can be broken up into two half reactions

The iron is oxidized while the cerium is reduced.

An oxidant species promotes oxidation and a reductant promotes reduction. The above reaction, cerium is the oxidant or oxidation agent while iron is the reductant or reducing agent.

Electrical charge, q, is measured in coulombs (C). The charge associated with chemical species is related to the number of moles through the Faraday constant, F=96,485.3 (~96,500) C/mole.

Electrical current, I, is measured in Amperes (A). Current is the amount of charge that passes in a unit time interval (seconds).

Ohm's law relates current to potential (E) through the resistance (R) of a circuit.

The potential is measured in Volts (V) and the resistance in Ohms (W).

Power (P) is measured in Watts (W = J/s) and is related to the current and potential by

The work is measured in Joules (J) and is related to the potential and the amount of charge

The relationship between the Gibb's free energy change, DG° (J/mole), and the electro-motive force (EMF), E° (V), is given by

where F (Coul/mole) is Faraday's constant (» 96,500) and n is the number of electrons transferred in the reaction.

The relationship between the Gibb's free energy change, DG°, and the equilibrium constant, K, is given by is given by

where k is Boltzmann's constant (8.313 J/mole K) and T (K) is the temperature.

The Nernst Equation gives the cell potentials, E (V),

where Q is the ratio of products over reactants typical of equilibrium calculations. For example, the generic reaction

has a Q

where the A are activities. For low concentration solutions

1) Calculations of equilibrium constants from tables of reduction potentials.

What is the K for ceric oxidation of ferrous iron?

The full reaction,

is comprised of the two half reactions

The E° are found from the table of reduction potentials in the back of the book. Notice that when the reaction is reversed, as for Fe²⁺ oxidation, the sign of E° is changed. The sum of the two half reactions yields the total reaction and the reaction potential is the sum of the two half reaction potentials

E° = 1.63 V - 0.77 V = +0.86 V

Using the formulas,

DG^o= -nFE°

DG° = -(1)(96,500 Coul/mole)(0.86 V)

DG° = -83 kJ/mole

DG° = -RT ln K

-83 kJ/mole = -(8.314 J/mole K)(298 K) ln K

Calculations of electrode potentials for electrodes using insoluble salts.

Q: What is the electrode potential for the electrode Ag/AgI(s)/I^- (0.01 M) ?

A: The overall reaction for this electrode is

This reaction cannot be found in most tables of reduction potentials. But the reaction is comprised of two components,

We can find the E° for the solubility product equilibrium from DG° =-kTlnK_sp and DG° =-nFE°

0.059log is valid at room temperature. In this case

E° = .059 log K_sp = .059 log 10^-16 = (.059)(-16) = -0.9 V

The electrode potential is E° =E° _red+E° _Ksp = 0.8V-0.9V = -0.1V

Now we use the Nernst Equation to find the electrode potential

E = E° - .059 log [I^-]

E = -0.1V - .059 log 10^-2

2) Another way to look at the electrode potential for Ag/AgI(s)/I^-(0.01 M)

The overall reaction for this electrode is

The reaction is comprised of two components,

Note the change of sign on the E° because it is an oxidation. We can initially ignore the fact that the electrode contains AgI and find E for the silver ion reduction.

The electrode potential is based on the concentration of Ag⁺ in solution. Since

and

In this case

In either calculation, the important point to recognize is that the electrode contains an insoluble salt.