UNIT 9

Applications of Thermodynamics

Examine entropy, spontaneity, Gibbs free energy, equilibrium, electrochemical cells, cell potential, and electrolysis.

Topics

Entropy and Spontaneity

Analyze entropy changes and determine how energy dispersal and particle arrangement influence whether a process is thermodynamically favored.

Gibbs Free Energy and Equilibrium

Use enthalpy, entropy, temperature, and equilibrium constants to determine whether reactions are thermodynamically favorable.

Electrochemistry

Explore galvanic and electrolytic cells, oxidation and reduction, cell potential, free energy, and quantitative electrolysis.

Entropy and Spontaneity

Entropy, represented by S, describes the dispersal of energy and the number of possible microscopic arrangements available to a system.

Entropy is sometimes described as disorder, but a more precise interpretation focuses on how many possible arrangements, or microstates, can produce the same observable state.

Microstates

A microstate is one possible arrangement of the particles and energy in a system. A system with more possible microstates has greater entropy.

S = k_B ln(W)

In this relationship:

S is entropy.
k_B is the Boltzmann constant.
W is the number of possible microstates.

Particle diagrams comparing a lower entropy ordered solid with a higher entropy dispersed gas — The gas has greater entropy because its particles can occupy many more positions and arrangements than particles fixed in a solid structure.

Factors That Affect Entropy

Change	General Effect on Entropy
Solid → Liquid → Gas	Entropy increases
Increasing temperature	Entropy increases
Increasing the number of gas particles	Entropy usually increases
Increasing particle complexity	Entropy generally increases
Mixing particles	Entropy generally increases
Expanding a gas into a larger volume	Entropy increases

For the same substance:

S_gas > S_liquid > S_solid

Predicting the Sign of ΔS

The entropy change of a reaction can often be predicted by comparing the physical states and numbers of particles on the reactant and product sides.

ΔS = S_products − S_reactants

Example: Decreasing Entropy

N₂(g) + 3H₂(g) → 2NH₃(g)

The reactants contain four total moles of gas, while the products contain two moles of gas.

Because the number of gas particles decreases, ΔS is negative.

Example: Increasing Entropy

CaCO₃(s) → CaO(s) + CO₂(g)

The reactant side contains only a solid, while the product side includes a gas.

The formation of gas greatly increases the number of accessible arrangements, so ΔS is positive.

Counting gas particles is useful when the physical states are otherwise similar. A change in physical state can be more important than simply counting total coefficients.

Standard Molar Entropy

Standard molar entropy, S°, is the entropy of one mole of a substance under standard-state conditions. Its common units are:

J/(mol·K)

Unlike standard enthalpies of formation, the standard molar entropy of an element in its standard state is not generally zero.

ΔS°_reaction = Sum of S° for products − Sum of S° for reactants

Each standard molar entropy must be multiplied by its coefficient in the balanced equation.

Example: Calculating Standard Entropy Change

H₂(g) + 1 2 O₂(g) → H₂O(l)

Use these standard molar entropies:

Substance	S° in J/(mol·K)
H₂(g)	130.7
O₂(g)	205.0
H₂O(l)	69.9

ΔS° = 69.9 − [130.7 + 0.5(205.0)]

ΔS° = −163.3 J/(mol·K)

The result is negative, which is consistent with gaseous reactants forming a liquid product.

Entropy Changes During Phase Transitions

At the temperature where two phases are in equilibrium, the entropy change for a phase transition can be calculated from its enthalpy change.

ΔS_transition = ΔH_transition T

Temperature must be measured in kelvin when it appears in thermodynamic equations.

Entropy of the Universe

The second law of thermodynamics describes thermodynamic favorability using the entropy change of the universe.

ΔS_universe = ΔS_system + ΔS_surroundings

Value	Meaning
ΔS_universe > 0	Process is thermodynamically favored
ΔS_universe = 0	System is at equilibrium
ΔS_universe < 0	Forward process is not thermodynamically favored

Entropy Change of the Surroundings

At constant temperature and pressure, the surroundings’ entropy change is related to the system’s enthalpy change.

ΔS_surroundings = −ΔH_system T

An exothermic process releases energy into the surroundings, increasing the surroundings’ entropy. An endothermic process removes energy from the surroundings, decreasing their entropy.

