AOT - Electrolysis (Lesson)
Electrolysis
Up to this point, voltaic cells have been the primary focus of the electrochemistry discussion. In these spontaneous processes, a redox reaction is utilized to produce electrical current which can be utilized for some useful purpose. In this lesson, attention will be turned to electrolysis processes where, instead of producing electrical current, an outside source of current is passed through a circuit to force an otherwise non-spontaneous process into occurring. These reactions serve a wide variety of useful purposes ranging from recharging reusable alkaline batteries to electroplating metals. They can also be utilized to produce chemical substances that would otherwise not occur in nature. One example is the industrial production of both sodium metal and chlorine gas.
2 NaCl (l) ⟶ 2 Na (s) + Cl2 (g)
Notice that a voltage source is indicated at the top of this diagram. The reaction of molten sodium chloride decomposing into its constituent elements is not a spontaneous process. In fact, the reverse reaction is highly spontaneous, and therefore, solid sodium metal and gaseous chlorine cannot be found naturally due to their high reactivity. In this process, the outside voltage source forces the reduction of sodium cations into solid sodium which deposits onto the cathode shown on the right. Simultaneously, this voltage forces the oxidation of chloride ions in the solution forming chlorine gas which bubbles out of the solution and is collected elsewhere. Electrochemical cells such as this are referred to as electrolytic cells.
Quantitative Electrolysis
When performing electrolysis, the amount of current that passes through the circuit ultimately determines the quantity of products that will be formed. For example, consider the electroplating (reduction) of aluminum ions
Al+3 (aq) + 3 e- ⟶ Al (s)
In simple stoichiometric terms, 1 mole of aluminum metal can be produced for every 3 moles of electrons that pass through this circuit. Three moles of electrons are required due to the charge of the aluminum ion. Determining the moles of electrons that pass through a circuit relies on knowing the current (amperes) passing through the circuit as well as how long this current was applied.
Q=I×t
This simple relationship allows the calculation of the charge, in Coulombs, that results from a certain current, I, passed through the circuit for a certain period of time, t, expressed in seconds. This charge can then be converted to the number of moles of electrons by using Faraday's constant as a conversion factor.
1 mol e- = 96,485 C
Take, for example, the production of cadmium metal from a CdCl2 solution that has had a current of 0.500 A pass through the solution for 79.0 minutes. What mass of Cd metal can be produced during this process? The reduction half-reaction would be as follows:
Cd+2 (aq) + 2 e- ⟶ Cd (s)
As this half-reaction suggests, 2 moles of electrons are required to reduce cadmium ions into the corresponding solid. First, however, the amount of charge that has passed through this solution must be determined using the relationship above and converting the time into seconds:
Q=I×t
Q=0.500×4,740
Q=2,370 C
From here basic stoichiometry can be used to determine the final answer.
You Try It!
In the following self-assessment activity, complete the calculation. Click on the plus sign to check your answer!
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