AP Chemistry — Practice
All practice problems and solutions for AP Chemistry, organized by unit. Worked examples stay on the unit pages.
Auto-collected from the practice sections of each unit’s notes (scripts/build_practice.py). Edit the source notes, not this page.
Unit 1: Atomic Structure and Properties
Section titled “Unit 1: Atomic Structure and Properties”Practice
Section titled “Practice”-
A neutral atom has the electron configuration . Which statement is correct?
(A) The atom is in group 4 and forms ions most often.
(B) The atom is in group 16 and has six valence electrons.
(C) The atom is a noble gas because the subshell is occupied.
(D) The atom has four valence electrons because the last exponent is .
The highest principal energy level is , and the atom has
as its valence-shell configuration. That is valence electrons, which places the atom in group .
-
Which set of particles is arranged in order of increasing radius?
(A)
(B)
(C)
(D)
The ions , , and are isoelectronic, each with electrons. In an isoelectronic series, radius decreases as nuclear charge increases. Potassium has the most protons, so is smallest; sulfur has the fewest, so is largest.
So the answer is
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Which photon has the greatest energy?
(A) A photon with wavelength
(B) A photon with wavelength
(C) A photon with wavelength
(D) A photon with wavelength
Photon energy is
Energy is inversely proportional to wavelength, so the shortest wavelength has the greatest energy.
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Which statement best explains why first ionization energy generally increases from left to right across a period?
(A) Atomic radius increases, so electrons are easier to remove.
(B) Effective nuclear charge increases, so valence electrons are held more strongly.
(C) Shielding increases sharply, so valence electrons are held more weakly.
(D) The number of occupied principal energy levels increases.
Across a period, protons are added to the nucleus while electrons are added to the same principal energy level. Shielding does not increase enough to cancel the increased nuclear attraction, so increases.
-
Which element has the electron configuration ?
(A) Mg
(B) Al
(C) Si
(D) P
After the neon core, gives three valence electrons in the third period. That is aluminum.
-
Which set of quantum numbers is not allowed for an electron in an atom?
(A)
(B)
(C)
(D)
For a given , the value of must be an integer from to . If , then is not allowed.
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A sample of chlorine contains atoms and atoms.
Calculate the average atomic mass of chlorine.
Explain why the average atomic mass is closer to than to .
A PES spectrum for chlorine shows peaks from core electrons and valence electrons. Explain why core-electron peaks appear at higher binding energy than valence-electron peaks.
Use a weighted average:
The percentages must be written as decimals because each isotope contributes only its fractional abundance to the average.
The average atomic mass is
The average is closer to because the isotope is much more abundant than . In a weighted average, the more abundant isotope pulls the average closer to its mass. Since about three-fourths of the atoms are , the average should sit much nearer than , which matches the calculated value.
Core electrons are closer to the nucleus and experience a larger effective nuclear attraction than valence electrons. They are also less shielded by other electrons. Because the attraction between the nucleus and a core electron is stronger, more energy is required to remove a core electron from the atom. Therefore, core-electron peaks appear at higher binding energy on a PES spectrum than valence-electron peaks.
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Sterling silver contains silver and copper. In a released AP Chemistry question, students compared atomic radii using Coulomb’s law. (Adapted from College Board, 2024 AP Chemistry FRQ 3.)
Identify which atom has the larger atomic radius: or .
Use shell structure and Coulomb’s law to justify your answer.
Explain why comparing only nuclear charge is not enough to predict the radius in this case.
has the larger atomic radius.
Silver’s valence electrons occupy a higher principal energy level than copper’s valence electrons. Copper is in period 4, while silver is in period 5, so the outer electrons in silver are farther from the nucleus. By Coulomb’s law, attraction decreases as distance increases:
Silver also has more inner electrons, which increases shielding. The greater distance and shielding make the attraction between the nucleus and valence electrons weaker, so the atomic radius is larger.
