Redox Reactions
Redox reactions, or oxidation-reduction reactions, have a number of similarities to acid-base
reactions. Fundamentally, redox reactions are a family of reactions that are concerned with the
transfer of electrons between species. Like acid-base reactions, redox reactions are a matched set
-- you don't have an oxidation reaction without a reduction reaction happening at the same time.
Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Each
reaction by itself is called a "half-reaction", simply because we need two (2) half-reactions to
form a whole reaction. In notating redox reactions, chemists typically write out the electrons
explicitly:
Cu (s) ----> Cu2+ + 2 eThis half-reaction says that we have solid copper (with no charge) being oxidized (losing
electrons) to form a copper ion with a plus 2 charge. Notice that, like the stoichiometry notation,
we have a "balance" between both sides of the reaction. We have one (1) copper atom on both
sides, and the charges balance as well. The symbol "e-" represents a free electron with a negative
charge that can now go out and reduce some other species, such as in the half-reaction:
2 Ag+ (aq) + 2 e- ------> 2 Ag (s)
Here, two silver ions (silver with a positive charge) are being reduced through the addition of
two (2) electrons to form solid silver. The abbreviations "aq" and "s" mean aqueous and solid,
respectively. We can now combine the two (2) half-reactions to form a redox equation:
We can also discuss the individual components of these reactions as follows. If a chemical
causes another substance to be oxidized, we call it the oxidizing agent. In the equation above,
Ag+ is the oxidizing agent, because it causes Cu(s) to lose electrons. Oxidants get reduced in the
process by a reducing agent. Cu(s) is, naturally, the reducing agent in this case, as it causes Ag+
to gain electrons.
As a summary, here are the steps to follow to balance a redox equation in acidic medium (add
the starred step in a basic medium):
1. Divide the equation into an oxidation half-reaction and a reduction half-reaction
2. Balance these
o Balance the elements other than H and O
o Balance the O by adding H2O
BY DR. J.J. GONGDEN
DEPT. OF CHEMISTRY, UNIJOS
www.vitalchemist.com
Balance the H by adding H+
Balance the charge by adding e3. Multiply each half-reaction by an integer such that the number of e- lost in one equals the
number gained in the other
4. Combine the half-reactions and cancel
5. **Add OH- to each side until all H+ is gone and then cancel again**
o
o
In considering redox reactions, you must have some sense of the oxidation number (ON) of the
compound. The oxidation number is defined as the effective charge on an atom in a compound,
calculated according to a prescribed set of rules. An increase in oxidation number corresponds to
oxidation, and a decrease to reduction. The oxidation number of a compound has some analogy
to the pH and pK measurements found in acids and bases -- the oxidation number suggests the
strength or tendency of the compound to be oxidized or reduced, to serve as an oxidizing agent
or reducing agent. The rules are shown below. Go through them in the order given until you have
an oxidation number assigned.
1.
2.
3.
4.
5.
For atoms in their elemental form, the oxidation number is 0
For ions, the oxidation number is equal to their charge
For single hydrogen, the number is usually +1 but in some cases it is -1
For oxygen, the number is usually -2
The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to
its total charge.
As a side note, the term "oxidation", with its obvious root from the word "oxygen", assumes that
oxygen has an oxidation number of -2. Using this as a benchmark, oxidation numbers were
assigned to all other elements. For example, if we look at H2O, and assign the value of -2 to the
oxygen atom, the hydrogens must each have an oxidation number of +1 by default, since water is
a neutral molecule. As an example, what is the oxidation number of sulfur in sulfur dioxide
(SO2)? Given that each oxygen atom has a -2 charge, and knowing that the molecule is neutral,
the oxidation number for sulfur must be +4. What about for a sulfate ion (SO4 with a total charge
of -2)? Again, the charge of all the oxygen atoms is 4 x -2 = -8. Sulfur must then have an
oxidation number of +6, since +6 + (-8) = -2, the total charge on the ion. Since the sulfur in
sulfate has a higher oxidation number than in sulfur dioxide, it is said to be more highly
oxidized.