Thermodynamic Favorability Versus Reaction Rate

Thermodynamic favorability describes whether a process has a tendency to occur under the stated conditions. It does not determine how quickly the process occurs.

Thermodynamics predicts whether a process is favored.

Kinetics determines how quickly the process occurs.

A thermodynamically favored reaction can still occur extremely slowly if it has a large activation-energy barrier.

Practice Questions

Question 1: Predicting the Sign of ΔS

Consider the reaction:

2SO₃(g) → 2SO₂(g) + O₂(g)

Predict whether ΔS is positive or negative and explain.

Show solution

The reactant side contains two total moles of gas. The product side contains three total moles of gas.

Increasing the number of gas particles increases the number of possible particle arrangements. Therefore, ΔS is positive.

Question 2: Ranking Entropy

Rank the following samples from lowest to highest entropy:

H₂O(s), H₂O(l), H₂O(g)

Show solution

H₂O(s) < H₂O(l) < H₂O(g)

Particles in a solid have relatively restricted positions. Liquid particles have more possible arrangements, while gas particles can occupy a much larger number of positions.

Question 3: Calculating Standard Entropy Change

Calculate ΔS° for:

N₂(g) + 3H₂(g) → 2NH₃(g)

Use the following data:

Substance	S° in J/(mol·K)
N₂(g)	191.5
H₂(g)	130.7
NH₃(g)	192.8

Show solution

ΔS° = Sum of S°_products − Sum of S°_reactants

Calculate the product total:

Products = 2(192.8) = 385.6 J/(mol·K)

Calculate the reactant total:

Reactants = 191.5 + 3(130.7) = 583.6 J/(mol·K)

ΔS° = 385.6 − 583.6

ΔS° = −198.0 J/(mol·K)

The negative result agrees with the decrease from four moles of gas to two moles of gas.

Question 4: Entropy of a Phase Transition

The molar enthalpy of fusion of water is 6.01 kJ/mol. Calculate ΔS_fusion at its normal melting temperature of 273.15 K.

Show solution

Convert the enthalpy from kilojoules to joules:

6.01 kJ/mol = 6010 J/mol

ΔS_fusion = ΔH_fusion T

ΔS_fusion = 6010 J/mol 273.15 K

ΔS_fusion = 22.0 J/(mol·K)

The value is positive because melting increases the number of accessible particle arrangements.

Question 5: Entropy Change of the Surroundings

A reaction has ΔH_system = −75.0 kJ/mol at 298 K. Calculate ΔS_surroundings.

Show solution

Convert the enthalpy to joules:

−75.0 kJ/mol = −75,000 J/mol

ΔS_surroundings = −ΔH_system T

ΔS_surroundings = −(−75,000 J/mol) 298 K

ΔS_surroundings = +252 J/(mol·K)

The surroundings gain entropy because the exothermic reaction releases energy into them.

Question 6: Entropy of the Universe

A process has:

ΔS_system = +120 J/(mol·K)
ΔS_surroundings = −80 J/(mol·K)

Calculate ΔS_universe and determine whether the process is thermodynamically favored.

Show solution

ΔS_universe = ΔS_system + ΔS_surroundings

ΔS_universe = +120 + (−80)

ΔS_universe = +40 J/(mol·K)

Because ΔS_universe is positive, the process is thermodynamically favored.

Question 7: Thermodynamics Versus Kinetics

A reaction is thermodynamically favored under a particular set of conditions, but almost no product forms during a short experiment.

Explain how both observations can be true.

Show solution

Thermodynamic favorability indicates that the reaction has a tendency to proceed toward products under the stated conditions.

It does not describe the reaction rate. If the reaction has a large activation-energy barrier, it may proceed extremely slowly.

Therefore, the reaction can be thermodynamically favored but kinetically slow.

Question 8: Comparing Gas Expansion

A gas initially occupies one side of an evacuated container. A divider is removed, allowing the gas to occupy the entire container at the same temperature.

What happens to the gas’s entropy?

Show solution

The entropy increases.

In the larger volume, each gas particle has more possible positions. The system therefore has more accessible microstates, even though the number of particles and temperature remain unchanged.