Silver has more protons than copper, which by itself would increase attraction. But the valence electrons in silver are also farther from the nucleus and more shielded. Radius depends on the balance of nuclear charge, shielding, and distance, not nuclear charge alone. On the AP exam, a complete explanation should explicitly compare both the attractive force from the nucleus and the distance/shielding effect.
Unit 2: Compound Structure and Properties
Section titled “Unit 2: Compound Structure and Properties”Practice
Section titled “Practice”-
Which molecule is polar?
(A)
(B)
(C)
(D)
has three N-H bonds and one lone pair on nitrogen, giving a trigonal pyramidal shape. The bond dipoles do not cancel.
-
Which compound should have the greatest lattice energy magnitude?
(A)
(B)
(C)
(D)
Lattice energy increases when ion charges are larger and ion radii are smaller. contains and , so the charge product is larger than in the salts with only or ions.
-
Which species has a tetrahedral molecular geometry?
(A)
(B)
(C)
(D)
has four bonding domains and no lone pairs on the central atom, so its molecular geometry is tetrahedral.
-
In a nitrate ion, , resonance means that
(A) the atoms repeatedly switch positions.
(B) one N-O bond is permanently double while the other two are permanently single.
(C) the actual ion has three equivalent N-O bonds with bond order between single and double.
(D) the ion violates conservation of charge.
Resonance structures are different valid Lewis structures with the same atom connectivity but different electron placement. In nitrate, the real ion is a resonance hybrid with equivalent N-O bonds.
-
Which molecule has a bent molecular geometry?
(A)
(B)
(C)
(D)
Water has two bonding domains and two lone pairs around oxygen, giving a bent molecular geometry.
-
Which bond is most polar?
(A) C-H
(B) C-C
(C) H-F
(D) Cl-Cl
The H-F bond has the largest electronegativity difference among the choices, so it is the most polar.
-
Consider the molecules and .
Draw a reasonable Lewis structure for each molecule.
Identify the hybridization of the carbon atom in each molecule.
Explain which molecule can form stronger intermolecular attractions with water.
In , carbon is bonded to two H atoms and double-bonded to O. In , carbon is bonded to three H atoms and single-bonded to O, while O is bonded to H and has two lone pairs.
Image placeholder: Lewis structures for and .
The carbon in has three electron domains, so it is
The carbon in has four electron domains, so it is
forms stronger attractions with water because it can both donate and accept hydrogen bonds through its group. The O atom has lone pairs that can accept hydrogen bonds, and the H attached to O can be donated into a hydrogen bond with water. can accept hydrogen bonds at oxygen, but it cannot donate hydrogen bonds because its H atoms are bonded to carbon, not to a highly electronegative atom. This gives methanol stronger overall interactions with water.
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The 2026 AP Chemistry exam asked students about chromate resonance and VSEPR geometry. (Adapted from College Board, 2026 AP Chemistry FRQ 2.)
Predict the molecular geometry around chromium in .
Explain why multiple resonance structures can be drawn for .
Explain why the four Cr-O bonds are expected to be equivalent in the resonance hybrid.
The chromium has four bonding regions around it, so the electron-domain geometry and molecular geometry are both
The double-bond character can be placed between chromium and different oxygen atoms while keeping the same atom connectivity and total number of valence electrons. These drawings differ only in electron placement, not in which atoms are bonded to which, so they are resonance structures rather than different compounds.
The resonance hybrid averages the valid resonance structures. Since no single resonance structure fully describes the ion, the Cr-O bonds have the same average bond order and are equivalent. A good particle-level explanation is that the extra electron density is delocalized over the Cr-O bonding framework rather than locked into one permanent double bond.
Unit 3: Substances and Mixtures
Section titled “Unit 3: Substances and Mixtures”Practice
Section titled “Practice”-
Which substance is expected to have the highest boiling point?