Working with redox reactions is fundamentally a bookkeeping issue. You need to be able to
account for all of the electrons as they transfer from one species to another. There are a number
of rules and tricks for balancing redox reactions, but basically they all boil down to dealing with
each of the two half-reactions individually. Consider for example the reaction of aluminum metal
to form alumina (Al2O3). The unbalanced reaction is as follows:
Looking at each half reaction separately:
BY DR. J.J. GONGDEN
DEPT. OF CHEMISTRY, UNIJOS
www.vitalchemist.com
This reaction shows aluminum metal being oxidized to form an aluminum ion with a +3 charge.
The half-reaction below shows oxygen being reduced to form two (2) oxygen ions, each with a
charge of -2.
If we combine those two (2) half-reactions, we must make the number of electrons equal on both
sides. The number 12 is a common multiple of three (3) and four (4), so we multiply the
aluminum reaction by four (4) and the oxygen reaction by three (3) to get 12 electrons on both
sides. Now, simply combine the reactions. Notice that we have 12 electrons on both sides, which
cancel out. The final step is to combine the aluminum and oxygen ions on the right side using a
cross multiply technique:
Taking care of the number of atoms, you should end up with:
One of the more useful calculations in redox reactions is the Nernst Equation. This equation
allows us to calculate the electric potential of a redox reaction in "non-standard" situations. There
exist tables of how much voltage, or potential, a reaction is capable of producing or consuming.
These tables, known as standard potential tables, are created by measuring potential at "standard"
conditions, with a pressure of 1 bar (≅1 atm), a temperature of 298° K (or 25° C, or room
temperature) and with a concentration of 1.0 M for each of the products. This standard potential,
or E°, can be corrected by a factor that includes the actual temperature of the reaction, the
number of moles of electrons being transferred, and the concentrations of the redox reactants and
products. The equation is:
BY DR. J.J. GONGDEN
DEPT. OF CHEMISTRY, UNIJOS
www.vitalchemist.com
Perhaps the best way of understanding this equation is through an example. Suppose we have
this reaction:
Fe(s) + Cd2+(aq) ------> Fe2+(aq) + Cd(s)
In this reaction iron (Fe) is being oxidized to iron(II) ion, while the cadmium ion (Cd2+) in
aqueous solution is being reduced to cadmium solid. The question is: how does this reaction
behave in "non-standard" conditions?
The first thing to answer is how does it behave in standard conditions? We need to look at the
standard potential for each half-reaction, then combine them to get a net potential for the
reaction. The two (2) half-reactions are:
Fe2+ (aq) + 2 e- ------> Fe (s), E° = -0.44 V
Cd2+ (aq) +2 e- ------> Cd (s), E° = -0.40 V
Notice that both half-reactions are shown as reductions -- the species gains electrons, and is
changed to a new form. But in the complete reaction above, Fe is oxidized, so the half-reaction
needs to be reversed. Quite simply, the potential for the half-reaction of iron is now 0.44 V. To
get the potential for the entire reaction, we add up the two (2) half-reactions to get 0.04 V for the
standard potential.
