Electroylsis of water and aqueous solutions

Up to now, we have considered galvanic cell run in the ``forward'' direction, i.e., such that they produce useful electricity. When operated in a ``reverse'' direction, a process called electrolysis, work must be input to drive a nonspontaneous reaction, and the products will have a higher free energy than the reactants. Such a process could be used, for example, to recharge a battery.

We will consider the electrolysis of water and aqueous solutions as simple examples. Consider, first, the electrolysis of water. The reaction corresponding to electrolysis of water is

This can be produced in an electrolytic cell for which the half-cell reactions, when inert electrodes, such as Pt are used, are:

What is the overall cell potential at 25 .

To calculate this, we use

The half-cell potential of the cathode is given by the Nernst equation:

where

Note that the half-cell potential is not the standard state value because the concentration of in pure water is 10 M rather than 1.0 M. Thus, at 25 , this becomes, assuming atm,

In order to calculate the half-cell potential of the anode, we first write it as a reduction reaction:

so that

From the data in Appendix E, the standard state half-cell potential is V, so by the Nernst equation:

assuming that atm. Thus, the overall cell potential is

The fact that it is negative means that , consistent with the nonspontaneous nature of the reaction.

The value 1.229 V is called the decomposition potential of water. If a potential source capable of delivering 1.229 volts is placed over such a cell, water will electrolyse.

Now consider the electrolysis of an aqueous solution. Suppose we have a 0.10 M NaCl solution. Two possible sets of half reactions can occur. At the cathode, the possible half-reactions are:

and at the anode:

Which of these reactions will occur if a voltage of 1.229 V is placed across the cell?

We already know the decomposition potential for pure water. We need to know the cell potential for the sodium and chloride half-cell reactions. These are given by the Nernst equation:

Note that this is more positive than the anode potential for pure water electrolysis, 0.815 V. Thus, Cl (g) has a greater tendency to be reduced than O , both of which occur spontaneously. But since oxidation occurs at the anode, Cl has a lesser tendency to be oxidized than . Thus, the anode product will be O (g). Even if the applied potential is increased above the decomposition potential 1.229 of water, the only thing that will happen is that water electrolysis will dominate.

However, is a 0.10 M NaI solution is used instead, then the anode half-cell potential for the reaction

can be computed using the Nernst equation is done above, and we find that

which is less positive than the anode reaction for water electrolysis. Thus, this reaction will dominate at the anode. At the cathode, the hydronium reaction still dominates, so that the overall cell potential in this case is

Although it is negative, indicating the nonspontaneous nature of the reaction, it is less than the decomposition potential of pure water. Thus, if an external potential larger than 1.008 is applied across the cell, H (g) and I (g) will be produced from the cathode and anode reactions, respectively. As [I ] decreases, O (g) will start to be produced, and the external potential will need to exceed 1.229 in order to maintain the electrolysis.

To summarize the results of electrolysis for neutral aqueous solutions:

1.

A species can be reduced only if its reduction potential is greater than -0.414 V.

2.

A species can be oxidized only its reduction potential is smaller tan 0.815 V.

In solutions with pH different from 7, i.e., acidic or basic solutions, the results will be modified according to the concentration.