(A)
(B)
(C)
(D)
can form hydrogen bonds because it has an O-H bond. The other choices rely mainly on London dispersion forces or dipole-dipole attractions.
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A solution is prepared by dissolving of in enough water to make of solution. Assuming complete dissociation, what is the approximate total particle concentration?
(A)
(B)
(C)
(D)
The formal concentration of is
Since dissociates into two ions, the total particle concentration is approximately
So the answer is
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Which pair is most likely to be miscible?
(A) Hexane and water
(B) Ethanol and water
(C) Sodium chloride and hexane
(D) Oil and water
Ethanol and water are both polar and can hydrogen bond with each other, so they mix well.
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If the pressure of a gas above a liquid is increased at constant temperature, the solubility of the gas in the liquid generally
(A) increases.
(B) decreases.
(C) remains exactly zero.
(D) becomes independent of gas identity.
Henry’s law says gas solubility increases as the partial pressure of the gas above the liquid increases.
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Which sample should behave most ideally as a gas?
(A) at high pressure and low temperature
(B) at high pressure and low temperature
(C) He at low pressure and high temperature
(D) vapor near condensation
Gases behave most ideally at low pressure and high temperature, especially when particles have weak intermolecular forces. Helium fits best.
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Which solution has the highest boiling point, assuming ideal behavior?
(A) glucose
(B)
(C)
(D) pure water
Boiling-point elevation depends on total dissolved particle concentration. produces about three ions per formula unit, the largest value among the choices.
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A student measures the absorbance of several solutions of at the same wavelength and path length.
Explain why absorbance can be used to determine concentration.
A solution has absorbance . A calibration line has equation , where is in . Calculate the concentration.
Explain why the wavelength should be chosen near the color most strongly absorbed by the solution.
Beer-Lambert law gives
If and are constant, absorbance is directly proportional to concentration. This means a calibration curve can be used because a solution with more absorbing particles in the same path length absorbs more light. The relationship is only reliable when all measurements use the same wavelength and cuvette path length.
Substitute into the calibration equation:
Thus
Choosing a strongly absorbed wavelength gives a larger absorbance change for a given concentration change, which makes the measurement more sensitive. If the wavelength is poorly absorbed, the absorbance values may be too small and close together, making it harder to distinguish concentrations accurately.
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The 2026 AP Chemistry exam included a spectrophotometry particle-diagram question about absorbance and ion concentration. (Adapted from College Board, 2026 AP Chemistry FRQ 6.)
Explain why a solution with greater concentration of colored ions has greater absorbance at a fixed wavelength.
If a calibration curve has equation and an unknown solution has , calculate .
In a particle diagram of equal volume, how should the number of colored ions compare between a solution and a solution?
More colored ions means more particles are available to absorb photons at that wavelength. For fixed path length, Beer-Lambert law says absorbance is proportional to concentration.
so
The solution has twice the concentration, so an equal-volume particle diagram should show twice as many colored ions. The volume of the box must stay the same; only the number of solute particles should change.
Image placeholder: Equal-volume particle diagrams showing the solution with twice as many colored ions as the solution.
Unit 4: Chemical Reactions
Section titled “Unit 4: Chemical Reactions”Practice
Section titled “Practice”-
What volume of is required to neutralize of ?
(A)
(B)
(C)
(D)
Moles of acid are
Each mole of requires moles of , so
The required volume is
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Which net ionic equation represents the precipitation reaction between aqueous and aqueous ?
(A)
(B)
(C)
(D)
Nitrates and sodium salts remain soluble, while is insoluble. The spectator ions are and .
So the answer is
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In the reaction , which species is oxidized?
(A)
(B)
(C)
(D)
Zinc goes from oxidation number in to in , so it loses electrons and is oxidized.
-
Which observation gives the strongest evidence that a precipitation reaction occurred?
(A) The solution remains clear and colorless.
(B) A solid appears after two aqueous solutions are mixed.
(C) The total volume of solution increases.
