🎯⭐ INTERACTIVE LESSON

Oxidation-Reduction (Redox) Reactions

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Oxidation-Reduction (Redox) Reactions - Complete Interactive Lesson

Part 1: Oxidation States

⚡ Oxidation States

Part 1 of 7 — Rules for Assigning Oxidation Numbers

Topics in This Part

Section
📏 Rules for Assigning Oxidation States
Rule 6 Is Your Calculation Tool
🧪 Worked Examples
Example 1: H₂SO₄
Example 2: MnO₄⁻ (permanganate ion)

🔑 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.

Rule	Description	Example
1	Free elements have oxidation state 0	Fe(s) = 0, O₂(g) = 0
2	Monoatomic ions = their charge	Na⁺ = +1, Cl⁻ = −1, Fe³⁺ = +3
3	Fluorine is always −1	HF: F = −1
4	Oxygen is usually −2	H₂O: O = −2
	Exception: peroxides (−1)	H₂O₂: O = −1
	Exception: OF₂ (+2)	OF₂: O = +2
5	Hydrogen is usually +1	HCl: H = +1
	Exception: metal hydrides (−1)

🧪 Worked Examples

Example 1: H₂SO₄

Problem: Find the oxidation state of sulfur in H₂SO₄.

Solution:

H = +1 (Rule 5), O = −2 (Rule 4)

$2(+1) + S + 4(-2) = 0$

$+2 + S - 8 = 0$

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 SO₄²⁻

2) Phosphorus in H₃PO₄

3) Manganese in MnO₂

Oxidation State Rules 🔽

Exit Quiz — Oxidation States ✅

Part 2: Identifying Redox Reactions

⚡ Identifying Redox Reactions

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

Topics in This Part

Section
📌 OIL RIG — The Key Mnemonic
How to Spot a Redox Reaction
Example
⚡ Oxidizing and Reducing Agents
Definitions

🔑 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.

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

Part 3: Oxidizing & Reducing Agents

⚡ Balancing Redox in Acidic Solution

Part 3 of 7 — The Half-Reaction Method

Topics in This Part

Section
⚗️ The Half-Reaction Method (Acidic Solution)
The 7 Steps
Key Principle
🧪 Worked Example
Step 1: Write half-reactions

🔑 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

Step	Action
1	Separate the equation into two half-reactions
2	Balance atoms other than O and H in each half-reaction
3	Balance O by adding H₂O
4

Part 4: Balancing Redox (Half-Reaction)

⚡ Balancing Redox in Basic Solution

Part 4 of 7 — Adding OH⁻ to Neutralize H⁺

Topics in This Part

Section
🧪 The Basic Solution Method
Strategy: Balance in Acid First, Then Convert
Why This Works
The Key Conversion
🧪 Worked Example

🔑 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⁻.

Step	Action
1–7	Balance as if in acidic solution (same 7 steps)
8	Add OH⁻ to — one OH⁻ for each H⁺

Part 5: Redox in Acidic & Basic Solutions

⚡ Activity Series and Predicting Redox

Part 5 of 7 — Metals Activity Series and Spontaneous Reactions

Topics in This Part

Section
📌 The Activity Series of Metals
Ranked from Most Active to Least Active
📌 Using the Activity Series
The Golden Rule
Examples

🔑 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

Rank	Metal	Oxidation	Notes
1	Li	Li → Li⁺ + e⁻	Most active — reacts with cold water
2	K	K → K⁺ + e⁻

Part 6: Problem-Solving Workshop

⚡ 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

Assign oxidation states — find which atoms change
Write half-reactions — one for oxidation, one for reduction
Check the medium:
- Acidic → use H₂O and H⁺
- Basic → balance in acid first, then add OH⁻
Balance each half-reaction (atoms, then charge with e⁻)
Equalize and add — cancel electrons
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.

Part 7: Synthesis & AP Review

⚡ 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.

Component	Role
Anode	Where oxidation occurs (negative terminal)
Cathode	Where reduction occurs (positive terminal)
Salt bridge	Allows ion flow to maintain charge balance
Wire

🔑 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:

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

Example 2: MnO₄⁻ (permanganate ion)

Problem: Find the oxidation state of manganese in MnO₄⁻.

