Chapter 1-7: Oxidation Numbers and Redox (Vocabulary Flashcards)

Nonmetal Oxides and Acid Formation

Nonmetal oxides: oxides of nonmetals (i.e., compounds containing oxygen bound to a nonmetal, such as CO₂).
Rule: Nonmetal oxides react with water to form acids.
- An acid typically begins with hydrogen (H) when formed from these reactions.
- Example:
- Carbon dioxide + water forms carbonic acid: $ \mathrm{CO2 + H2O \rightarrow H2CO3} $
- Concept mentioned in class: these acids can relate to carbonate chemistry (carbonate ion $\mathrm{CO_3^{2-}}$ ) and balancing considerations with charges (hydrogen’s +1 charge vs carbonate’s 2− charge).
Note on terminology:
- When you hear “acid” formed from a nonmetal oxide and water, think of an H-containing species (like H₂CO₃ in the CO₂ example).
- “Hydroxides” are mentioned as a related concept to hydroxide-containing species, but the main rule here is nonmetal oxides + water → acids.

Redox Reactions: Oxidation Numbers and Nomenclature

Redox (oxidation-reduction) reactions involve a transfer of electrons between species.
- Oxidation = loss of electrons.
- Reduction = gain of electrons.
Two common memory aids:
- Leo says GER: gain Electron Reduction; Lose Electron oxidation (GER: Gain, Electron, Reduction).
- Oil Rig: Oxidation is Loss, Reduction is Gain.
Oxidation numbers (ON) are used to track electron transfer across a reaction.
In redox problems, you often see many ON values; you should write down the numbers that you know and work out the unknowns using the rules (see the five rules below).
Clues to identify redox from a formula:
- Look for changes in oxidation numbers between reactants and products.
- Example discussion from the transcript: Chromium changes from ON +6 in a reactant to ON +3 in a product.
- If chromium goes +6 → +3, that is a decrease in oxidation state (reduction). Because electrons are negatively charged, reducing Cr means Cr has gained electrons.
- In this example, the other species must balance the electron gain by losing electrons (oxidation).
Worked example context in transcript (long, complex balancing with polyatomic ions, water, etc.) is used to illustrate how ONs are assigned and how redox direction is inferred before coefficients are balanced.

Five Oxidation Number Rules (as introduced in the lecture)

1) Rule 1: Elements in their natural state have oxidation number 0.

Natural state means the element is not an ion (e.g., Zn, Cu, Al, etc., as neutral atoms). For diatomic elements (e.g., O₂, N₂, H₂), the ON is 0 for the molecule in its elemental form.
Note: Ions by themselves are not in a natural state; ions carry charge and have ONs corresponding to that charge (e.g., Zn^{2+} is ON +2).
2) Rule 2: Oxygen in a compound is almost always ON = −2.
The exception is peroxides (e.g., $\mathrm{H2O2}$ ), where oxygen has ON = −1.
Generally: for any oxide in a compound, O contributes −2 per atom.
Example shown: in many oxides, you will assign O as −2 unless you’re specifically dealing with a peroxide.
3) Rule 3: The oxidation number of metals in a compound equals the charge that metal would have as a cation (based on its group in the periodic table).
Example: Calcium in compounds tends to be Ca^{2+} → ON(Ca) = +2.
If a metal is present as a neutral atom (not an ion) in a compound, use the charge implied by its role in that compound (e.g., Ca in CaCl₂ has ON = +2).
Distinction: if a metal is by itself (not in a compound) and is in natural state, its ON is 0; if it is an ion, use the ionic charge.
4) Rule 4: The oxidation number of hydrogen in a compound is normally +1.
The exception is metal hydrides (e.g., in a metal hydride like NaH, or CaH₂), where hydrogen has ON = −1.
Important caution: a metal hydride contains two elements only, with H behaving as −1.
5) Rule 5: The sum of all oxidation numbers in a compound or ion equals the compound’s or ion’s overall charge.
For neutral molecules, the sum is 0.
For ions, the sum equals the ion’s charge (e.g., NO₃⁻ has ONs summing to −1).
- Additional clarifications from the lecture:
- An oxidation state can be 0 even when the atom is not in its neutral, natural form (e.g., carbon in some molecules within a molecule can have ON = 0). This is noted as possible even when not in a pure elemental form.
- If an atom appears in the same compound on the same side of the equation, the ON assigned to those atoms is the same (e.g., carbons on the same side of a neutral equation will share the same ON if they are in the same oxidation environment).

