Electrolytic Cells

The concept of reversing the direction of the spontaneous reaction in a galvanic cell through the input of electricity is at the heart of the idea of electrolysis. See for a comparison of galvanic and electrolytic cells. If you would like to review your knowledge of galvanic cells (which I strongly suggest) before learning about electrolytic cells, click here.

Electrolytic cells, like galvanic cells, are composed of two half-cells--one is a reduction half-cell, the other is an oxidation half-cell. Though the direction of electron flow in electrolytic cells may be reversed from the direction of spontaneous electron flow in galvanic cells, the definition of both cathode and anode remain the same--reduction takes place at the cathode and oxidation occurs at the anode. When comparing a galvanic cell to its electrolytic counterpart, as is done in , occurs on the right-hand half-cell. Because the directions of both half-reactions have been reversed, the sign, but not the magnitude, of the cell potential has been reversed. Note that copper is spontaneously plated onto the copper cathode in the galvanic cell whereas it requires a voltage greater than 0.78 V from the battery to plate iron on its cathode in the electrolytic cell.

You should be asking yourself at this point how it is possible to make a non-spontaneous reaction proceed. The answer is that the electrolytic cell reaction is not the only one occurring in the system-the battery is a spontaneous redox reaction. By Hess's Law, we can sum the DG of the battery and the electrolytic cell to arrive at the DG for the overall process. As long as that DG for the overall reaction is negative, the system of the battery and the electrolytic cell will continue to function. The condition for DG being negative for the system (you should prove this for yourself) is that E_battery is greater than - E_cell.

During the early history of the earth, hydrogen and oxygen gasses spontaneously reacted to form the water in the oceans, lakes, and rivers we have today. That spontaneous direction of reaction can be used to create water and electricity in a galvanic cell (as it does on the space shuttle). However, by using an electrolytic cell composed of water, two electrodes and an external source emf one can reverse the direction of the process and create hydrogen and oxygen from water and electricity. shows a setup for the electrolysis of water.

The reaction at the anode is the oxidation of water to O₂ and acid while the cathode reduces water into H₂ and hydroxide ion. That reaction has a potential of -2.06 V at standard conditions. However, this process is usually performed with [H⁺] = 10^-7 M and [OH^-] = 10^{- 7} M, the concentrations of hydronium and hydroxide in pure water. Applying the Nernst Equation to calculate the potentials of each half-reaction, we find that the potential for the electrolysis of pure water is -1.23 V. To make the electrolysis of water occur, one must apply an external potential (usually from a battery of some sort) of greater than or equal to 1.23 V. In practice, however, it is necessary to use a slightly larger voltage to get the electrolysis to occur on a reasonable time scale.

Pure water is impractical to use in this process because it is an electrical insulator. That problem is circumvented by the addition of a minor amount of soluble salts that turn the water into a good conductor (as noted in ). Such salts have subtle effects on the electrolytic potential of water due to their ability to change the pH of water. Such effects from the salts are generally so small that they are usually ignored.

Electroplating allows the production of metal coatings of such desirable commodities as silver and gold. People make fortunes gold or silver plating junk metal (usually aluminum) because they can sell gold plated necklaces for a comparable price to the real thing (or even pass them off as being solid gold). That's how electrochemistry can be used to rip you off! In our discussion of electroplating, we will discuss how you can set up a cell for electroplating, how you can calculate the amount of precious material consumed, and various other calculations you can perform with electroplating. In terms of the variety of electrochemistry problems possible to ask, this section, perhaps rivaled by Thermodynamics, is the richest.

The setup for electroplating is quite simple and the entire cell is usually conducted in a single solution as shown in .

The gold from the anode is oxidized and dissolves in solution as Au³⁺. The electrons arriving at the aluminum glasses frame cathode reduce the Au³⁺ in solution to Au (s) on the surface of the frame cathode. We can calculate how long we should have our glasses frame in solution if we want a certain amount of gold to be plated.

Let's assume it takes 1.0 g of gold to provide an adequate coating for our glasses and also assume that we are using an emf sufficient to produce 10 amperes (A) of current (1 A = 1 coulomb per second). how long it will take to plate that 1.0 g of gold.

As you can see from the , such a problem only involves the use of unit cancellation. To calculate the time needed to deposit a certain amount of material, you need to start with the amount, converted to moles. Then, multiply by the number of electrons consumed in the reduction (in this case 3). Using the definition of a faraday, 96500 C per mole of electrons, you can convert between moles and charge. Finally, by using the definition of an ampere, 1 C per second, you can convert the amount of charge required to deposit the material into a time in seconds. There are various ways of phrasing this same problem such as "how much gold is deposited in 146 seconds at 10 A" or "what current is required to deposit 1.0 g of gold in 146 seconds." Don't be fooled by those permutations of the same problem, they all boil down to simple unit cancellation which you have been doing since you learned how to do stoichiometry. Also note that in these problems, you do not need to know the cell potential. Students often try, incorrectly, to use the cell potential somewhere in that calculation. Furthermore, you need only know the number of electrons transferred--you could solve the same problem without even knowing what material was being plated (as long as you know its molar mass).