Oxidizing and Reducing Agents

Definitions

The Tricky Part

⚠️ Warning: The names seem backwards! The agent is named for what it does to the other species, not what happens to itself.

• The oxidizing agent is the one that takes electrons (gets reduced)
• The reducing agent is the one that gives electrons (gets oxidized)

Example (continued)

$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

• Zn is the reducing agent — it gives up electrons (gets oxidized: 0 → +2)
• $Cu^{2+}$ is the oxidizing agent — it takes electrons (gets reduced: +2 → 0)

Common Oxidizing Agents

Common Reducing Agents

⚗️ Redox vs. Non-Redox Reactions

Not All Reactions Are Redox!

Quick Test

💡 Tip: If elements appear as reactants or products (in their free state, oxidation state = 0), the reaction is almost certainly redox.

Identifying Redox Quiz 🎯

Identify the Redox Components 🧮

For the reaction: $\text{2Al}(s) + 3\text{Cl}_2(g) \rightarrow 2\text{AlCl}_3(s)$

1) What element is oxidized? (type the element symbol)

2) What element is reduced? (type the element symbol)

3) How many electrons are transferred per Al atom?

Redox Terminology 🔽

Exit Quiz — Identifying Redox ✅

⚡ Balancing Redox in Acidic Solution

Part 3 of 7 — The Half-Reaction Method

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

• Understanding the core concepts covered in Part 3
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

⚗️ The Half-Reaction Method (Acidic Solution)

The 7 Steps

Key Principle

🔑 Key Concept: Electrons lost in oxidation must equal electrons gained in reduction — electrons are neither created nor destroyed.

🧪 Worked Example

Problem: Balance in acidic solution:

$\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$

Step 1: Write half-reactions

Reduction: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$

Step 2: Balance atoms (non-O, non-H)

Already balanced (1 Mn each side, 1 Fe each side).

Step 3: Balance O with $H_{2}O$

$\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Step 4: Balance H with $H^{+}$

$8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Step 5: Balance charge with $e^{-}$

Reduction: Left charge: 8(+1) + (−1) = +7. Right charge: +2. Need $5e^{-}$ on left. $5e^- + 8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Oxidation: Left charge: +2. Right charge: +3. Need $1e^{-}$ on right. $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$

Step 6: Equalize electrons (multiply oxidation by 5)

$5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-$

Step 7: Add and cancel $e^{-}$

$\boxed{5\text{Fe}^{2+} + 8\text{H}^+ + \text{MnO}_4^- \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}}$

✅ Verify

⚠️ Warning: Always verify both atoms AND charge — a common mistake is balancing atoms but not charge!

• Atoms: 5 Fe ✓, 1 Mn ✓, 4 O ✓, 8 H ✓
• Charge: Left: 5(+2) + 8(+1) + (−1) = +17. Right: 5(+3) + (+2) + 0 = +17 ✓

Half-Reaction Method Quiz 🎯

Half-Reaction Practice 🧮

For the half-reaction in acidic solution: $\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$

1) How many $H_{2}O$ molecules are needed (and on which side)? Type the coefficient only.

2) How many $H^{+}$ ions are needed? Type the coefficient only.

3) How many electrons are needed? Type the coefficient only.

Acidic Solution Balancing Concepts 🔽

Exit Quiz — Balancing Redox in Acidic Solution ✅

⚡ Balancing Redox in Basic Solution

Part 4 of 7 — Adding $OH^{-}$ to Neutralize $H^{+}$

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

• Understanding the core concepts covered in Part 4
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

🧪 The Basic Solution Method

Strategy: Balance in Acid First, Then Convert

🔑 Key Concept: Always balance in acidic solution first (Steps 1–7), then convert to basic by adding $OH^{-}$.

Why This Works

⚠️ Warning: In basic solution, free $H^{+}$ ions don't exist! If your final equation still has $H^{+}$, you haven't finished converting.

By adding $OH^{-}$ to neutralize every $H^{+}$, we convert to a form appropriate for basic conditions.

The Key Conversion

$\boxed{\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}}$

If there are 6 $H^{+}$ in your acidic-balanced equation, add 6 $OH^{-}$ to both sides.