Worked Example: Dichromate and Polyatomic Ions (illustrative ON solving)

Polyatomic dichromate ion: $\mathrm{Cr2O7^{2-}}$
- Oxygen ON is −2 per O, with 7 O atoms → total from oxygen = $7\times(-2) = -14.$
- Let the oxidation state of chromium be x for each Cr; there are 2 Cr atoms, so total from Cr = $2x.$
- The overall charge of the ion is −2, so:
 $2x + (-14) = -2 \Rightarrow 2x = 12 \Rightarrow x = +6.$
- Result: Each Cr in $\mathrm{Cr2O7^{2-}}$ has ON = +6.
Ammonium ion $\mathrm{NH_4^+}$
- Hydrogen ON = +1; there are 4 hydrogens → total from H = $4(+1) = +4.$
- The overall charge is +1, so the nitrogen ON must satisfy: $xN + 4 = +1 \Rightarrow xN = -3.$
- Result: N in ammonium has ON = −3.
Water, $\mathrm{H_2O}$
- Each hydrogen ON = +1; for 2 H, total = +2.
- Oxygen ON = −2; total = −2.
- Sum = 0 (neutral molecule).
Key takeaway from this part: ONs can be assigned to polyatomic ions and neutral molecules by applying the rules and using the overall charge constraint.

Practical approach to a big redox problem (transcript-based strategy)

Step 1: List known ONs that you can confidently assign from the formula:
- H in most compounds: +1; in water: H = +1.
- O in most compounds: −2 (except peroxides).
- N in ammonium: −3 (since NH₄⁺ overall charge is +1).
- Metals in ionic form: ON equals their common ionic charge (e.g., Cr in charged species may have ONs like +3, +6, etc.).
Step 2: For polyatomic ions with a known total charge, use the equation:
- Sum of ONs in the ion equals its charge.
- Example: For $\mathrm{Cr2O7^{2-}}$ , use $2x + 7(-2) = -2$ to solve for x (ON of Cr).
Step 3: For a neutral compound, sum ONs to 0; for an ion, sum ONs to the ion’s charge.
Step 4: Determine which elements are oxidized and which are reduced by comparing ONs from reactants to products:
- If ON increases → oxidation (lose electrons).
- If ON decreases → reduction (gain electrons).
Step 5: Use the comparison to identify the redox couple(s) and then, if needed, balance using a systematic redox method (half-reaction method), noting that the transcript emphasizes identifying oxidation/reduction directions before coefficients.
Step 6: Note practical tips from the example:
- The overall reaction may involve multiple species (polyatomic ions, water, etc.).
- Carbons in the same molecule on the same side will share a charge situation that helps determine whether a given C is oxidized or reduced (e.g., carbon ON changing from −2 to 0 indicates oxidation).
- In the example provided, chromium moved from +6 to +3 (reduced), and carbon moved from −2 to 0 (oxidized).
Summary from the example reaction (harmonized view):
- Chromium: ON changes from +6 to +3 → reduction (gains electrons).
- Carbon: ON changes from −2 to 0 → oxidation (loses electrons).
- The balancing of coefficients is not the focus here; the emphasis is recognizing which species are oxidized vs reduced based on ON changes.

Quick reference: ON changes and what they imply

Positive change in ON for an element in a reaction → oxidation (loss of electrons).
Negative change in ON for an element in a reaction → reduction (gain of electrons).
For example: $\mathrm{Cr^{6+} \rightarrow Cr^{3+}}$ implies gain of 3 electrons by chromium (reduction).
For example: $\mathrm{C^{-2} \rightarrow C^{0}}$ implies loss of 2 electrons by carbon (oxidation).
Always check the overall charge balance and the sum of ONs to ensure consistency with the reaction’s charge and mass balance.

AP Exam administrative note (as mentioned in the transcript)

Undecided option for AP exam status must be set by 09/26/2025 in the College Board account.
Exam order must be placed by 10/03; the status cannot be changed after 09/26/2025.
Students should log into their College Board AP account to indicate whether they will take the AP exam.
If you don’t have an AP account, you should create one; a class code is provided for access to the AP Chemistry course.
The class code and instructions include what to do if you already have an account (log in and select the class code) vs. creating a new account and selecting undecided/take/not take status.
There is a test fee; even if you decide not to take the test, you may still receive a grade bump depending on the school policy.

Connections to broader principles and real-world relevance

Redox concepts underpin many real-world processes: battery operation, corrosion, metabolism, and industrial redox processes.
Understanding oxidation states helps in predicting reaction directions, balancing redox reactions, and rationalizing why certain species behave as oxidizing or reducing agents.
Recognizing polyatomic ions and their charges is essential for solving complex redox problems without getting lost in coefficients.
The balancing approach and ON-tracking connect to foundational ideas in chemistry: conservation of charge, conservation of mass, and electron bookkeeping.

Summary tips for studying this material

Memorize the five oxidation-number rules as a quick reference during problems:
- Natural state ON = 0.
- Oxygen in compounds is −2 (peroxides exception: −1).
- Metals in compounds adopt the charge typical for their group (and ions in ionic form have that charge as ON).
- Hydrogen is +1 in most compounds; hydrides have H = −1.
- Sum of ONs in a molecule/ion equals its overall charge.
Use the “identify the knowns” step when confronted with large molecules to anchor the ON assignment.
Practice with polyatomic ions to reinforce how charges constrain ONs (e.g., Cr in $\mathrm{Cr2O7^{2-}}$ is +6).
When in doubt, start with the easy ONs (H, O, known ions) and work toward the unknowns.
Keep in mind the common redox direction rules (oxidation = loss, reduction = gain) and verify by checking ON changes.