VIDEO EXPLANATION

Watch: Entropy and Spontaneity

Watch a worked example involving entropy predictions, standard molar entropy, or the entropy change of the universe.

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Gibbs Free Energy and Equilibrium

Gibbs free energy combines enthalpy, entropy, and temperature into a single quantity that predicts whether a process is thermodynamically favored under a specified set of conditions.

Gibbs Free Energy

ΔG = ΔH − TΔS

In this equation:

ΔG is the Gibbs free-energy change.
ΔH is the enthalpy change.
T is the absolute temperature in kelvin.
ΔS is the entropy change.

Before using ΔG = ΔH − TΔS, make sure ΔH and ΔS use compatible energy units.

If ΔH is in kJ/mol, convert ΔS from J/(mol·K) to kJ/(mol·K).

Interpreting the Sign of ΔG

Value of ΔG	Meaning
ΔG < 0	Forward process is thermodynamically favored
ΔG = 0	System is at equilibrium
ΔG > 0	Forward process is not thermodynamically favored

A positive ΔG for the forward reaction means that the reverse reaction has a negative ΔG and is thermodynamically favored under the same conditions.

Example: Calculating ΔG

A reaction has ΔH = −85.0 kJ/mol and ΔS = −150 J/(mol·K). Calculate ΔG at 298 K.

Convert ΔS to kilojoules:

−150 J/(mol·K) = −0.150 kJ/(mol·K)

ΔG = ΔH − TΔS

ΔG = −85.0 kJ/mol − (298 K)(−0.150 kJ/(mol·K))

ΔG = −40.3 kJ/mol

Because ΔG is negative, the forward reaction is thermodynamically favored at 298 K.

Temperature Dependence

Whether a reaction is thermodynamically favored can depend on the signs of ΔH and ΔS and on the temperature.

Sign of ΔH	Sign of ΔS	Temperature Dependence
Negative	Positive	Favored at all temperatures
Positive	Negative	Not favored at any temperature
Negative	Negative	Favored at lower temperatures
Positive	Positive	Favored at higher temperatures

Finding the Transition Temperature

The temperature at which a process changes between favored and not favored occurs when ΔG = 0.

0 = ΔH − TΔS

T = ΔH ΔS

Example: Temperature Threshold

A process has ΔH = +50.0 kJ/mol and ΔS = +120 J/(mol·K). Calculate the temperature at which ΔG = 0.

Convert ΔS to kilojoules:

+120 J/(mol·K) = +0.120 kJ/(mol·K)

T = 50.0 kJ/mol 0.120 kJ/(mol·K)

T = 417 K

Because both ΔH and ΔS are positive, the process is favored at temperatures above 417 K.

Standard Free Energy and Equilibrium

The standard free-energy change, ΔG°, is related to the equilibrium constant.

ΔG° = −RT ln(K)

In this equation:

R = 8.314 J/(mol·K)
T is measured in kelvin
K is the equilibrium constant

ΔG°	K	Equilibrium Position
Negative	K > 1	Products favored
Zero	K = 1	Neither side strongly favored
Positive	K < 1	Reactants favored

Example: Finding K from ΔG°

A reaction has ΔG° = −10.0 kJ/mol at 298 K. Calculate K.

Convert ΔG° to joules:

−10.0 kJ/mol = −10,000 J/mol

−10,000 = −(8.314)(298)ln(K)

ln(K) = 4.04

K = e^4.04 = 56.6

Because K is greater than 1, products are favored under standard-state equilibrium conditions.

Free Energy under Nonstandard Conditions

The actual free-energy change depends on the current composition of the reaction mixture.

ΔG = ΔG° + RT ln(Q)

Q is the reaction quotient calculated using the current concentrations or partial pressures.

Comparison	Sign of ΔG	Favored Direction
Q < K	Negative	Toward products
Q = K	Zero	At equilibrium
Q > K	Positive	Toward reactants

ΔG° describes standard-state conditions.

ΔG describes the current conditions.

A reaction with positive ΔG° can still proceed toward products under nonstandard conditions if Q is sufficiently small.

