GENERAL CHEMISTRY · OXIDATION–REDUCTION REACTIONS

Close the ledger.
Then close the circuit.

Learn all four official topics through atom, charge, oxidation-state, potential, current, pathway, and notation ledgers, then retrieve with sixteen original five-choice questions.

Start the lessons ↓Open all problems →

4guided lessons

16practice questions

12study objectives

5choices per item

A repeatable electrochemistry routine

Inventory. Balance. Model. Verify.

01Inventory the system
Name every species, phase, charge, coefficient, medium, temperature, concentration, current, time, and cell mode supplied.
02Balance electron transfer
Track oxidation states, construct half-reactions, equalize electrons, and conserve atoms plus net charge.
03Select the cell model
Identify anode and cathode, standard versus nonstandard potential, galvanic versus electrolytic operation, or Faraday stoichiometry.
04Verify the boundary
Check signs, intensive potential, Q versus q, seconds, electron ratios, pathways, phase boundaries, and model assumptions.

Four prerequisite-aware lessons

Let conservation choose the model.

Each lesson pairs two semantic ledgers with a fully worked example and closed-note retrieval before multiple-choice practice.

LESSON 1 · 24 MIN

Study + retrieve

Balance the electron ledger

Build and verify net ionic redox equations in acidic or basic solution while conserving every atom, net charge, and transferred electron.

ESSENTIAL QUESTIONWhich species belong to the stated medium, and does each half-reaction close its atom and charge ledgers before the electrons cancel?

Redox balancing ledger

Half-reaction order in acidic solution
Step	Balancing species	Verification
1 · Split	Oxidation and reduction halves	Changing atoms appear in the correct half
2 · Atoms	Non-H/O, then H₂O and H⁺	Every element total matches
3 · Charge	Electrons on one side	Net charge matches
4 · Combine	Scale to equal electrons	Electrons cancel completely

Acidic-to-basic conversion and final audit
Operation	Safe move	Failure cue
Neutralize H⁺	Add equal OH⁻ to both sides	Direct H⁺ → OH⁻ symbol swap
Form water	H⁺ + OH⁻ → H₂O	Different additions on the two sides
Cancel	Remove identical H₂O across sides	Canceling unlike phases or species
Final form	Atoms, charge, simplest ratio	Free H⁺ left in a stated basic medium

REDOX / ELECTROCHEMICAL LEDGERS · CAPTIONS INCLUDED

Split by oxidation-state change

Write one half-reaction for the species losing electrons and one for the species gaining them. Balance non-hydrogen and non-oxygen atoms first, then use the medium-specific species and electrons to close each ledger.

Oxidation produces electrons
Reduction consumes electrons

Respect the aqueous medium

In acidic solution, use H₂O for oxygen and H⁺ for hydrogen before balancing charge with electrons. For basic solution, first obtain a valid acidic ledger, add equal OH⁻ to both sides, form water, and cancel only identical species.

Acidic: H₂O, then H⁺
Basic: neutralize H⁺ with OH⁻

Recombine and audit

Scale the half-reactions to equal electron counts, add them, cancel electrons and unchanged species, and reduce to the simplest whole-number ratio. A plausible-looking equation still fails if either element totals or net charge differ.

Equal electron transfer
Atoms + charge + simplest ratio

Worked example

Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.

1
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Oxidation: Fe²⁺ → Fe³⁺ + e⁻.
2
Multiply the iron half-reaction by 5 so both halves transfer 5 electrons.
3
Add and cancel electrons: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Both sides contain net charge +17.

ConclusionThe final equation conserves Mn, O, H, Fe, and net charge, contains no electrons, and is already in the simplest whole-number ratio.

Close the notes first

Retrieve the ledger.

01On which side do electrons appear in an oxidation half-reaction?

The product side.

Oxidation is electron loss.

02How is a valid acidic equation converted to basic conditions?

Add equal OH⁻ to both sides, form water with H⁺, then cancel water where allowed.

Equal addition preserves the equation while removing free H⁺ from the final basic form.

03What three checks must the final net equation pass?

Every element, net charge, and complete electron cancellation.

Atom balance without charge balance is not a valid redox equation.

LESSON 2 · 21 MIN

Study + retrieve

Assign states and name the agents

Solve oxidation numbers from a complete species-charge ledger, apply common rule exceptions, and use before/after changes to label processes and agents.

ESSENTIAL QUESTIONWhich oxidation-number rules control this formula, and which reacting species undergoes the change that defines each agent?

