Redox Reactions

Oxidation-Reduction Reactions

How do we determine whether a given chemical reaction is an oxidation-reduction reaction? We can do so by keeping track of the oxidation numbers of all the elements involved in the reaction. This procedure tells us which elements (if any) are changing oxidation state. For example, take the following reaction:

Zn (s) + 2 H⁺ (aq) ----> Zn²⁺ (aq) + H₂ (g)

We can see that the Zn (s) oxidation number has changed from 0 to +2, producing Zn²⁺ (aq) and H⁺ (aq) oxidation number has changed fron +1 to 0, producing H₂ (g).

When for an element, its oxidation number increases, we can say that the element was oxidized. If on the other hand, its oxidation number decreased in the course of the reaction, we can say that the substance was reduced. In the above presented reaction, we can conclude that Zn (s) was oxidized and H⁺ (aq) was reduced. It is very important to realize that the substance either oxidized or reduced is always found in the reactants side of the equation. Conversely, since Zn (s) was oxidized, it is causing the reduction of H⁺ (aq) and it is therefore the reducing agent. Likewise, H⁺ (aq) is causing the oxidation of Zn (s), making it the oxidizing agent. This is clear when we break the reaction into half-reactions, one for oxidation and one for reduction.

Oxidation: Zn (s) -----> Zn²⁺ (aq) + 2 e^-

Reduction: 2 H⁺ (aq) + 2e^- ----> H₂ (g)

This allow us to clearly see that an oxidation is basically a loss of electrons and a reduction is the process by which a substance gains electrons. It is important to realize that the driving force for such a reaction is the transfer of electrons between substances. Therefore, without an oxidation there cannot be a reduction and viceversa. Another important fact is that when we add the half-reaction of oxidation with the half-reaction of reduction, we obtain the initial reaction. We must never have extra electrons free when we add both reactions together. The number of electrons lost by a substance in an oxidation must be gained by the substance being reduced.

Consider then the following half-reactions:

Oxidation: Sn²⁺ (aq) ------> Sn⁴⁺(aq) + 2 e^-

Reduction: Fe³⁺ (aq) + e^- ------> Fe²⁺

If we add both reactions together without first balancing the second reaction, we obtain:

Sn²⁺ (aq) + Fe³⁺ (aq) ------> Sn⁴⁺ (aq) + Fe²⁺ (aq) + e^-

This is clearly wrong, since we have a free electron, we must therefore multiply the reduction reaction times two, balancing the numbers of electrons transferred during the reaction. When we do this, we obtain the correct balanced oxidation-reduction reaction:

Sn²⁺(aq) + 2 Fe³⁺ (aq) -------> Sn⁴⁺ (aq) + 2 Fe²⁺ (aq)

What is the importance of redox reactions? Their importance lies in the fact that we can use the transfer of electrons between species to do useful work. This is accomplished by constructing a voltaic cell. A voltaic cell relies in separating the reactants into two chambers and therefore forcing the electrons to travel through electrodes to perform the oxidation and the reduction. While the electrons are traveling from one electrode to another, we can make them do work. A sketch of a voltaic cell is shown in the next page.

Diagram of a Voltaic Cell:

From the sketch we can see that the chamber were the oxidation takes place, is called the anode and that the chamber were the reduction takes place is called the cathode. The electrons leave the anode and enter the cathode. By leaving the anode, Zn is being oxidized and released into the electrolyte solution. To keep the charges balanced, anions from the salt bridge migrate toward the anode. This is the reason that the anode got its name, since it attracts anions. When the electrons enter the cathode, Cu²⁺ in the electrolyte solution are attracted to it, where they are reduced. To keep the charges balanced, cations from the salt bridge migrate toward the cathode. Just like the anode got its name by attracting anions, the cathode got its name because it attracts cations. The reaction taking place is then obtaining by combining the oxidation and reduction reaction.

Zn (s) + Cu²⁺ (aq) ------> Zn²⁺ (aq) + Cu (s).

The reaction will continue until all the Zn in the anode is depleted or all the Cu²⁺ is removed from solution and deposited as Cu (s) in the cathode. This reaction will develop a potential, that can be calculated from the equation:

E^o_cell = E^o_cathode - E^o_anode

Looking at the reduction potential table (in your book), you find that the reduction potential value of Cu²⁺ + 2 e^- ----> Cu is +0.34 V. For the Zn²⁺ + 2 e^- -----> Zn, the value is -0.76 V. You use the values as given in the table. DO NOT CHANGE THE SIGN OF THE E^o VALUE FOR THE ANODE, THE NEGATIVE SIGN IN FRONT OF THIS VALUE IN THE ABOVE PRESENTED EQUATION WILL DO IT FOR YOU. Therefore, for the above presented cell, the E^o_cellvalue is:

E^o_cell = +0.34 - (-0.76) = +1.10 V

Remember that the E^o values mean that the concentrations of ions in solution are 1.0 M, if a gas is present, the pressure is 1.0 atm and the temperature at which the potential was measured was 25 ^o C.

You should practice as much as possible in calculating cell potentials and identifying redox reactions. Use your book as a resource and if any questions arise, please be sure that you see me as soon as possible.