🎯⭐ INTERACTIVE LESSON

Oxidation-Reduction (Redox) Reactions

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Oxidation-Reduction (Redox) Reactions - Complete Interactive Lesson

Part 1: Oxidation States

⚡ Oxidation States

Part 1 of 7 — Rules for Assigning Oxidation Numbers

Oxidation-reduction (redox) reactions involve the transfer of electrons between species. To track where electrons go, we assign oxidation states (also called oxidation numbers) to every atom. These are not always real charges — they're a bookkeeping tool that lets us identify which atoms gain or lose electrons.

📏 Rules for Assigning Oxidation States

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

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)	NaH: H = −1
6	Sum of oxidation states = charge of species	Neutral compound: sum = 0
		Ion: sum = ion charge

Rule 6 Is Your Calculation Tool

For any compound or polyatomic ion:

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

🧪 Worked Examples

Example 1: H₂SO₄

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

Now that you can assign oxidation states, it's time to use them to identify redox reactions and determine which species is oxidized, which is reduced, and who the oxidizing and reducing agents are.

📌 OIL RIG — The Key Mnemonic

$\text{OIL RIG}$

	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

Part 3: Oxidizing & Reducing Agents

⚡ Balancing Redox in Acidic Solution

Part 3 of 7 — The Half-Reaction Method

Balancing redox equations is more complex than balancing regular equations because you must balance both atoms and charge. The half-reaction method breaks the problem into two manageable pieces: one for oxidation and one for reduction.

⚗️ 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	Balance H by adding H⁺
5	Balance charge by adding electrons (e⁻)
6	Equalize electrons — multiply half-reactions so e⁻ cancel
7	Add half-reactions together and simplify

Key Principle

Electrons lost in oxidation must equal electrons gained in reduction. This is why we equalize in Step 6.

Part 4: Balancing Redox (Half-Reaction)

⚡ Balancing Redox in Basic Solution

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

Many redox reactions occur in basic (alkaline) solution — for example, in batteries and biological systems. The method is almost identical to the acidic method, with one extra step at the end: we neutralize H⁺ by adding OH⁻.

🧪 The Basic Solution Method

Strategy: Balance in Acid First, Then Convert

Step	Action
1–7	Balance as if in acidic solution (same 7 steps)
8	Add OH⁻ to both sides — one OH⁻ for each H⁺
9	Combine H⁺ + OH⁻ → H₂O on the appropriate side
10	Cancel any H₂O that appears on both sides

Why This Works

In basic solution, free H⁺ ions don't exist — they would react with the abundant OH⁻. By adding OH⁻ to neutralize every H⁺, we convert to a form appropriate for basic conditions.

The Key Conversion

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

Part 5: Redox in Acidic & Basic Solutions

⚡ Activity Series and Predicting Redox

Part 5 of 7 — Metals Activity Series and Spontaneous Reactions

Not every possible redox reaction actually occurs. The activity series ranks metals (and hydrogen) by their tendency to lose electrons. This ranking lets you predict whether a single-replacement reaction will happen spontaneously.

📌 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⁻	Reacts violently with water
3	Ba	Ba → Ba²⁺ + 2e⁻	Reacts with water
4	Ca	Ca → Ca²⁺ + 2e⁻	Reacts with water
5	Na	Na → Na⁺ + e⁻	Reacts with cold water
6	Mg	Mg → Mg²⁺ + 2e⁻

Part 6: Problem-Solving Workshop

⚡ Problem-Solving Workshop

Part 6 of 7 — Mixed Redox Balancing Practice

This workshop brings together all the redox skills: assigning oxidation states, identifying oxidized/reduced species, balancing in acidic solution, and balancing in basic solution. Work through these problems systematically using the half-reaction method.

🛠️ 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

Common Patterns to Recognize

Species	Typical Behavior	Product
MnO₄⁻ (acidic)	Strong oxidizer	Mn²⁺
MnO₄⁻ (basic)	Moderate oxidizer	MnO₂
Cr₂O₇²⁻ (acidic)	Strong oxidizer	Cr³⁺
NO₃⁻ (acidic, dilute)	Oxidizer	NO

Part 7: Synthesis & AP Review

⚡ Synthesis & AP Review

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

Redox reactions are the foundation of electrochemistry — the study of how chemical energy and electrical energy interconvert. This final lesson connects the redox concepts you've learned to galvanic cells, electrolysis, and the types of problems you'll see on the AP exam.

🔗 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	Carries electrons from anode to cathode

Memory Aid

AN OX and a RED CAT:

Anode = Oxidation
Reduction = Cathode

Cell Notation

Example 2: MnO₄⁻ (permanganate ion)

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

Example 3: Cr₂O₇²⁻ (dichromate ion)

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

Example 4: Na₂O₂ (sodium peroxide)

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

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

The names seem backwards! Remember:

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

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 ✅

🧪 Worked Example

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⁻

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

Verify

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 ✅

If there are 6 H⁺ in your acidic-balanced equation, add 6 OH⁻ to both sides.

🧪 Worked Example

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:

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

✅ 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

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

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

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

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:

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

Metals below hydrogen (Cu, Ag, Pt, Au) do NOT react with dilute HCl or H₂SO₄.

🔧 Practical Applications

Why Gold Doesn't Corrode

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 ✅

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 ✅

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

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

Key Standard Reduction Potentials

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

Spontaneity

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

Relationship to Free Energy

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

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

📏 Rules for Assigning Oxidation States

Rule 6 Is Your Calculation Tool

🧪 Worked Examples

Example 1: H₂SO₄

Part 2: Identifying Redox Reactions

⚡ Identifying Redox Reactions

📌 OIL RIG — The Key Mnemonic

How to Spot a Redox Reaction

Part 3: Oxidizing & Reducing Agents

⚡ Balancing Redox in Acidic Solution

⚗️ The Half-Reaction Method (Acidic Solution)

The 7 Steps

Key Principle

Part 4: Balancing Redox (Half-Reaction)

⚡ Balancing Redox in Basic Solution

🧪 The Basic Solution Method

Strategy: Balance in Acid First, Then Convert

Why This Works

The Key Conversion

Part 5: Redox in Acidic & Basic Solutions

⚡ Activity Series and Predicting Redox

📌 The Activity Series of Metals

Ranked from Most Active to Least Active

Part 6: Problem-Solving Workshop

⚡ Problem-Solving Workshop

🛠️ Problem-Solving Strategy

Decision Flowchart

Common Patterns to Recognize

Part 7: Synthesis & AP Review

⚡ Synthesis & AP Review

🔗 Redox ↔ Electrochemistry Connection

Galvanic (Voltaic) Cells

Memory Aid

Cell Notation

Example 2: MnO₄⁻ (permanganate ion)

Example 3: Cr₂O₇²⁻ (dichromate ion)

Example 4: Na₂O₂ (sodium peroxide)

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

🧪 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

🧪 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

Why Gold Doesn't Corrode

Galvanized Steel

Copper Pennies in Silver Nitrate

Dissolving Gold

🔋 Standard Cell Potential

Calculating Ecell∘E^\circ_{\text{cell}}Ecell∘​

Key Standard Reduction Potentials

Spontaneity

Relationship to Free Energy

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