Oxidation-number ledger

Introductory assignment rules and priority
Context	Usual value	Boundary
Elemental substance	0	Applies to the uncombined element
Monatomic ion	Ionic charge	Not the charge of a polyatomic ion
Fluorine	−1 in compounds	Close the complete species sum
O / H	Usually −2 / +1	Peroxides −1; metal hydrides −1

Change, electron, and agent labels
Observed change	Electron language	Agent identity
Oxidation number rises	Oxidation · electron loss	Reducing agent
Oxidation number falls	Reduction · electron gain	Oxidizing agent
Magnitude	\|Δ state\| × atom count	Must match after coefficients
No state change	No redox evidence	May be spectator or unchanged

REDOX / ELECTROCHEMICAL LEDGERS · CAPTIONS INCLUDED

Make the signed sum close

Elemental substances are 0 and monatomic ions equal their charge. In a compound or polyatomic ion, multiply each oxidation number by its atom count so the total equals the species charge.

Neutral sum = 0
Ion sum = net charge

Apply rule priority and exceptions

Fluorine is −1 in compounds; group 1 and 2 metals are commonly +1 and +2. Oxygen is usually −2 but is −1 in a peroxide, while hydrogen is usually +1 with nonmetals but −1 in a binary metal hydride.

Peroxide oxygen = −1
Metal-hydride hydrogen = −1

Label the change undergone

An oxidation-number increase is oxidation and electron loss; a decrease is reduction and electron gain. The species oxidized is the reducing agent, while the species reduced is the oxidizing agent.

Reducing agent is oxidized
Oxidizing agent is reduced

Worked example

In Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), identify both agents and the electron count.

1
Cu changes from 0 to +2, so it loses 2 electrons and is oxidized.
2
Each Ag changes from +1 to 0; two Ag⁺ ions gain 2 electrons total and are reduced.
3
Cu is the reducing agent because it is oxidized. Ag⁺ is the oxidizing agent because it is reduced.

ConclusionAgent labels follow the change undergone by that species—not the process it causes in the other species.

Close the notes first

Retrieve the ledger.

01What must the oxidation-number sum equal for SO₄²⁻?

−2.

The coefficient-weighted oxidation-number sum equals the ion’s net charge.

02What is oxygen’s oxidation number in H₂O₂?

−1.

Hydrogen peroxide contains a peroxide linkage, a standard oxygen exception.

03Which species is the oxidizing agent?

The species that is reduced.

It accepts electrons while causing another species to oxidize.

LESSON 3 · 26 MIN

Study + retrieve

Calculate potential and electrical yield

Calculate standard and nonstandard cell potential, connect E° to ΔG° and K, and convert current through Faraday’s constant to a chemical amount.

ESSENTIAL QUESTIONIs the quantity intensive or extensive, are conditions standard or current, and how many moles of electrons belong to the balanced reaction?

Electrochemical calculation ledger

Potential, Gibbs energy, and equilibrium
Relationship	Required input	Interpretation
E°cell	E°cathode − E°anode	Potential is intensive
ΔG°	−nFE°	n comes from canceled electrons
K	ln K = nFE°/(RT)	E° > 0 means K > 1
Current E	E° − RT ln Q/(nF)	Use current Q and stated T

Current, charge, and product amount
Conversion	Equation	Unit or boundary
Current → charge	q = It	A = C/s; time in seconds
Charge → electrons	mol e⁻ = q/F	F = C/mol e⁻
Electrons → product	Use electrode stoichiometry	Ion charge sets the ratio
Actual yield	Apply stated efficiency	Assume 100% only when supported

REDOX / ELECTROCHEMICAL LEDGERS · CAPTIONS INCLUDED

Subtract reduction potentials

Choose the cathode as the reduction and the anode as the oxidation, then use E°cell = E°cathode − E°anode with tabulated reduction potentials. Scaling a half-reaction never scales its intensive potential.

Cathode minus anode
Do not multiply E°

Connect standard-state signs

For the balanced reaction, ΔG° = −nFE°. At a stated temperature, ln K = nFE°/(RT). Thus positive E° corresponds to negative ΔG° and K greater than 1 for the reaction as written.

n comes from canceled electrons
E° > 0 ↔ ΔG° < 0 ↔ K > 1

Separate Q from electrical charge

The Nernst equation corrects E° with the dimensionless reaction quotient Q under current conditions. Electrolysis instead uses electrical charge q = It, moles of electrons q/F, and the electrode half-reaction’s electron-to-product ratio.

