Electro-Chemistry
Key Ideas

9.4: Thermodynamic and Kinetic Control

CED/Enduring Learning Objectives: Explain in terms of kinetics why a thermodynamically favored reaction might not occur at a measurable rate

Rxns can be thermodynamically favorable (ΔG is negative) but do not make products, processes are under “kinetic control” and have large ass activation energies equating a slow reaction

Just because the reaction proceeds slowly does not mean it is at equilibrium

Kinetic control = thermodynamically favored + not at a measurable rate 

Catalysts have NO effect on thermodynamic favorability but can be used to decrease activation energy + reaction rate

Many thermodynamically favored processes usually occur at extremely slow rates

9.5: Free Energy and Equilibrium

CED/Enduring Learning Objectives:Explain whether a process is thermodynamically favored using the relationships between K, ΔG, and T

“Thermodynamically favored” (ΔG < 0) products are favored at equilibrium (K > 1)

At equilibrium, there is no net change in concentration, so K = e⁻Δᴳ/ᴿᵀ and ΔG = -RTlnK

K and ΔG can be assumed through estimating when ΔG is negative, K > 1 and rxn favors products, when ΔG is positive, K < 1 and rxn favors reactants, when ΔG is 0 rxn is at equilibrium

Breaking of bonds lewis/rxn = (+)ΔH, Forming of bonds lewis/rxn = (-)ΔH

ΔG = -RTlnK → K = e⁻Δᴳ/ᴿᵀ (because u divide ΔG by -RT, then ln both sides to solve for )

Make sure all units are the same

Temperature (T) has to be in units of Kelvin

K = e⁻Δᴳ/ᴿᵀ

If given K equilibrium, and temperature convert temperature to Kelvin and plug in ΔG = -RTlnK

9.6: Coupled Reactions

Thermodynamically UNFAVORABLE processes (+ΔG value) are unlikely but can occur through external sources of energy (eg. sunlight, electricity), or can be “coupled” to another FAVORABLE rxn

Coupled rxn’s are two rxn’s that share a common intermediate

Hess’s Law (rules) are applied to determine the ΔG for the coupled rxn

The sum of the rxn’s ΔG values is negative, making the overall process thermodynamically favorable

Reactions need to share a common intermediate (basically something formed in one eqn and used in the following eqn)

Combining these eq's can “coupled” and make the new eqn thermodynamically favorable

Share a common intermediate

Produce a thermodynamically favorable overall rxn when combined

Typically used to drive unfavorable rxn that is industrially important such as extraction of metals from ores

Hess’s Law is used to calculate the ΔG values for the overall reaction

An external energy source can also drive a thermodynamically unfavorable process

Hess Law Review

flipping the equation = flipping the sign (+/-)

multiplying the equation by coefficients = multiplying the ΔG values by that number

combine the equations (basically add them together) as long as their is a common intermediate

for this one make sure ΔG value becomes positive (+) that is the whole point

9.7: Galvanic (Voltaic) and Electrolytic Cells

Electrochemical cell converts energy released by a thermodynamically favorable redox reaction to electrical energy or uses electrical energy to drive an unfavorable redox rxn

Galvanic (or voltaic) cells involve a favorable rxn

Electrolytic cells involve an unfavorable rxn

Both contain an anode, where oxidation occurs, and a cathode where reduction occurs

Galvanic Cell Components

Anode can be a reactant (eg. metals) that gets oxidized or an inert solid (eg. platinum/graphite)

Anode (reactant) mass decreases as the metal (reactant) is converted to ions, allowing for the movement of charged particles, electrons flowing from the anode to the cathode through the wire

Cathode can be a product (eg. metal) that is deposited or an inert solid (eg. platinum/graphite) 

Cathode (product) mass of the cathode increases, as the solid is deposited on the cathode, which allows for the movement of charged particles, and electrons flow into the cathode

Salt bridge is a device that allows for the movement of ions between half-cells

Contains an inert ionic solution

Allows the movement of cations into the reduction half-cell and anions into the oxidation half-cell making charge-balanced, necessary for current to flow into the circuit

No salt bridge, no reaction occurs

Tips: mneumonic  An Ox, Red Cat

ANODE = OXIDATION

CATHODE = REDUCTION

Always anode to cathode

E (V values) always stay the same Hess’s Law does not apply to V values

Electrolytic Cell

Thermodynamically unfavorable rxn, a power source is needed (eg. battery source)

The reaction often occurs in a single chamber (no salt bridge required)

Electrons flow from the anode to the power source and from the power source to the cathode

Occurs in an ionic solution/liquid (Similar to Galvanic Cells)

Cations (+) flow to the cathode / Anions (-) flow to the anode,  to maintain charge balance

9.8: Cell Potential and Free Energy

Reactions occurring in the electrochemical cell has an electric potential difference (voltage) between the oxidation and reduction in the half-cells, symbols for reduction potential (E°), and units are volts (V)

