Unit 9 Beta

Thermodynamics and Electrochemistry: the one-page sheet

9.1 Introduction to Entropy

Entropy measures how many ways a system’s particles and energy can be arranged. It rises when matter spreads out (more gas, more volume, solid to liquid to gas, dissolving) and when energy spreads out (higher temperature). Predict the sign of ΔS from moles of gas and phases, and justify it with dispersal, not "disorder".

Entropy measures the number of microstates: the ways the particles and their energy can be arranged.
Entropy increases when matter disperses: solid → liquid → gas, a gas expanding, a solute dissolving, or a reaction making more moles of gas.
Entropy increases when energy disperses: at a higher temperature the energy is spread over a wider range of speeds.
To predict the sign of ΔS for a reaction, compare the moles of gas first, then the phases.
Justify with dispersal and microstates, never with "disorder".

a system can be in a huge number of microstates that look the same from outside the system ends up in the condition that has the most microstates entropy rises when matter disperses entropy rises when energy disperses

entropy: A measure of the number of ways (microstates) the particles of a system and their energy can be arranged; it rises when matter or energy spreads out.

9.2 Absolute Entropy and Entropy Change

Standard molar entropies S° are measured up from zero for a perfect crystal at 0 K, so every one is positive. The entropy change of a reaction is the sum of nS° for the products minus the sum for the reactants, in J/(mol·K). Its sign should agree with the change in moles of gas.

A perfect crystal at 0 K has S = 0; every pure substance at 298 K has a positive S°, elements included (unlike ΔH°f).
S° values: gases ≫ liquids > solids; among similar substances, more atoms per molecule means a larger S°.
ΔS° = Σ nS°(products) − Σ nS°(reactants), each multiplied by its coefficient. Units: J/(mol·K).
Reverse a reaction: flip the sign. Double it: double ΔS°.
Check the sign of the answer against the change in moles of gas.

its entropy is zero, the starting point for measuring every S° the standard molar entropy S° of every pure substance at 298 K is positive, elements included ΔS° of a reaction is Σ nS°(products) − Σ nS°(reactants) checking its sign catches missing coefficients and sign slips

standard molar entropy: The entropy of one mole of a substance in its standard state, S°, in J/(mol·K); it is positive for every pure substance above 0 K. ΔS° of a reaction is Σ nS°(products) − Σ nS°(reactants).

9.3 Gibbs Free Energy and Thermodynamic Favorability

Gibbs free energy, ΔG° = ΔH° − TΔS°, decides thermodynamic favorability: a process is favored when ΔG° < 0. Match the units (kJ and J) and use kelvin. The signs of ΔH° and ΔS° tell you whether a process is favored at all temperatures, none, or only above or below T = ΔH°/ΔS°.

ΔG° = ΔH° − TΔS°. A process is thermodynamically favored when ΔG° < 0.
Make units match: ΔS° from J/(mol·K) to kJ/(mol·K) (÷ 1000), and T in kelvin.
ΔH° < 0, ΔS° > 0: favored at all T. ΔH° > 0, ΔS° < 0: not favored at any T. Same signs: temperature decides.
When the signs agree, the crossover is T = ΔH°/ΔS°, where ΔG° = 0.
ΔG°_rxn = Σ nΔG°f(products) − Σ nΔG°f(reactants); ΔG°f of an element in its standard state is 0.

a negative ΔH° makes a process more likely to be favored a positive ΔS° also makes a process more likely to be favored a process is thermodynamically favored when ΔG° < 0 temperature decides which term wins when the two disagree, and T = ΔH°/ΔS° is where they balance

Gibbs free energy: Gibbs free energy, G; its change ΔG = ΔH − TΔS combines enthalpy and entropy into one test of whether a process is favored. ΔG° is the value under standard conditions.
thermodynamically favored: A process is thermodynamically favored when ΔG < 0: it can proceed on its own, without a continuous input of energy, though possibly very slowly.

9.4 Thermodynamic and Kinetic Control

A negative ΔG° means a process is thermodynamically favored, not that it is fast. When the activation energy is very large, a favored reaction can be too slow to observe; it is under kinetic control. Rate is set by Ea; favorability by ΔG°.

Thermodynamics (ΔG°) answers can it go? Kinetics (Ea, rate) answers how fast? They are independent.
A thermodynamically favored process with a very large activation energy may show no measurable change: it is under kinetic control.
Examples: diamond → graphite, paper and gasoline in air, H₂ and O₂ without a spark.
A spark, heat or a catalyst can start such a reaction; none of them changes ΔG°.
No visible reaction does not prove ΔG° > 0.

it says whether a reaction can go, not how fast a large Ea makes the rate tiny, however negative ΔG° is it is under kinetic control: the reactants are kinetically stable the favored reaction then runs at a measurable rate; ΔG° is unchanged

kinetic control: The situation in which a thermodynamically favored process (ΔG° < 0) is too slow to observe because its activation energy is very large; the reactants are kinetically stable.