Free Energy at Equilibrium

At equilibrium:

ΔG = 0

Q = K

The forward and reverse reactions continue, but neither direction has a net thermodynamic driving force.

Coupled Reactions

A thermodynamically unfavorable reaction can be combined with a sufficiently favorable reaction so that the overall process has a negative ΔG.

ΔG_overall = ΔG_{reaction 1} + ΔG_{reaction 2}

Example: Reaction Coupling

One reaction has ΔG = +35 kJ/mol. It is coupled to a second reaction with ΔG = −50 kJ/mol.

ΔG_overall = +35 kJ/mol + (−50 kJ/mol)

ΔG_overall = −15 kJ/mol

The combined process is thermodynamically favored because its overall ΔG is negative.

Manipulating Free-Energy Changes

Free-energy changes are manipulated in the same way as reaction enthalpies and equilibrium constants.

Change to Reaction	Change to ΔG°
Reverse the reaction	Reverse the sign of ΔG°
Multiply the equation by a factor	Multiply ΔG° by the same factor
Add reactions	Add their ΔG° values

Thermodynamic and Kinetic Control

A product can form more quickly because its pathway has a lower activation energy, while a different product may be more thermodynamically stable.

The kinetic product forms through the faster pathway.

The thermodynamic product has the lower final free energy.

Reaction rate and thermodynamic favorability answer different questions.

Practice Questions

Question 1: Calculating ΔG

A reaction has ΔH = +42.0 kJ/mol and ΔS = +180 J/(mol·K). Calculate ΔG at 298 K and determine whether the reaction is thermodynamically favored.

Show solution

Convert ΔS to kilojoules:

+180 J/(mol·K) = +0.180 kJ/(mol·K)

ΔG = ΔH − TΔS

ΔG = +42.0 − (298)(0.180)

ΔG = −11.6 kJ/mol

Because ΔG is negative, the reaction is thermodynamically favored at 298 K.

Question 2: Predicting Temperature Dependence

A reaction has a negative ΔH and a negative ΔS. At what general range of temperatures is the reaction most likely to be thermodynamically favored?

Show solution

ΔG = ΔH − TΔS

The negative ΔH favors a negative ΔG. Because ΔS is negative, the term −TΔS is positive and opposes thermodynamic favorability.

At lower temperatures, the positive −TΔS term is smaller. Therefore, the reaction is most likely favored at lower temperatures.

Question 3: Finding a Temperature Threshold

A reaction has ΔH = −60.0 kJ/mol and ΔS = −150 J/(mol·K).

Calculate the temperature at which ΔG = 0 and state whether the reaction is favored above or below that temperature.

Show solution

Convert ΔS to kilojoules:

−150 J/(mol·K) = −0.150 kJ/(mol·K)

T = ΔH ΔS

T = −60.0 kJ/mol −0.150 kJ/(mol·K)

T = 400 K

Because both ΔH and ΔS are negative, the reaction is favored at temperatures below 400 K.

Question 4: Finding K from ΔG°

A reaction has ΔG° = +12.0 kJ/mol at 298 K. Calculate K. Use R = 8.314 J/(mol·K).

Show solution

Convert ΔG° to joules:

+12.0 kJ/mol = +12,000 J/mol

ΔG° = −RT ln(K)

12,000 = −(8.314)(298)ln(K)

ln(K) = −4.84

K = e^−4.84 = 7.9 × 10⁻³

K is less than 1, which is consistent with the positive ΔG° and a reactant-favored equilibrium.

Question 5: Finding ΔG° from K

A reaction has K = 2.5 × 10⁴ at 298 K. Calculate ΔG°. Use R = 8.314 J/(mol·K).

Show solution

ΔG° = −RT ln(K)

ΔG° = −(8.314)(298) ln(2.5 × 10⁴)

ΔG° = −25,100 J/mol

ΔG° = −25.1 kJ/mol

The negative value agrees with the large equilibrium constant and product-favored equilibrium.

Question 6: Nonstandard Conditions

A reaction has ΔG° = +5.00 kJ/mol at 298 K. The current reaction quotient is Q = 0.0100.

Calculate ΔG and determine the favored direction. Use R = 8.314 J/(mol·K).