Solution:

O = −2 (Rule 4)

$\text{Mn} + 4(-2) = -1$ (charge of ion)

$\text{Mn} - 8 = -1$

$\boxed{\text{Mn} = +7}$

Example 3: Cr₂O₇²⁻ (dichromate ion)

Problem: Find the oxidation state of chromium in Cr₂O₇²⁻.

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: Na₂O₂ (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!)

	Meaning	Electrons	Oxidation State
Oxidation	Is	Loss (of electrons)	Increases (more positive)
Reduction	Is	Gain (of electrons)	Decreases (more negative)

How to Spot a Redox Reaction

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

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

Atom	Reactant	Product	Change	Process
Zn	0	+2	↑ +2	Oxidized (lost 2e⁻)
Cu	+2	0	↓ −2	Reduced (gained 2e⁻)

⚡ Oxidizing and Reducing Agents

Definitions

Agent	What It Does	What Happens to It
Oxidizing agent	Causes oxidation in another species	Gets reduced itself
Reducing agent	Causes reduction in another species	Gets oxidized itself

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)
Cu²⁺ is the oxidizing agent — it takes electrons (gets reduced: +2 → 0)

Common Oxidizing Agents

Agent	Why
KMnO₄ (Mn = +7)	Mn is easily reduced
K₂Cr₂O₇ (Cr = +6)	Cr is easily reduced
HNO₃ (concentrated)	NO₃⁻ is a strong oxidizer
O₂	Oxygen readily gains electrons
Halogens (F₂, Cl₂)	Very electronegative

Common Reducing Agents

Agent	Why
Active metals (Na, Mg, Zn)	Easily lose electrons
H₂	Can donate electrons
C (carbon/coke)	Commonly reduces metal ores

⚗️ Redox vs. Non-Redox Reactions

Not All Reactions Are Redox!

Reaction Type	Redox?	Why
Combustion	✅ Yes	Carbon/hydrogen oxidized, oxygen reduced
Synthesis (metal + nonmetal)	✅ Yes	Metal loses e⁻, nonmetal gains e⁻
Single replacement	✅ Yes	One element displaces another
Double replacement	❌ No	Ions just swap partners — no electron transfer
Acid-base (neutralization)	❌ No	Proton transfer, not electron transfer
Precipitation	❌ No	Ions combine to form solid — no e⁻ transfer

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 ✅

🔑 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 H₂O

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

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 H₂O 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 ✅

⚠️ 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.

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

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

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

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

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 H₂O: $2\text{H}^+ + 2\text{MnO}_4^- + \text{Br}^- \rightarrow 2\text{MnO}_2 + \text{H}_2\text{O} + \text{BrO}_3^-$

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 H₂O 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 H₂O molecules appear on the LEFT side after combining H⁺ + OH⁻?

3) After canceling H₂O, how many H₂O remain on the product side? (Hint: 8 H₂O form on the left, 4 H₂O already on right)

Acidic vs. Basic Balancing 🔽

Exit Quiz — Balancing Redox in Basic Solution ✅

📌 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) + CuSO₄(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) + ZnSO₄(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, H₂SO₄) to produce H₂ 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 H₂SO₄.

🔧 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 AgNO₃ 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 (Cu²⁺).

Dissolving Gold

Gold requires aqua regia (a mixture of HNO₃ and HCl) — ordinary acids cannot oxidize it.

Activity Series Quiz 🎯

Predict the Reaction 🧮

Will a reaction occur? Type yes or no.

1) Ag(s) + CuSO₄(aq) → ?

2) Mg(s) + FeCl₂(aq) → ?

3) Fe(s) + HCl(aq) → ?