BY DR. J.J. GONGDEN
DEPT. OF CHEMISTRY, UNIJOS
www.vitalchemist.com
The question now is: what is the total potential (in volts) for a nonstandard reaction? Suppose
again that we have the same reaction, except now we have 0.0100 M Fe2+ instead of the standard
1.0 M. We need to use the Nernst equation to help us calculate that value. If you go to the Redox
Half-Reaction Calculator, you should notice that the reaction is selected and the appropriate
values are entered into the boxes. Since we don't have any species "B" or "D", we have entered
zero for their concentrations. The concentration of the solid Fe is 1.0 M (actually, concentrations
of solids and solvents (liquids) don't enter into the Nernst equation, but we set them to 1.0 so that
the mathematics works out). If you click on the "Evaluate" button, you should learn that the
standard potential is -0.44 V, while the nonstandard potential is -0.5 V. If you scroll down on the
calculator, you can enter 0.5 as the first half-reaction. We again change the sign since we're
actually reversing the Fe reaction
Using the calculator again, we calculate the nonstandard potential of the Cd reaction. Suppose
we now have a concentration of Cd2+ of 0.005 M, what is its potential? The calculator should
return a standard potential of -0.4 V and a nonstandard potential of -0.47 V. Place this value in
the box for the second half-reaction, then click on "Evaluate". You should learn that the net
nonstandard potential is 0.03 V, slightly less than the value of the net standard potential. Since
this value is less than the net standard potential of 0.04 V, there is less of a tendency for this
reaction to transfer electrons from reactants to products. In other words, less iron will be
oxidized and cadmium will be reduced than at standard conditions.
Test your use of the redox calculator by calculating the net standard potential for this reaction:
2 Ag+ (aq, 0.80 M) + Hg (l)------> 2 Ag (s) + Hg2+ (aq, 0.0010M)
Answer: 0.025 V. Since the value is positive, the reaction will work to form the products
indicated. Negative values of the potential indicate that the reaction tends to stay as reactants and
not form the products. The net standard potential for this reaction is 0.01 V -- since the
nonstandard potential is higher, this reaction will form products than the standard reaction.
Free energy and the standard potential can also be related through the following equation:
Where:
ΔG = change in free energy
n = number of moles
If a reaction is spontaneous, it will have a positive Eo, and negative ΔG, and a large K value
(where K is the equilibrium constant-this is discussed more in the kinetics section).
The energy released in any spontaneous redox reaction can be used to perform electrical work
using an electrochemical cell (a device where electron transfer is forced to take an external
pathway instead of going directly between the reactants. Think of the reaction between zinc and
copper. Instead of placing a piece of zinc directly into a solution containing copper, we can form
BY DR. J.J. GONGDEN
DEPT. OF CHEMISTRY, UNIJOS
www.vitalchemist.com
a cell where solid pieces of zinc and copper are placed in two different solutions such as sodium
nitrate. The two solids are called electrodes. The anode is the electrode where oxidation occurs
and mass is lost where as the cathode is the electrode where reduction occurs and mass is gained.
The two electrodes are connected by a circuit and the two (2) solutions are connected by a "salt
bridge" which allows ions to pass through. The anions are the negative ions and they move
towards the anode. The cations are the positive ions and they move towards the cathode.
The following is a diagram of an electrochemical cell with zinc and copper acting as the
electrodes.
An external electric current hooked up to an electrochemical cell will make the electrons go
backwards. This process is called electrolysis. This is used, for example, to make something gold
plated. You would put the copper in a solution with gold and add a current which causes the gold
ions to bond to the copper and therefore coating the copper. The time, current, and electrons
needed determine how much "coating" occurs. The key to solving electolysis problems is
learning how to convert between the units. Useful information: 1 A=1 C/sec; 96,500 coulombs
can produce one (1) mole of e-; the electrons needed is determined by the charge of the ion
involved
Example Problem: If you are trying to coat a strip with aluminum and you have a current of 10.0
A (amperes) running for one hour, what mass of Al is formed?
The solution of this problem involves a lengthly unit conversion process:
BY DR. J.J. GONGDEN
DEPT. OF CHEMISTRY, UNIJOS
www.vitalchemist.com
Practice Redox Problem: balance the following redox reaction in acidic solution:
S(s) + NO3-(aq) --> SO2(g) + NO(g)
The redox solution is available.
Practice Electrolysis Problem: It takes 2.30 min using a current of 2.00 A to plate out all of the
silver from 0.250 L of a solution containing Ag+. What was the original concentration of Ag+ in
solution?
BY DR. J.J. GONGDEN
DEPT. OF CHEMISTRY, UNIJOS
www.vitalchemist.com