Q changes current potential
q in coulombs drives amount

Worked example

For Zn(s) | Zn²⁺ || Cu²⁺ | Cu(s), E°red(Cu²⁺/Cu) = +0.34 V and E°red(Zn²⁺/Zn) = −0.76 V. Find E°cell and ΔG° for n = 2.

1
Copper is the cathode reduction and zinc is the anode oxidation.
2
E°cell = +0.34 − (−0.76) = +1.10 V. Do not multiply either potential by 2.
3
ΔG° = −(2)(96485 C/mol e⁻)(1.10 J/C) = −2.12 × 10⁵ J/mol = −212 kJ/mol.

ConclusionThe positive potential and negative standard Gibbs energy support the forward galvanic direction under standard conditions; they do not by themselves state reaction speed.

Close the notes first

Retrieve the ledger.

01Does doubling a balanced half-reaction double its electrode potential?

No.

Potential is energy per charge and is intensive.

02At equilibrium, how do E and ΔG compare with zero?

Both are zero for the current reaction conditions.

Q equals K, so there is no net thermodynamic driving force.

03What converts current and time to moles of electrons?

q = It, followed by moles e⁻ = q/F.

An ampere is one coulomb per second, and F is coulombs per mole of electrons.

LESSON 4 · 21 MIN

Study + retrieve

Read the operating cell

Trace electron and ion pathways, preserve anode/cathode reaction roles across cell modes, and translate a galvanic cell into conventional line notation.

ESSENTIAL QUESTIONWhat process occurs at each electrode, how is charge transported, and which interfaces must the notation preserve?

Cell roles and notation ledger

Electrode roles, signs, and charge paths
Feature	Galvanic	Electrolytic
Anode	Oxidation · negative	Oxidation · positive
Cathode	Reduction · positive	Reduction · negative
Electron path	External circuit: anode → cathode
Internal path	Ions in electrolyte or separator

Conventional galvanic line notation
Symbol	Meaning	Safe reading
\|	One phase boundary	Electrode–solution or gas–solution interface
\|\|	Ionic junction	Salt bridge or porous separator
,	Same-phase species	No phase boundary between them
Pt(s)	Inert conductor	Needed when no conducting solid redox species exists

REDOX / ELECTROCHEMICAL LEDGERS · CAPTIONS INCLUDED

Keep reaction roles invariant

Oxidation always occurs at the anode and reduction always occurs at the cathode. In a galvanic cell the anode is negative and cathode positive; in an electrolytic cell external power makes the anode positive and cathode negative.

AnOx · RedCat
Signs depend on cell mode

Trace two different pathways

Electrons travel through the external conductor from anode to cathode. Ions—not electrons—carry charge internally: in the introductory separated-cell model, anions migrate toward the anode compartment and cations toward the cathode compartment.

Wire carries electrons
Bridge carries ions

Write boundaries, not a picture

Conventional galvanic notation places the anode on the left and cathode on the right. A single line marks a phase boundary, a double line the ionic junction, commas separate same-phase species, and an inert conductor is named when no conducting solid redox species is present.

Anode | solution || solution | cathode
Pt supplies a surface, not a reactant

Worked example

Translate the spontaneous zinc–copper cell into roles, pathways, and notation.

1
Zn(s) oxidizes at the negative anode; Cu²⁺ reduces at the positive cathode.
2
Electrons move through the wire from zinc to copper. Bridge anions migrate toward the zinc compartment and cations toward the copper compartment.
3
Write Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).

ConclusionThe double line represents the ionic junction, not the electron wire; the external circuit and internal ion path complete different parts of the circuit.

Close the notes first

Retrieve the ledger.

01Where does oxidation occur in an electrolytic cell?

At the anode.

The anode/cathode definitions do not change with cell mode.

02What crosses a salt bridge?

Ions.

Electrons use the external conductor.

03What does a double vertical line mean in cell notation?

The salt bridge or porous ionic separator.

A single line marks one phase boundary.

All sixteen Oxidation–Reduction problems

Balance the claim before selecting it.

Question order and all five answer options shuffle each time. Reports automatically include the exact question, content version, skill, and seed.

16 PRACTICE QUESTIONS

Retrieve before you review.

Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.

Interpret results carefully

Raw accuracy directs review—not score prediction.

These original questions are draft and have not been calibrated to the official score scale. Use each explanation to repair the exact balancing, oxidation-state, potential, Nernst, Faraday, pathway, sign, or notation decision.

DAT TRAIN does not claim topic quotas because the official manual does not publish them. Stoichiometry owns general mole ratios; Thermodynamics owns the broader Gibbs framework; Chemical Equilibria owns generalized Q/K composition.