⚡ 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 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)
• $2(+1) + S + 4(-2) = 0$
• $+2 + S - 8 = 0$
• $S = +6$

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!)

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 ✅

⚡ 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

$\text{OIL RIG}$

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

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

Common Reducing Agents

Redox vs. Non-Redox Reactions

Not All Reactions Are Redox!

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 ✅

⚡ 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

Key Principle

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

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 ✅

⚡ 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

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

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

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

⚡ 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

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 ✅

⚡ 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

1. Assign oxidation states — find which atoms change
2. Write half-reactions — one for oxidation, one for reduction
3. Check the medium:
   • Acidic → use H₂O 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

Common Patterns to Recognize

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

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.

Memory Aid

AN OX and a RED CAT:

• Anode = Oxidation
• Reduction = Cathode

Cell Notation

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

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

Spontaneity

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