🧪 Worked Example

Problem: Balance in basic solution:

$\text{MnO}_4^- + \text{Br}^- \rightarrow \text{MnO}_2 + \text{BrO}_3^-$

Steps 1–7: Balance in acidic solution first

Reduction: $\text{MnO}_4^- \rightarrow \text{MnO}_2$

• Balance O: $\text{MnO}_4^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$
• Balance H: $4\text{H}^+ + \text{MnO}_4^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$
• Balance charge: $3e^- + 4\text{H}^+ + \text{MnO}_4^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$

Oxidation: $\text{Br}^- \rightarrow \text{BrO}_3^-$

• Balance O: $3\text{H}_2\text{O} + \text{Br}^- \rightarrow \text{BrO}_3^-$
• Balance H: $3\text{H}_2\text{O} + \text{Br}^- \rightarrow \text{BrO}_3^- + 6\text{H}^+$
• Balance charge: $3\text{H}_2\text{O} + \text{Br}^- \rightarrow \text{BrO}_3^- + 6\text{H}^+ + 6e^-$

Equalize electrons: Multiply reduction by 2: $6e^- + 8\text{H}^+ + 2\text{MnO}_4^- \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O}$

Add: $8\text{H}^+ + 2\text{MnO}_4^- + 3\text{H}_2\text{O} + \text{Br}^- \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O} + \text{BrO}_3^- + 6\text{H}^+$

Simplify $H^{+}$ and $H_{2}O$: $2\text{H}^+ + 2\text{MnO}_4^- + \text{Br}^- \rightarrow 2\text{MnO}_2 + \text{H}_2\text{O} + \text{BrO}_3^-$

Steps 8–10: Convert to basic

Add 2 $OH^{-}$ to both sides (to neutralize 2 $H^{+}$):

$2\text{H}_2\text{O} + 2\text{MnO}_4^- + \text{Br}^- \rightarrow 2\text{MnO}_2 + \text{H}_2\text{O} + \text{BrO}_3^- + 2\text{OH}^-$

Cancel 1 $H_{2}O$ from both sides:

$\boxed{\text{H}_2\text{O} + 2\text{MnO}_4^- + \text{Br}^- \rightarrow 2\text{MnO}_2 + \text{BrO}_3^- + 2\text{OH}^-}$

✅ No $H^{+}$ remains — appropriate for basic solution!

Basic Solution Balancing Quiz 🎯

Basic Solution Conversion 🧮

An equation balanced in acidic solution is:

$3\text{Cu}(s) + 8\text{H}^+ + 2\text{NO}_3^- \rightarrow 3\text{Cu}^{2+} + 2\text{NO}(g) + 4\text{H}_2\text{O}$

Convert to basic solution:

1) How many $OH^{-}$ must be added to both sides?

2) How many $H_{2}O$ molecules appear on the LEFT side after combining $H^{+}$ + $OH^{-}$?

3) After canceling $H_{2}O$, how many $H_{2}O$ remain on the product side? (Hint: 8 $H_{2}O$ form on the left, 4 $H_{2}O$ already on right)

Acidic vs. Basic Balancing 🔽

Exit Quiz — Balancing Redox in Basic Solution ✅

⚡ Activity Series and Predicting Redox

Part 5 of 7 — Metals Activity Series and Spontaneous Reactions

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

• Understanding the core concepts covered in Part 5
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 The Activity Series of Metals

Ranked from Most Active to Least Active

📌 Using the Activity Series

The Golden Rule

🔑 Key Concept: A metal can displace (replace) any metal below it in the activity series from a solution of that metal's ions.

$\boxed{\text{More active metal} + \text{Less active metal ion} \rightarrow \text{Reaction occurs!}}$

$\boxed{\text{Less active metal} + \text{More active metal ion} \rightarrow \text{No reaction (NR)}}$

Examples

Zn(s) + $CuSO_{4}(aq)$ → ?

• Zn is ABOVE Cu in the series → reaction occurs $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

Cu(s) + $ZnSO_{4}(aq)$ → ?

• Cu is BELOW Zn in the series → no reaction (NR)

Metals and Acids

Metals above hydrogen in the activity series react with dilute acids (HCl, $H_{2}SO_{4}$) to produce $H_{2}$ gas:

$\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$

⚠️ Warning: Metals below hydrogen (Cu, Ag, Pt, Au) do NOT react with dilute HCl or $H_{2}SO_{4}$.

🔧 Practical Applications

Why Gold Doesn't Corrode

💡 Tip: Gold (Au) is at the bottom of the activity series — it cannot be oxidized by water, air, or common acids. This is why gold jewelry stays shiny for thousands of years.

Galvanized Steel

Steel (mostly Fe) is coated with zinc (Zn). Since Zn is more active than Fe, the zinc corrodes preferentially, protecting the iron underneath. This is called sacrificial protection.

Copper Pennies in Silver Nitrate

When a copper penny is placed in $AgNO_{3}$ solution: $\text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)$

Cu is above Ag → reaction occurs. Silver crystals grow on the penny while the solution turns blue $(Cu^{2+})$.

Dissolving Gold

Gold requires aqua regia (a mixture of $HNO_{3}$ and HCl) — ordinary acids cannot oxidize it.