9.5 Free Energy and Equilibrium

The standard free energy change and the equilibrium constant are linked by ΔG° = −RT ln K. Negative ΔG° means K > 1 and products are favored; positive means K < 1. Because the link is exponential, small changes in ΔG° make big changes in K. Use R = 8.314 J/(mol·K), kelvin and ln.

ΔG° = −RT ln K, with R = 8.314 J/(mol·K), T in kelvin, and ΔG° in J/mol.
ΔG° < 0 ⇔ K > 1 (products favored). ΔG° > 0 ⇔ K < 1 (reactants favored). ΔG° = 0 ⇔ K = 1.
To find K: K = e^(−ΔG°/RT). Use ln and e, not log and 10.
ΔG° near zero means K near 1: both reactants and products are present in large amounts.
Reverse a reaction: ΔG° changes sign and K inverts. Double it: ΔG° doubles and K is squared.

it measures how far the reaction goes before reaching equilibrium a negative ΔG° gives ln K > 0 and K > 1 a change of a few kJ/mol in ΔG° changes K by a large factor ΔG = 0, while ΔG° keeps its fixed value

ΔG° = −RT ln K: The link between standard free energy change and the equilibrium constant, ΔG° = −RT ln K: a negative ΔG° means K > 1 and a positive ΔG° means K < 1.

9.6 Free Energy of Dissolution

Dissolving a solid absorbs energy to separate solute and solvent particles and releases energy as solute-solvent attractions form. The balance, ΔH°soln, can have either sign. ΔS°soln is usually positive but can be negative when water is ordered around small, highly charged ions. The solid dissolves readily when ΔG°soln < 0.

Dissolving = separate the solute particles (absorbs energy) + separate the solvent particles (absorbs) + form solute-solvent attractions (releases).
ΔH°soln is the small difference between large numbers, so it can be positive (cold pack) or negative (hot pack).
ΔS°soln is usually positive (ions disperse), but small, highly charged ions order the water around them and can make it negative.
A solid dissolves readily when ΔG°soln = ΔH°soln − TΔS°soln < 0; ΔG°soln = −RT ln Ksp links it to solubility.

these steps absorb energy (Coulombic attractions and hydrogen bonds are overcome) this step releases energy, and the balance of the three steps is ΔH°soln ΔS°soln is usually positive, but can be negative for small, highly charged ions a solid dissolves readily when ΔG°soln < 0, whether it is driven by ΔH° or by ΔS°

enthalpy of solution: The enthalpy change when a solute dissolves in a solvent, ΔH°soln: the energy absorbed to separate solute and solvent particles minus the energy released as solute-solvent attractions form. With ΔS°soln it gives the free energy of dissolution, ΔG°soln.

9.7 Coupled Reactions

An unfavored reaction can be made to proceed by coupling it to a strongly favored one through a shared intermediate. The ΔG° values add (and K values multiply), so the overall process is favored if the total ΔG° is negative. External energy such as electricity or light can also drive an unfavored process.

A coupled reaction pairs an unfavored reaction (ΔG° > 0) with a favored one (ΔG° < 0) through a shared intermediate.
When reactions add, their ΔG° values add and their K values multiply.
The overall process is favored if ΔG°(total) < 0. Neither step’s own ΔG° changes.
Examples: Cu₂S + O₂ (smelting), ATP hydrolysis driving cell reactions.
External energy (electric current, light) can also drive a process with ΔG° > 0.

it does not proceed to a useful extent on its own the two reactions add into one overall process the ΔG° values add, and the K values multiply the overall ΔG° is negative and the coupled process is favored

coupled reaction: A pair of reactions linked by a shared intermediate, so a thermodynamically favored reaction drives an unfavored one; their ΔG° values add and their K values multiply.

9.8 Galvanic (Voltaic) and Electrolytic Cells

An electrochemical cell separates oxidation (at the anode) from reduction (at the cathode), so electrons flow through a wire from anode to cathode while ions move through a salt bridge. A galvanic cell runs a favored reaction and produces a current; an electrolytic cell uses an outside power supply to drive an unfavored one.