Show solution

Convert ΔG° to joules:

+5.00 kJ/mol = +5000 J/mol

ΔG = ΔG° + RT ln(Q)

ΔG = 5000 + (8.314)(298)ln(0.0100)

ΔG = 5000 − 11,400

ΔG = −6400 J/mol = −6.40 kJ/mol

Under the current nonstandard conditions, ΔG is negative, so the reaction proceeds toward products.

Question 7: Reaction Coupling

Reaction 1 has ΔG = +28 kJ/mol. Reaction 2 has ΔG = −45 kJ/mol. The two reactions are coupled.

Calculate the overall ΔG and determine whether the combined process is thermodynamically favored.

Show solution

ΔG_overall = ΔG₁ + ΔG₂

ΔG_overall = +28 + (−45)

ΔG_overall = −17 kJ/mol

The combined process is thermodynamically favored because its overall ΔG is negative.

Question 8: Manipulating ΔG°

The following reaction has ΔG° = −18.0 kJ:

A → 2B

Determine ΔG° for:

4B → 2A

Show solution

First reverse the original reaction. Reversing it changes the sign of ΔG°.

2B → A ΔG° = +18.0 kJ

Then multiply the reversed equation by two.

4B → 2A ΔG° = 2(+18.0)

ΔG° = +36.0 kJ

Question 9: Standard Versus Current Conditions

A reaction has a positive ΔG° but a negative ΔG under the current conditions. Explain why this is possible.

Show solution

ΔG° applies to standard-state conditions, while ΔG depends on the current reaction quotient.

ΔG = ΔG° + RT ln(Q)

If Q is sufficiently small, the RT ln(Q) term is negative enough to make the actual ΔG negative. The reaction can therefore proceed toward products even though it is not product-favored under standard conditions.

VIDEO EXPLANATION

Watch: Gibbs Free Energy

Watch a worked example involving ΔG, temperature, equilibrium constants, nonstandard conditions, or coupled reactions.

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Electrochemistry

Electrochemistry examines the relationship between oxidation– reduction reactions and electrical energy. Electrochemical cells separate oxidation and reduction into two half-reactions.

Oxidation and Reduction

Oxidation: Loss of electrons

Reduction: Gain of electrons

Oxidation and reduction always occur together because electrons lost by one substance must be gained by another.

Anodes and Cathodes

Anode: Oxidation occurs

Cathode: Reduction occurs

These definitions remain true for both galvanic and electrolytic cells.

Anode → Oxidation

Cathode → Reduction

Galvanic Cells

A galvanic cell, also called a voltaic cell, uses a thermodynamically favored redox reaction to produce electrical energy.

The Zinc–Copper Cell

At the zinc anode, oxidation occurs:

Zn(s) → Zn²⁺(aq) + 2e⁻

At the copper cathode, reduction occurs:

Cu²⁺(aq) + 2e⁻ → Cu(s)

The overall cell reaction is:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

In this galvanic cell:

• Electrons flow from Zn to Cu

• The zinc electrode loses mass

• The copper electrode gains mass

• [Zn²⁺] increases

• [Cu²⁺] decreases

Electron Flow and Conventional Current

Electrons travel through the external circuit from the anode to the cathode.

Electron flow: Anode → Cathode

Conventional current is defined in the direction opposite to electron flow.

The Salt Bridge

The salt bridge completes the electrical circuit and prevents charge from building up in either half-cell.

At the anode, oxidation produces positive ions. Anions from the salt bridge migrate toward the anode compartment.

At the cathode, positive ions are consumed. Cations from the salt bridge migrate toward the cathode compartment.

Salt-bridge anions migrate toward the anode.

Salt-bridge cations migrate toward the cathode.

Electrons do not travel through the salt bridge.

Cell Notation

Cell notation provides a compact representation of an electrochemical cell.

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

The anode is written on the left and the cathode on the right. A single vertical line represents a phase boundary, while the double line represents the salt bridge.

Standard Reduction Potentials

Standard reduction potentials measure the tendency of species to gain electrons under standard-state conditions.

A more positive reduction potential indicates a greater tendency to be reduced.

Standard cell potential is calculated using:

E°_cell = E°_cathode − E°_anode

Both values are taken from a standard reduction-potential table. The anode value is subtracted because the listed half-reaction is being reversed to represent oxidation.