Activity Series Quiz 🎯

Predict the Reaction 🧮

Will a reaction occur? Type yes or no.

1) Ag(s) + $CuSO_{4}(aq)$ → ?

2) Mg(s) + $FeCl_{2}(aq)$ → ?

3) Fe(s) + HCl(aq) → ?

Activity Series Concepts 🔽

Exit Quiz — Activity Series ✅

⚡ Problem-Solving Workshop

Part 6 of 7 — Mixed Redox Balancing Practice

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

• Working through complete multi-step problems from start to finish
• Building problem-solving strategies you can apply on the AP exam
• Identifying which concepts to apply and in what order

🛠️ Problem-Solving Strategy

Decision Flowchart

1. Assign oxidation states — find which atoms change
2. Write half-reactions — one for oxidation, one for reduction
3. Check the medium:
   • Acidic → use $H_{2}O$ and $H^{+}$
   • Basic → balance in acid first, then add $OH^{-}$
4. Balance each half-reaction (atoms, then charge with $e^{-}$)
5. Equalize and add — cancel electrons
6. Verify — atoms AND charge must balance

💡 Tip: Always verify BOTH atoms and charge in your final answer — a common source of lost points on the AP exam.

Common Patterns to Recognize

🔑 Key Concept: Memorize these common species and their typical products — they appear frequently on the AP exam.

Balancing Practice — Acidic Solution 🎯

Balancing Practice — Basic Solution 🎯

Quick Oxidation State Check 🧮

Determine the oxidation state change for the underlined element in each half-reaction.

1) $\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$: Each Cr changes from ____ to +3 (give initial oxidation state with sign)

2) $\text{I}^- \rightarrow \text{I}_2$: Each I changes from −1 to ____ (give final oxidation state with sign)

3) $\text{SO}_3^{2-} \rightarrow \text{SO}_4^{2-}$: S changes from ____ to +6 (give initial oxidation state with sign)

Redox Balancing Strategy 🔽

Exit Quiz — Problem-Solving Workshop ✅

⚡ Synthesis & AP Review

Part 7 of 7 — Connecting Redox to Electrochemistry and AP-Style Problems

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

• Solving AP-style questions that integrate multiple concepts from this unit
• Writing clear, concise explanations using proper chemistry terminology
• Identifying and avoiding common AP exam traps and mistakes

🔗 Redox ↔ Electrochemistry Connection

Galvanic (Voltaic) Cells

A galvanic cell converts chemical energy → electrical energy using a spontaneous redox reaction.

Memory Aid

🔑 Key Concept: AN OX and a RED CAT — Anode = Oxidation, Reduction = Cathode.

• Anode = Oxidation
• Reduction = Cathode

Cell Notation

$\boxed{\text{Anode} | \text{Anode ion} || \text{Cathode ion} | \text{Cathode}}$

Example: $\text{Zn}(s) | \text{Zn}^{2+}(aq) || \text{Cu}^{2+}(aq) | \text{Cu}(s)$

This represents: Zn is oxidized at the anode, $Cu^{2+}$ is reduced at the cathode.

🔋 Standard Cell Potential

Calculating $E^\circ_{\text{cell}}$

$\boxed{E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}}$

Key Standard Reduction Potentials

Spontaneity

💡 Tip: Positive $E^\circ_{\text{cell}}$ means the reaction runs on its own (galvanic cell). Negative means you must force it (electrolysis).

• $E^\circ_{\text{cell}} > 0$ → spontaneous (galvanic cell)
• $E^\circ_{\text{cell}} < 0$ → non-spontaneous (requires electrolysis)

Relationship to Free Energy

$\boxed{\Delta G^\circ = -nFE^\circ_{\text{cell}}}$

Where $n$ = moles of electrons transferred, $F$ = Faraday's constant (96,485 C/mol).

AP-Style Redox Questions — Set 1 🎯

Cell Potential Calculations 🧮

Use the reduction potentials: $Ag^{+}/Ag$ = +0.80 V, $Fe^{2+}/Fe$ = −0.45 V, $Cu^{2+}/Cu$ = +0.34 V

1) Calculate $E^\circ_{\text{cell}}$ for Fe | $Fe^{2+}$ || $Ag^{+}$ | Ag (in V, to 3 significant figures)

2) Calculate $E^\circ_{\text{cell}}$ for Fe | $Fe^{2+}$ || $Cu^{2+}$ | Cu (in V, to 3 significant figures)

3) Is the cell Cu | $Cu^{2+}$ || $Fe^{2+}$ | Fe spontaneous? Type yes or no.

AP Redox Review 🔽

AP-Style Questions — Set 2 🏆

Final Exit Quiz — Redox Mastery ✅