Fundamentals of Electrochemistry

14.1 Basic Concepts

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It involves the study of galvanic cells, standard potentials, the Nernst equation, equilibrium constants, and the use of cells as chemical probes.

14.2 Galvanic Cells

Galvanic cells are electrochemical cells that convert chemical energy into electrical energy via spontaneous redox reactions. Each cell consists of two half-cells, a positive electrode (cathode) and a negative electrode (anode), connected by an external circuit and an electrolyte.

14.3 Standard Potentials

The standard potential (E°) is the measure of the intrinsic ability of a half-reaction to gain electrons and act as a reductant. It is measured under standard conditions (1 M concentration, 1 atm pressure, and 25°C). Positive E° values indicate a stronger oxidizing agent, while negative values suggest a stronger reducing agent.

14.4 Nernst Equation

The Nernst equation describes how the electric potential of a cell changes with the concentrations of the reactants and products. For a half-cell reaction involving silver as an example, the reaction can be expressed as follows:

AgCl(s) + e ̄ ⇌ Ag(s) + Cl ̄ (E° = +0.222 V)

Here, the chloride ion's concentration affects the cell potential. It is noted that the concentration of chloride ions is [Cl^{-}] = 2 imes 0.0167 M = 0.0334 M. Using the Nernst equation:

E₊ = E° - 0.05916 log[Cl−]

Substituting in the values gives:

E₊ = 0.222 - 0.05916 log(0.0334) = 0.309 V

Using the alternative half-reaction:

Ag+ + e ̄ ⇌ Ag(s) (E° = +0.799 V)

Results in:

E₊ = 0.799 - 0.05916 log(1/[Ag+])

The concentration of [Ag+] can be derived from the solubility product (Ksp) as:

[Ag+] = Ksp/[Cl−] = [Ag^{+}] = 1.8 imes 10^{-10}/0.0334 = 5.4 imes 10^{-9} M

Calculating the electric potential:

E₊ = 0.799 - 0.05916 log(1/(5.4 × 10⁻⁹)) = +0.309 V

Regardless of the half-reaction chosen, the resulting potential will yield the same E₊ value.

Key tips in analyzing galvanic cells include identifying elements in two oxidation states, exemplified by lead and copper in the following cell:

Pb(s)|PbF2(s)|F ̄(aq)ǁCu2+(aq)|Cu(s)

Rules for Assigning Oxidation Numbers

Elements: Oxidation number of free elements is zero (e.g., H₂, O₂).
Monoatomic Ions: Charge of the ion is the oxidation number; e.g., Na⁺ (+1), P³⁻ (-3).
Polyatomic Ions: The sum of oxidation numbers equals the charge on the ion (e.g., SO₄²⁻).
Neutral Compounds: The total oxidation number of the atoms equals zero.
Oxygen: Usually -2 (with exceptions: peroxides -1 and bonded to fluorine +2).
Hydrogen: +1 unless bonded to metals (hydrides).
Halogens: Fluorine always -1; others typically -1 unless bonded to more electronegative elements.

Identifying Half-Reactions

The half-reactions for the lead-acid battery's cell can be expressed as:

Right half-cell: Cu²⁺ + 2e ̄ ⇌ Cu(s)
Left half-cell: PbF₂(s) + 2e ̄ ⇌ Pb(s) + 2F ̄ or Pb²⁺ + 2e ̄ ⇌ Pb(s)

Your choice does not impact the overall reaction but may depend on the ease of finding necessary concentrations for calculations.

14.5 E° and the Equilibrium Constant

Galvanic cells generate electricity due to non-equilibrium states. The equilibria described occur within half-cells and on a larger scale between them. When a battery discharges fully, it reaches equilibrium, resulting in E becoming zero, and the cell is termed 'dead.'

The relationship between cell potential (E) and reaction quotient (Q) can be expressed as:

E = E° - 0.05916 n log Q

At equilibrium, E = 0 and Q = K, leading to:

E° = (0.05916/n) log K

Finding K from E° can be derived from the relationship linking Gibbs free energy

ΔG° = -nFE° and ΔG° = -RT ln K

Thus:

E° = (RT/nF) ln K

Example Problem

To find the equilibrium constant for:

Cu(s) + 2Fe³⁺ ⇌ 2Fe²⁺ + Cu²⁺

First, identify the half-reactions:

Half-Reaction 1: 2Fe³⁺ + 2e ̄ ⇌ 2Fe²⁺ (E° = +0.771 V)
Half-Reaction 2: Cu²⁺ + 2e ̄ ⇌ 2Cu(s) (E° = +0.339 V)

Combining, we get:

2Fe³⁺ + 2Cu(s) ⇌ 2Fe²⁺ + Cu²⁺

Calculating E° gives:

E° = +0.771 - +0.339 = 0.432 V

Using this, the equilibrium constant K can be calculated:

K = 10^(2 × 0.432/0.05916) = 4 × 10¹⁴

Redox Reactions

The relation between E° and free energy extends beyond redox reactions, allowing calculations for others, such as:

FeCO₃(s) + 2e ̄ ⇌ Fe(s) + CO₃²⁻

Here, you can deduce Ksp by identifying relevant half-reactions and corresponding E° values leading to:

K = 10^(nE°/0.05916) = 10^(2(-0.316)/0.05916) = 10⁻¹¹

Example Problem: Complexation

For the complexation of Cu⁺ by ethylenediamine:

Cu⁺ + 2en ⇌ Cu(en)₂⁺

Using provided E° values, we can find:

E° = +0.637 V and formation constant β₂ = 5.9 × 10¹⁰

14.6 Cells as Chemical Probes

Galvanic cells can also function as probes for chemical equilibria. The equilibrium states can be established both between half-cells and within each cell individually. A cell operating at a non-zero voltage indicates the net reaction is not at equilibrium, while prolonged standing leads to equilibrium in each half-cell.

In the right-hand half-cell, we have:

AgCl(s) ⇌ Ag⁺ + Cl ̄

And in the left, reach equilibrium with:

CH₃COOH ⇌ CH₃COO ̄ + H⁺

The redox reaction occurring in the right half-cell is:

AgCl(s) + e ̄ ⇌ Ag(s) + Cl ̄ (aq, 0.10 M) (E° = 0.222 V)

The left half-cell can be presented as:

2H⁺(aq, ? M) + 2e ̄ ⇌ H₂(g, 1 bar) (E° = 0 V)

As the cells are not in equilibrium, a voltage is measured (0.503 V). The Nernst equation for the described reaction is formulated as:E = E₊ - E₋ = 0.222−0.05916 log[Cl−]−0−0.05916 log([H⁺]²)

Substituting in parameters gives:

0.503 = 0.222−0.05916 log[0.10]−0−0.05916 log(1/[H⁺]²)

Solving yields:

[H⁺] = 1.8 × 10⁻⁴ M

This calculated [H⁺] allows for evaluating the CH₃COOH/CH₃COO⁻ equilibrium.

Kₐ = (CH₃COO⁻)[H⁺]/[CH₃COOH] = 0.0050[1.8 × 10⁻⁴]/0.050 = 1.8×10⁻⁵

This shows how the cell serves as a probe to measure [H⁺] in the left half-cell.

Guidelines for Equilibrium and E Calculations

Write the two half-reactions along with their standard potentials.
Formulate the Nernst equation for the net reaction, inserting known values.
Identify the single unknown variable in the Nernst equation for resolution.
Solve for this unknown concentration, addressing the related chemical equilibrium problem.

Fundamentals of Electrochemistry Overview

Fundamentals of Electrochemistry Overview