Example: Standard Cell Potential

Use the following reduction potentials:

Cu²⁺ + 2e⁻ → Cu E° = +0.34 V

Zn²⁺ + 2e⁻ → Zn E° = −0.76 V

Copper has the more positive reduction potential, so it is reduced at the cathode. Zinc is oxidized at the anode.

E°_cell = +0.34 V − (−0.76 V)

E°_cell = +1.10 V

The positive cell potential indicates that the cell reaction is thermodynamically favored under standard-state conditions.

Do not multiply standard electrode potentials by reaction coefficients. Potential is not an extensive quantity.

Cell Potential and Free Energy

ΔG° = −nFE°_cell

In this equation:

n is the number of moles of electrons transferred.
F is Faraday’s constant: 96,485 C/mol e⁻.
E°_cell is measured in volts.

E°_cell	ΔG°	Meaning
Positive	Negative	Forward reaction is thermodynamically favored
Zero	Zero	System is at equilibrium
Negative	Positive	Forward reaction is not thermodynamically favored

Cell Potential and the Equilibrium Constant

ΔG° = −RT ln(K)

nFE°_cell = RT ln(K)

A positive standard cell potential corresponds to K greater than 1 for the written cell reaction.

Nonstandard Cell Potentials

Cell potential changes when ion concentrations or gas pressures differ from standard-state conditions.

E = E° − RT nF ln(Q)

At 25°C, this can be written as:

E = E° − 0.0592 n log(Q)

This is the Nernst equation. As the cell operates and approaches equilibrium, Q approaches K and the cell potential approaches zero.

Galvanic and Electrolytic Cells

Property	Galvanic Cell	Electrolytic Cell
Reaction	Thermodynamically favored	Driven by external electrical energy
E_cell	Positive for written reaction	Negative without external energy
ΔG	Negative	Positive for the driven reaction
Anode sign	Negative	Positive
Cathode sign	Positive	Negative
Oxidation location	Anode	Anode
Reduction location	Cathode	Cathode

The electrode signs change between galvanic and electrolytic cells, but oxidation always occurs at the anode and reduction always occurs at the cathode.

Quantitative Electrolysis

The amount of substance produced during electrolysis depends on the total electrical charge transferred.

Charge = Current × Time

q = It

Moles of electrons = Charge Faraday’s constant

Example: Mass Deposited during Electrolysis

A current of 2.00 A transfers charge for 1930 seconds through a system in which:

Cu²⁺ + 2e⁻ → Cu

Calculate the mass of copper produced.

q = (2.00 C/s)(1930 s) = 3860 C

Moles e⁻ = 3860 C 96,485 C/mol e⁻ = 0.0400 mol e⁻

Two moles of electrons produce one mole of copper:

0.0400 mol e⁻ × 1 mol Cu 2 mol e⁻ = 0.0200 mol Cu

Mass Cu = (0.0200 mol)(63.55 g/mol) = 1.27 g

Practice Questions

Question 1: Identifying the Electrodes

Use the following standard reduction potentials:

Ag⁺ + e⁻ → Ag E° = +0.80 V

Al³⁺ + 3e⁻ → Al E° = −1.66 V

For a galvanic cell, identify the anode, cathode, and direction of electron flow.

Show solution

Ag⁺ has the more positive reduction potential, so it is reduced at the silver cathode.

Ag⁺ + e⁻ → Ag

Aluminum is oxidized at the aluminum anode.

Al → Al³⁺ + 3e⁻

Electrons flow through the wire from the aluminum anode to the silver cathode.

Question 2: Calculating Standard Cell Potential

Using the reduction potentials from Question 1, calculate E°_cell.

Show solution

E°_cell = E°_cathode − E°_anode

E°_cell = +0.80 V − (−1.66 V)

E°_cell = +2.46 V

The positive cell potential indicates that the written galvanic-cell reaction is thermodynamically favored under standard-state conditions.

The potentials are not multiplied by the coefficients used to balance the electrons.

Question 3: Electrode Mass and Ion Concentration

Consider the galvanic-cell reaction:

Mg(s) + Ni²⁺(aq) → Mg²⁺(aq) + Ni(s)

Identify which electrode gains mass, which loses mass, and how each ion concentration changes as the cell operates.