Comparing the reduction potentials of different oxidation/reduction standard reduction potentials (E red) are used, potentials are only accurate under standard conditions (1 M solutions, 1 atm pressure, 25°C)

Tables only list reduction half-reactions and standard potentials, oxidation processes, the reduction half-rxn equation, and the sign of the voltage are both reversed

Cell’s Standard potential (E°cell) is calculated from the reduction half-rxn standard potential for the processes of the cathode and anode (half-reactions occurring in the cell)

E°cell = E°cathode - E°anode

The half-reaction at the anode is oxidation, minus sign in in the above formula accounts for the reversal of the reduction potential for this process

Electrical potential does not depend on the amount of substance reacting, do not multiply the values of the half-rxn potentials by the coefficients in the overall cell reaction

Thermodynamic Favorability

Favorable reactions have (+) overall cell potentials (Voltaic cells = positive cell potentials)

Unfavorable reactions have (-) overall cell potentials (Electrolytic cells = negative cell potentials)

Do NOT multiply standard reduction potentials by stoichiometric coefficients

The half-reaction equation and the sign of the standard reduction potential must be reversed for oxidation

1 volt = 1 joule/coulomb

ΔG° (standard Gibbs free energy) is proportional to:

the negative of the cell potential for the redox rxn from which is constructed

the moles of electrons (n) transferred in the redox rxn

Faraday’s constant (F) is the amount of charge (q) in coulombs (C) per mole of electrons

(+) E°꜀ₑₗₗ → (-) ΔG° (voltaic cells, thermodynamically favorable rxn)

(-) E°꜀ₑₗₗ → (+) ΔG° (electrolytic cells, thermodynamically unfavorable rxn)

Key equation ΔG° = -nFE꜀ₑₗₗ (n is the number of electrons in the balanced half-reactions for a process)

1 volt = 1 joule/Coulomb

E cell and ΔG are inversely proportional

Basically, for questions be able to write in cell notation

Identify the anode/cathode

Multiply coefficients, and electrons to cancel out (take the most negative E/V value)

Do not multiply the E°꜀ₑₗₗ/V value Hess’s Law doesn't matter (the E/V value given is the value constant)

Then solve by using E°꜀ₑₗₗ equation

(V values stay the same despite you rewriting the equations to cancel out the electrons)

Always flip the most negative rxn V value in this case 0.54 volts is the smallest number so you flip the top rxn ((the equations are always given in the same direction (e^- on the same side)) 

9.9: Cell Potential Under Nonstandard Conditions

Cell potential (E°꜀ₑₗₗ) depends on the concentration (or pressures) of reactants/products

Standard conditions are 1.0M solutions and 1.0 atm gasses, corresponding to Q = 1 Standard cell potentials (E°꜀ₑₗₗ) calculated from standard reduction potentials only apply under these conditions

As a cell operates, reactants are consumed and products are produced meaning their concentrations change, thus cell potential changes

Changing the mass of solids does not affect the cell potential because it does not affect Q

Under standard conditions (Q = 1) the chemical system in the cell is not at equilibrium

As the system approaches equilibrium the magnitude (absolute value) of the cell potential decreases, reaching zero at equilibrium (when Q = K) 

Voltaic cells, K > 1, thus when Q is increased above 1 the system moves closer to equilibrium and the cell potential decreases relative to standard conditions

Conversely, when Q is decreased below 1, the cell is further from equilibrium, and the cell potential increases relative to standard conditions

The opposite effects are observed in electrolytic cells because K <  1 “Increasing the magnitude” of the cell potential in electrolytic cells means making it more negative

In a voltaic cell ↑Q means ↓E°꜀ₑₗₗ and ↓Q means ↑E°꜀ₑₗₗ

The opposite relationships are seen in electrolytic cells (this refers to [E°꜀ₑₗₗ] because cell potentials are negative

Never use a Le Chatelier argument when discussing nonstandard cells because they are not at equilibrium

Changing the mass of the solid does not affect cell potential

9.10: Electrolysis and Faraday’s Law

Charge Flow in Electrolysis

The amount of charge (q) flowing through an electrolytic cell is a function of the current (l) and the time that the cell operates (t)

The equation is l = q / t

q is related to changes in amounts of reactants and products in the cell, stoichiometry is used to analyze

Attentive to units and use of dimensional analysis is important

Current is in a unit of Amperes (A), equivalent to Coulombs/second (C/s)

Time needs to be in seconds when using Faraday’s law equation

Faraday’s Constant (F), is the charge per mole of electrons (96,485 C/mol e-)

Moles of electrons per mole of reaction is determined by the chemical equation

Moles/molar mass is used to convert between grams and moles

Changes in amounts of reactants/products in a cell are related to charge flow

(basically a lot of dimensional analysis/unit conversions)