Activity Series Concepts 🔽

Exit Quiz — Activity Series ✅

Species	Typical Behavior	Product
MnO₄⁻ (acidic)	Strong oxidizer	Mn²⁺
MnO₄⁻ (basic)	Moderate oxidizer	MnO₂
Cr₂O₇²⁻ (acidic)	Strong oxidizer	Cr³⁺
NO₃⁻ (acidic, dilute)	Oxidizer	NO
NO₃⁻ (acidic, conc.)	Oxidizer	NO₂
H₂O₂	Can oxidize or reduce	O₂ or H₂O

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 ✅

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

Anode = Oxidation
Reduction = Cathode

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

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

This represents: Zn is oxidized at the anode, Cu²⁺ 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

Half-Reaction	$E^\circ$ (V)
$\text{F}_2 + 2e^- \rightarrow 2\text{F}^-$

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)
$E^\circ_{\text{cell}} < 0$ → (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, Fe²⁺/Fe = −0.45 V, Cu²⁺/Cu = +0.34 V

1) Calculate $E^\circ_{\text{cell}}$ for Fe | Fe²⁺ || Ag⁺ | Ag (in V, to 3 significant figures)

2) Calculate $E^\circ_{\text{cell}}$ for Fe | Fe²⁺ || Cu²⁺ | Cu (in V, to 3 significant figures)

3) Is the cell Cu | Cu²⁺ || Fe²⁺ | Fe spontaneous? Type yes or no.

AP Redox Review 🔽

AP-Style Questions — Set 2 🏆

Final Exit Quiz — Redox Mastery ✅

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}

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^-

Oxidation-Reduction (Redox) Reactions

Oxidation-Reduction (Redox) Reactions - Complete Interactive Lesson

Part 1: Oxidation States

⚡ Oxidation States

Topics in This Part

What You'll Master in Part 1

📏 Rules for Assigning Oxidation States

🧪 Worked Examples

Example 1: H₂SO₄

Part 2: Identifying Redox Reactions

⚡ Identifying Redox Reactions

Topics in This Part

What You'll Master in Part 2

📌 OIL RIG — The Key Mnemonic

Part 3: Oxidizing & Reducing Agents

⚡ Balancing Redox in Acidic Solution

Topics in This Part

What You'll Master in Part 3

⚗️ The Half-Reaction Method (Acidic Solution)

The 7 Steps

Part 4: Balancing Redox (Half-Reaction)

⚡ Balancing Redox in Basic Solution

Topics in This Part

What You'll Master in Part 4

🧪 The Basic Solution Method

Strategy: Balance in Acid First, Then Convert

Part 5: Redox in Acidic & Basic Solutions

⚡ Activity Series and Predicting Redox

Topics in This Part

What You'll Master in Part 5

📌 The Activity Series of Metals

Ranked from Most Active to Least Active

Part 6: Problem-Solving Workshop

⚡ Problem-Solving Workshop

Practice Makes Perfect

What You'll Master in Part 6

🛠️ Problem-Solving Strategy

Decision Flowchart

Common Patterns to Recognize

Part 7: Synthesis & AP Review

⚡ Synthesis & AP Review

Bringing It All Together

What You'll Master in Part 7

🔗 Redox ↔ Electrochemistry Connection

Galvanic (Voltaic) Cells

Rule 6 Is Your Calculation Tool

Example 2: MnO₄⁻ (permanganate ion)

Example 3: Cr₂O₇²⁻ (dichromate ion)

Example 4: Na₂O₂ (sodium peroxide)

How to Spot a Redox Reaction

Example

⚡ Oxidizing and Reducing Agents

Definitions

The Tricky Part

Example (continued)

Common Oxidizing Agents

Common Reducing Agents

⚗️ Redox vs. Non-Redox Reactions

Not All Reactions Are Redox!

Quick Test

Key Principle

🧪 Worked Example

Step 1: Write half-reactions

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

Step 3: Balance O with H₂O

Step 4: Balance H with H⁺

Step 5: Balance charge with e⁻

Step 6: Equalize electrons (multiply oxidation by 5)

Step 7: Add and cancel e⁻

✅ Verify

Why This Works

The Key Conversion

🧪 Worked Example

Steps 1–7: Balance in acidic solution first

Steps 8–10: Convert to basic

📌 Using the Activity Series

The Golden Rule

Examples

Metals and Acids

🔧 Practical Applications

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