Show solution

Magnesium is oxidized:

Mg(s) → Mg²⁺(aq) + 2e⁻

The magnesium electrode loses mass, and [Mg²⁺] increases.

Nickel ions are reduced:

Ni²⁺(aq) + 2e⁻ → Ni(s)

The nickel electrode gains mass, and [Ni²⁺] decreases.

Question 4: Cell Potential and Free Energy

A galvanic cell has E°_cell = +1.10 V and transfers two moles of electrons per mole of reaction. Calculate ΔG°.

Use F = 96,485 C/mol e⁻.

Show solution

ΔG° = −nFE°_cell

ΔG° = −(2 mol e⁻) (96,485 C/mol e⁻) (1.10 J/C)

ΔG° = −212,000 J/mol

ΔG° = −212 kJ/mol

The negative ΔG° agrees with the positive cell potential.

Question 5: Cell Potential and K

A cell transfers two moles of electrons and has E°_cell = +0.300 V at 298 K. Calculate K.

Use F = 96,485 C/mol e⁻ and R = 8.314 J/(mol·K).

Show solution

nFE°_cell = RT ln(K)

ln(K) = nFE°_cell RT

ln(K) = (2)(96,485)(0.300) (8.314)(298)

ln(K) = 23.4

K = e^23.4 = 1.4 × 10¹⁰

The very large K is consistent with the positive standard cell potential.

Question 6: The Nernst Equation

For the cell reaction:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

E° = 1.10 V. Calculate E at 25°C when [Zn²⁺] = 0.100 M and [Cu²⁺] = 1.00 M.

Show solution

Write the reaction quotient:

Q = [Zn²⁺] [Cu²⁺]

Q = 0.100 1.00 = 0.100

Two electrons are transferred, so n = 2.

E = E° − 0.0592 2 log(0.100)

E = 1.10 − (0.0296)(−1.00)

E = 1.13 V

Because Q is less than 1, the current conditions provide a slightly larger cell potential than the standard-state value.

Question 7: Salt-Bridge Ion Movement

In a galvanic cell, oxidation at the anode produces additional metal cations in the anode solution. Explain which salt-bridge ions migrate toward that half-cell and why.

Show solution

Anions from the salt bridge migrate toward the anode compartment.

Oxidation produces additional positive ions in that solution. Anion migration prevents excess positive charge from accumulating and maintains electrical neutrality.

Question 8: Quantitative Electrolysis

A current of 1.50 A transfers charge for 1930 seconds in a system described by:

Ag⁺ + e⁻ → Ag

Calculate the mass of silver produced. Use F = 96,485 C/mol e⁻ and a molar mass of 107.87 g/mol for Ag.

Show solution

Calculate the charge transferred:

q = It

q = (1.50 C/s)(1930 s) = 2895 C

Convert charge to moles of electrons:

Moles e⁻ = 2895 C 96,485 C/mol e⁻

Moles e⁻ = 0.0300 mol

The half-reaction uses one electron per silver ion, so 0.0300 mol Ag forms.

Mass Ag = (0.0300 mol)(107.87 g/mol)

Mass Ag = 3.24 g

Question 9: Galvanic Versus Electrolytic Cells

Compare the anode and cathode in galvanic and electrolytic cells. Which properties remain the same, and which signs change?

Show solution

In both types of cells:

Oxidation occurs at the anode.
Reduction occurs at the cathode.
Electrons flow from anode to cathode.

In a galvanic cell:

The anode is negative.
The cathode is positive.

In an electrolytic cell:

The anode is positive.
The cathode is negative.

Question 10: Cell Potential at Equilibrium

What happens to Q and E_cell as a galvanic cell reaches equilibrium?

Show solution

As the cell reaction proceeds, Q approaches K. At equilibrium:

Q = K

ΔG = 0

E_cell = 0

At this point, the cell no longer has a net thermodynamic driving force to produce electrical work.

VIDEO EXPLANATION

Watch: Electrochemical Cells

Watch a worked example involving electrode identification, cell potential, free energy, the Nernst equation, or electrolysis.

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