(D) The beaker is made of glass.
A precipitate is an insoluble solid that forms from ions in solution.
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What is the oxidation number of sulfur in ?
(A)
(B)
(C)
(D)
Oxygen is usually . Let sulfur be :
So .
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Which pair of aqueous solutions will produce a precipitate when mixed?
(A) and
(B) and
(C) and
(D) and
and form , which is insoluble.
- A sample of impure reacts with excess according to
The reaction produces of .
Calculate the moles of that reacted.
Calculate the mass of in the sample.
Calculate the percent by mass of in the impure sample.
The balanced equation has a mole ratio between and . The problem says the acid is in excess, so all of the carbonate that can react is converted to products, and the moles of produced directly equal the moles of that reacted:
Using for ,
The answer has three significant figures because the measured amount of is given as .
The mass percent is
This means of the impure sample was reactive , and the remaining was impurity that did not produce .
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A released AP Chemistry question asked students to identify a limiting reactant from experimental data. (Adapted from College Board, 2024 AP Chemistry FRQ 2.)
In a trial, of reacts with of according to . Identify the limiting reactant.
Calculate the theoretical moles of produced.
Explain why the excess reactant remains after the limiting reactant is consumed.
Compare how much is needed for the available Al:
Only is available, which is less than the required to consume all of the aluminum. Therefore, is the limiting reactant.
Use the limiting reactant:
The mole ratio comes from the balanced equation: moles of produce moles of .
The balanced reaction requires fixed mole ratios. Once is used up, no more can form, even though some Al remains. The excess reactant remains because there are no longer enough particles of the limiting reactant available to collide and react in the required stoichiometric ratio.
Unit 5: Kinetics
Section titled “Unit 5: Kinetics”Practice
Section titled “Practice”- For the reaction , the initial-rate data below are collected.
| Trial | Initial rate | ||
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 |
What is the rate law?
(A)
(B)
(C)
(D)
Compare trials 1 and 2. Doubling while holding constant makes the rate four times larger, so the reaction is second order in .
Compare trials 2 and 3. Doubling while holding constant does not change the rate, so the reaction is zero order in .
So the answer is
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A catalyst increases the rate of a reaction by
(A) increasing
(B) lowering the activation energy
(C) increasing the equilibrium constant
(D) increasing the enthalpy change of the reaction
A catalyst provides an alternate pathway with lower activation energy. It does not change , , or .
-
For a first-order reaction, which plot should be linear?
(A) versus time
(B) versus time
(C) versus time
(D) rate versus
For a first-order reaction,
so versus time is linear.
-
If the rate law is , what happens to the rate when is doubled and is held constant?
(A) It doubles.
(B) It triples.
(C) It quadruples.
(D) It stays the same.
The rate depends on . Doubling multiplies the rate by
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A reaction has a rate law . What are the units of if rate is measured in ?
(A)
(B)
(C)
(D)
For ,
-
Increasing temperature usually increases reaction rate because
(A) the activation energy becomes zero.
(B) more collisions have energy greater than or equal to .
(C) the equilibrium constant must become larger.
(D) the reaction mechanism cannot change.
At higher temperature, particles have greater kinetic energy on average, so a larger fraction of collisions can overcome the activation energy.
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A first-order decomposition has rate constant .
Calculate the half-life.
If the initial concentration is , calculate the concentration after .
Explain how the slope of a graph of versus time is related to .
For a first-order reaction,
Thus
The units are seconds because the rate constant has units of for a first-order reaction.
Use the integrated rate law:
Equivalently,
This is less than the initial , which is reasonable because the reactant is decomposing over time.
For a first-order reaction, a graph of versus time is linear with slope
The negative slope shows that decreases as time increases. The magnitude of the slope gives the rate constant, so a steeper negative line means a faster first-order reaction.