Oxidation at the anode, reduction at the cathode, in every cell. Electrons flow through the wire from anode to cathode.
A galvanic (voltaic) cell runs a thermodynamically favored reaction and produces a current.
An electrolytic cell uses an outside power supply to drive a reaction that is not favored.
The salt bridge carries ions, not electrons: anions toward the anode, cations toward the cathode, keeping each side neutral.
Evidence from a cell: the anode metal loses mass; metal plates onto the cathode.

electrons released by oxidation at the anode must travel through the wire to the cathode ions move through the salt bridge to keep each compartment neutral the free energy of the reaction does electrical work: a galvanic cell an unfavored reaction is driven: an electrolytic cell

galvanic cell: An electrochemical cell (also called a voltaic cell) in which a thermodynamically favored redox reaction, split into two half-cells, produces an electric current.
electrode: A conductor where a half-reaction takes place: oxidation at the anode, reduction at the cathode.
salt bridge: A tube or porous barrier of an inert electrolyte that lets ions move between the half-cells, keeping each electrically neutral and completing the circuit.
electrolytic cell: An electrochemical cell in which an outside power supply drives a redox reaction that is not thermodynamically favored.

9.9 Cell Potential and Free Energy

Standard reduction potentials rank how readily species gain electrons. In a galvanic cell the half-reaction with the more positive E° is the cathode, and E°cell = E°(cathode) − E°(anode). E° is intensive and is never multiplied by coefficients. ΔG° = −nFE°, so a positive E°cell means a favored reaction with K > 1.

A standard reduction potential E° measures how readily a species gains electrons. More positive: stronger oxidizing agent.
E°cell = E°(cathode) − E°(anode), using both values as reduction potentials. Positive E°cell: favored (galvanic).
E° is intensive: never multiply it by a coefficient when you balance electrons.
ΔG° = −nFE°, with n the mol e⁻ transferred in the balanced equation and F = 96,485 C/mol e⁻ (1 J = 1 C·V).
E° > 0 ⇔ ΔG° < 0 ⇔ K > 1. A negative E°cell means the reaction needs an electrolytic cell.

the half-reaction with the more positive E° is reduced at the cathode E°cell = E°(cathode) − E°(anode), positive for a galvanic cell balancing electrons never multiplies an E° ΔG° = −nFE°, so a positive E°cell means ΔG° < 0 and K > 1

cell potential: The voltage of an electrochemical cell, E. Under standard conditions E°cell = E°(cathode) − E°(anode), where each E° is a standard reduction potential; a positive E°cell means a favored reaction.
Faraday constant: F = 96,485 C/mol e⁻, the charge on one mole of electrons. It links cell potential to free energy, ΔG° = −nFE°.

9.10 Cell Potential Under Nonstandard Conditions

Away from standard conditions, a cell’s potential depends on Q: it is above E° when Q < 1, equal to E° at Q = 1, below E° when Q > 1, and zero at equilibrium (Q = K). The Nernst equation expresses this. A concentration cell has E° = 0 and runs only on a concentration difference.

Q < 1 → E > E°. Q = 1 → E = E°. Q > 1 → E < E°. Q = K → E = 0 (equilibrium, a dead battery).
The Nernst equation, E = E° − (RT/nF) ln Q, describes this; the exam focuses on the direction of the change.
Q uses only dissolved species and gases: solid electrodes and spectator ions do not affect E.
A concentration cell has the same half-reaction on both sides (E° = 0) and runs until the concentrations are equal.

its actual drive, ΔG, differs from ΔG°, so E differs from E° the reaction is further from equilibrium and E > E° Q rises and E falls E = 0 and the battery is dead

Nernst equation: E = E° − (RT/nF) ln Q: the cell potential under nonstandard conditions. For a galvanic cell (E° > 0), E > E° when Q < 1, E < E° when Q > 1, and E = 0 at equilibrium (Q = K).

9.11 Electrolysis and Faraday's Law

In electrolysis a current drives an unfavored redox reaction. The amount of product follows a chain: charge q = It; moles of electrons = q/F; moles of product from the half-reaction’s electrons per ion; mass from the molar mass. The same chain run backward gives the current or time needed.

Electrolysis uses an electric current to drive an unfavored redox reaction, such as plating a metal or splitting a molten salt.
Charge: q = It (coulombs = amperes × seconds). Convert minutes and hours to seconds.
Moles of electrons: q ÷ F, with F = 96,485 C/mol e⁻.
Use the half-reaction: Ag⁺ needs 1 e⁻, Cu²⁺ needs 2, Al³⁺ needs 3. Then moles × molar mass.
The same charge through cells in series gives the same mol e⁻ but different masses of metal.

a charge q = It, in coulombs, passes through the cell moles of electrons = q ÷ F moles of product = mol e⁻ ÷ electrons per ion mass = moles × molar mass

electrolysis: Using an electric current from an outside source to drive a redox reaction that is not thermodynamically favored, as in electroplating or producing a metal from its molten salt.
Faraday's law: The amount of substance produced at an electrode is proportional to the charge passed: q = It, moles of electrons = q/F, and the half-reaction converts moles of electrons to moles of product.