ELECTROCHEMISTRY
Before we can talk reactions…
Oxidation State (or Oxidation Number): real or apparent charge of an atom in a compound; used for keeping
track of electrons in a redox reaction
*In an ionic compound, the charges are real (actual ionic charges). In a molecular compound, the charges are
apparent (charges that atoms would have if electrons were completely transferred; of course, they are not!).
Rules for Assigning Oxidation Numbers:
1. In free elements, each atom has an oxidation number of 0.
2. For monatomic ions, the oxidation number is equal to the charge of the ion.
3. The oxidation number of oxygen in most compounds is –2. However, in hydrogen peroxide (H2O2) and
peroxide ion (O22-), the oxidation number of oxygen is –1.
4. The oxidation number of hydrogen is +1, except when it is bonded to metals of binary compounds (in
which it has an oxidation number of –1).
5. Fluorine has an oxidation number of –1 in all its compounds. When Cl, Br, and I occur as halide ions,
they also have oxidation numbers of –1.
6. In a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero. In a polyatomic
ion, the sum of the oxidation numbers of all the atoms must be equal to the net charge of the ion.
Oxidation State Examples: N2, P4, Fe3+, H2O, H2O2, CH4, NiH2, KCl, CO2, NO3–, MnO4–
Electrochem Lingo
Oxidation: loss of electrons by an atom in a chemical reaction
Reduction: gain of electrons by an atom in a chemical reaction
“LEO says GER”
Redox Reaction: chemical reaction in which electrons are transferred from one atom to another (number of
electrons lost = number of electrons gained)
Oxidizing Agent: substance that causes oxidation; it’s the stuff that gets reduced!
Reducing Agent: substance that causes reduction; it’s the stuff that gets oxidized!
Now let’s apply our understanding of redox terminology to the chemical reaction that occurs in alkaline
batteries. Which substance was oxidized? Which substance was reduced? Which is the oxidizing agent? Which
is the reducing agent?
Zn (s) + 2 MnO2 (s) → ZnO (s) + Mn2O3 (s)
Practice Redox Reactions: Identify the elements that undergo oxidation and reduction in each reaction.
Which substance is the oxidizing agent? The reducing agent?
Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
4Al (s) + 3 O2 (g) → 2 Al2O3 (s)
Cu (s) + 2 AgNO3 (aq) → Cu(NO3)2 (aq) + 2 Ag (s)
CH4 (s) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
4 I– (aq) + 2 Cu2+ (aq) → I2 (s) + 2 CuI (aq)
Balancing Redox Reactions
It’s not always possible to balance redox reactions by inspection. Take, for example, the reaction below:
Au3+ (aq) + I– (aq) → Au (s) + I2 (s)
At first glance, it seems that this equation can be balanced by placing a 2 in front of the I–.
Au3+ (aq) + 2 I– (aq) → Au (s) + I2 (s)
Note, however, that although the atoms are now balanced, the charge is not. The sum of the charges on the left
is +1, and the sum of the charges on the right is zero, as if the products could somehow have one more electron
than the reactants. To correctly balance this equation, it helps to look more closely at the oxidation and
reduction that occur in the reaction. The iodine atoms are changing their oxidation number from -1 to 0, so
each iodide ion must be losing one electron. The Au3+ is changing to Au, so each gold(III) cation must be
gaining three electrons. The unbalanced half-reactions are:
Oxidation: 2 I– (aq) → I2 (s) + 2eReduction: Au3+ (aq) + 3e- → Au (s)
We know that in redox reactions, the number of electrons lost by the reducing agent must be equal to the
number of electrons gained by the oxidizing agent; thus, for each Au3+ that gains three electrons, there must be
three I- ions that each lose one electron. Therefore, the balanced redox reaction is
2 Au3+ (aq) + 6 I– (aq) → 2 Au (s) + 3 I2 (s)
How to Balance Redox Reactions: “The Half-Reaction Method”
Step 1: Separate reaction into two half reactions (oxidation and reduction).
Step 2: Balance the atoms in each half reaction. First, balance all atoms except H and O. Next, in
acidic solution, add H2O to balance O atoms and then add H+ to balance H atoms. (In basic solution, use
OH- and H2O to balance H and O atoms.)
Step 3: Use electrons to balance the charges in each half reaction.
Step 4: Ensure that both half reactions have the same number of electrons. If not, multiply by the
equation(s) by the appropriate coefficient(s).
Step 5: Recombine the half reactions.
Practice Balancing Redox Reactions: Now you try it!
Cu + Al3+ → Cu2+ + Al
Fe3+ + Sn2+ → Fe2+ + Sn4+
MnO4- + I- → I2 + Mn2+ (Assume an acidic solution)
Predicting the Spontaneity of Redox Reactions
As you already know, removing electrons from any atom/ion requires energy. Adding electrons to any atom/ion releases energy.
We quantify the energy absorbed/released by individual oxidation/reduction processes relative to the standard
hydrogen electrode (SHE):
2 H+ (1.00 M) + 2e– → H2 (1 atm)
Eo = 0 V
See table of Standard Reduction Potentials.
E° = standard half-cell voltage; measures likelihood of reaction to occur
spontaneously in the direction it is written at standard conditions
(25°C, 1 atm, 1 M concentrations for all aqueous solutions)
o
o
Ecell
= Eoxo + Ered
Note: Half-cell potentials are
intensive variables and thus do
NOT depend on the quantity
of substance oxidized/reduced.
1 Volt = 1 Joule/Coulomb
o
If Ecell
> 0 , the reaction is spontaneous in the direction written.
o
If Ecell
< 0 , the reaction is not spontaneous in the direction written.
Practice Problems: Predict whether the following reactions are spontaneous.
Cu + 2 Ag+ → 2 Ag + Cu2+
Oxidation:
Eoxo =
Reduction:
o
=
Ered
o
=
Ecell
3 Cu + 2 Al3+ → 3 Cu2+ + 2 Al
Oxidation:
Eoxo =
Reduction:
o
=
Ered
o
=
Ecell
Yes, indeed! E° and ΔG are related!
o
ΔG = −nFEcell
n = number of moles of electrons exchanged
F = Faraday’s constant = 96,487 C/mol e or 96,487 J/(V mol e)
*A Coulomb is a unit of electrical charge. The charge carried by one electron is 1.60 X 10-19 C.
Electrochemical Cells
Separating a redox reaction into separate reduction and oxidation parts is not purely a theoretical exercise. In
real life, two half reactions can be carried out in separate compartments (as in the alkaline battery) or beakers (as
we used in lab) and coupled so that electrons flow through an external circuit.
Anode: attracts anions (site of oxidation)
Cathode: attracts cations (site of reduction)
A. Galvanic/Voltaic Cells or Batteries
A redox reaction occurring in a galvanic cell has a negative ΔG and is therefore a spontaneous reaction. These
reactions “push electrons” and are used to do work.
Oxidation: Zn (s) → Zn2+ (aq) + 2e-
Eoxo = 0.76 V
o
= 0.34 V
Ered
Reduction: Cu2+ (aq) + 2e- → Cu (s)
_________________________________________________
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
o
= 1.10 V
Ecell
B. Electrolytic Cells
A redox reaction occurring in an electrolytic cell has a positive ΔG
and is therefore a nonspontaneous reaction. In electrolysis, electrical
energy is required to induce reaction; i.e., we need a source of
electrical power (a battery!) to force a reaction to take place.
Net Reaction: 2 H2O (l) → 2 H2 (g) + O2 (g)
What are the two half-reactions? How can you rationalize these
reactions with your observations?