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The 2026 AP Chemistry exam included a kinetics question using concentration-time data and a graph of natural log of concentration. (Adapted from College Board, 2026 AP Chemistry FRQ 2.)
Explain how concentration-time data can support that a reaction is first order in a reactant.
If a plot of versus time has slope , identify .
Calculate the half-life for the reaction.
If a plot of versus time is linear, the data support a first-order relationship in . Equivalently, the concentration should decrease by the same fraction over equal time intervals. A plot of versus time or versus time would be used to test zero-order or second-order behavior, so the linear graph identifies the order.
For a first-order reaction, the slope is , so
The sign of is positive; the negative sign belongs to the slope because the concentration is decreasing.
For a first-order reaction, this half-life is constant, meaning every the concentration is cut in half regardless of the starting concentration.
Unit 6: Thermochemistry
Section titled “Unit 6: Thermochemistry”Practice
Section titled “Practice”-
A reaction has . Which statement is correct?
(A) The reaction absorbs heat from the surroundings.
(B) The products have greater enthalpy than the reactants.
(C) The reaction is exothermic.
(D) The reaction must be spontaneous at all temperatures.
A negative means heat is released by the system. The reaction is exothermic.
-
A sample of water is warmed from to . Using , how much heat is absorbed by the water?
(A)
(B)
(C)
(D)
Use
Here , so
-
Which process is endothermic?
(A) Freezing water
(B) Condensing steam
(C) Burning methane
(D) Vaporizing liquid water
Vaporization requires energy to overcome intermolecular attractions in the liquid.
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In a coffee-cup calorimeter, the solution temperature increases. What is true about the reaction occurring in the solution?
(A) The reaction releases heat to the solution.
(B) The reaction absorbs heat from the solution.
(C) The reaction has no enthalpy change.
(D) The reaction must have .
If the solution temperature increases, the solution absorbed heat. The reaction released that heat, so the reaction is exothermic.
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Which equation correctly represents Hess’s law?
(A)
(B)
(C)
(D)
Hess’s law says enthalpy is a state function, so the enthalpy change for an overall reaction equals the sum of the enthalpy changes for the steps.
-
If bonds broken require and bonds formed release , what is the approximate ?
(A)
(B)
(C)
(D)
Use
- A reaction is represented by
with for the reaction as written.
Calculate the enthalpy change for forming of .
Calculate the enthalpy change when of reacts completely.
Explain why breaking bonds is endothermic even when the overall reaction is exothermic.
The reaction forms moles of water, so the enthalpy change must be divided by to find the value per mole of water:
The negative sign means heat is released when water forms from hydrogen and oxygen.
The reaction as written consumes moles of . Consuming moles doubles the reaction, so the enthalpy change also doubles:
Energy must be added to separate bonded atoms, so bond breaking is endothermic. The overall reaction is exothermic because forming the O-H bonds in water releases more energy than was required to break the H-H and O=O bonds. In bond-enthalpy language, is negative because the energy released by bonds formed is greater than the energy absorbed by bonds broken.
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The 2026 AP Chemistry exam included a sodium oxide thermochemistry problem using formation enthalpy and limiting reactants. (Adapted from College Board, 2026 AP Chemistry FRQ 7.)
For with , calculate for .
If Na reacts completely with excess oxygen, calculate the heat released.
Explain why elements in their standard states have .
The reaction forms moles of from elements in their standard states, so the reaction enthalpy is twice the molar enthalpy of formation:
The reaction releases per moles Na.
So of heat is released.
Formation enthalpy measures formation from elements in their standard states. An element already in its standard state requires no formation reaction, so its assigned value is zero. This is a reference convention that lets formation enthalpies be added and subtracted consistently in Hess’s law calculations.
Unit 7: Equilibrium
Section titled “Unit 7: Equilibrium”Practice
Section titled “Practice”-
For the reaction , which expression is ?
(A)
(B)
(C)
(D)
Gas and aqueous equilibrium expressions use concentrations raised to stoichiometric coefficients:
-
For a system at equilibrium, adding a catalyst will
(A) increase
(B) decrease
(C) shift the equilibrium toward products
(D) leave the equilibrium composition unchanged
A catalyst speeds both forward and reverse reactions. It helps the system reach equilibrium faster, but it does not change or the equilibrium composition.
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If for a reaction mixture, the reaction will proceed
(A) toward products until equilibrium is reached.
(B) toward reactants until equilibrium is reached.
(C) in neither direction because it is already at equilibrium.
(D) only if a catalyst is added.
If , the mixture has too much product relative to equilibrium, so it shifts toward reactants.
-
Which species is omitted from the equilibrium expression for ?
(A) only
(B) and
(C) and
(D) all species
Pure solids are omitted from equilibrium expressions, so both solids are omitted.
-
For , what is the expression for ?
(A)
(B)
(C)
(D)
Pure solids are omitted from equilibrium expressions, so
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If an exothermic equilibrium reaction is heated, the equilibrium shifts
(A) toward products.
(B) toward reactants.
(C) nowhere, because temperature does not affect equilibrium.
(D) only if a catalyst is present.
For an exothermic reaction, heat behaves like a product. Adding heat shifts the equilibrium toward reactants.
- At a certain temperature, for
A sealed container initially has and no .
Write the equilibrium-constant expression.
Set up an ICE table using for the amount of consumed.
Calculate the equilibrium concentrations of and .
The coefficient in front of becomes the exponent in the equilibrium expression. There are no solids or liquids to omit in this reaction.
The ICE setup is
| Initial | ||
| Change | ||
| Equilibrium |
The appears because every mole of that reacts produces moles of .
Substitute into :
So
and
Thus
The positive root is . Therefore,
and
The negative root is rejected because it would make no physical sense for the reaction progress variable in this setup. Both equilibrium concentrations are positive, which is a useful check.
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The 2024 AP Chemistry exam included an equilibrium particle-diagram question for . (Adapted from College Board, 2024 AP Chemistry FRQ 5.)
Write the expression for .
If , , and , calculate .
If , predict the direction the system shifts.
The coefficient on becomes the exponent .
Since , the system has too little product relative to equilibrium. It shifts toward products, forming more and consuming and . The value of does not change during the shift because temperature is not changed.
Image placeholder: Particle diagram for showing the mixture shifting toward more .
Unit 8: Acid-Base Equilibrium
Section titled “Unit 8: Acid-Base Equilibrium”Practice
Section titled “Practice”-
What is the of a solution of at ?
(A)
(B)
(C)
(D)
is a strong acid, so
Thus
-
A buffer contains equal concentrations of and . If for is , what is the of the buffer?
(A)
(B)
(C)
(D)
When , the Henderson-Hasselbalch equation gives
-
Which solution has the greatest ?
(A)
(B)
(C)
(D)
is a strong base, so it produces the greatest and therefore the greatest .
-
At the half-equivalence point in a weak acid-strong base titration,
(A) .
(B) for every weak acid.
(C) all weak acid has been converted to conjugate base.
(D) .
At the half-equivalence point, . Henderson-Hasselbalch becomes
-
Which species is the conjugate base of ?
(A)
(B)
(C)
(D)
A conjugate base is formed by removing one proton. Removing from gives .
-
Which mixture is a buffer?
(A) and
(B) and
(C) and
(D) and
A buffer contains a weak acid and its conjugate base, or a weak base and its conjugate acid. and form a buffer pair.
-
A solution of acetic acid, , has .
Write the acid-ionization equation.
Calculate using the small- approximation.
Calculate the .
Explain what happens to the percent ionization if sodium acetate is added.
Water is included in the chemical equation because it accepts the proton, but liquid water is omitted from the expression.
Let at equilibrium. Then
Using the small- approximation,
Thus
This value is small compared with , so the small- approximation is reasonable:
Sodium acetate adds the common ion , shifting the acid ionization left. Since less acetic acid ionizes, decreases and the percent ionization decreases. This is the common-ion effect.
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The 2026 AP Chemistry exam included a nitrous acid titration and indicator question. (Adapted from College Board, 2026 AP Chemistry FRQ 3.)
Explain why the equivalence point of a weak acid-strong base titration has .
A sample of is titrated to equivalence with of . Calculate the molarity of .
Explain why an indicator should change color near the steep part of the titration curve.
At equivalence, the weak acid has been converted mostly into its conjugate base. The conjugate base reacts with water to produce :
Because is produced, the solution is basic and the equivalence-point pH is greater than .
At equivalence,
Thus
The steep part of the titration curve is where a tiny volume change causes a large pH change, so the color change most closely marks the equivalence point. If the indicator changes color far from that steep region, it will signal the endpoint too early or too late and create systematic error.
Image placeholder: Weak acid-strong base titration curve with equivalence point above and an indicator transition range near the steep region.
Unit 9: Thermodynamics and Electrochemistry
Section titled “Unit 9: Thermodynamics and Electrochemistry”Practice
Section titled “Practice”-
For a spontaneous galvanic cell under standard conditions, which statement is correct?
(A) and
(B) and
(C) and
(D) and
For a spontaneous galvanic cell,
Spontaneous means , which requires .
-
Which change always increases the entropy of the system?
(A)
(B)
(C)
(D)
Sublimation changes a solid directly into a gas. Gas particles have many more accessible microstates than particles in a solid.
-
A reaction has and . The reaction is most likely thermodynamically favorable at
(A) low temperatures only.
(B) high temperatures only.
(C) all temperatures.
(D) no temperatures.
Use
When both and are positive, high temperature makes the term large and negative.
-
In an electrolytic cell, oxidation occurs at the
(A) anode.
(B) cathode.
(C) salt bridge.
(D) voltmeter.
Oxidation always occurs at the anode, in both galvanic and electrolytic cells.
-
If for a reaction under standard conditions, which statement is true?
(A)
(B)
(C)
(D) for the corresponding galvanic cell
Since
a negative means , so .
-
What mass of Ag is plated by of electrons from ?
(A)
(B)
(C)
(D)
For , of electrons plates of Ag.
- A galvanic cell is based on the reaction
Use and .
Identify the anode and cathode.
Calculate .
Calculate for the reaction.
Explain the direction of electron flow in the external circuit.
Zinc is oxidized:
so zinc is the anode. Copper(II) is reduced at the cathode:
This follows the rule that oxidation occurs at the anode and reduction occurs at the cathode.
The positive cell potential is consistent with the reaction being spontaneous as a galvanic cell under standard conditions.
Here .
So
Electrons are produced at the zinc anode and consumed at the copper cathode, so electrons flow from Zn to Cu through the external circuit. Ions move through the salt bridge to maintain charge balance, but electrons do not travel through the salt bridge.
Image placeholder: Galvanic cell diagram with Zn anode, Cu cathode, electron flow from Zn to Cu, and ion flow through the salt bridge.
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The 2024 AP Chemistry exam included electroplating and nonspontaneous electrochemistry in a sterling silver context. (Adapted from College Board, 2024 AP Chemistry FRQ 3.)
Explain why an electrolytic plating process requires an external power source.
Calculate the time required to plate of from using a current of .
Identify whether reduction or oxidation occurs at the object being plated.
Electroplating is nonspontaneous as written, so an external power source is needed to drive electron flow and force reduction of metal ions onto the object. In other words, electrical energy is used to make a thermodynamically unfavorable redox process occur.
Moles of copper plated:
For ,
Charge required:
Since ,
This is about if converted to minutes.
Metal ions gain electrons and become solid metal on the object, so reduction occurs at the object being plated. The object being plated acts as the cathode in the electrolytic cell.