Chemistry 30 Redox Reactions Unit: Electrolysis The Electrolytic Process

• One job in the chemical industry is to make elements out of ionic solids. The most abundant ionic solid on the earth is sodium chloride (NaCl) which is made up of Na+ ions and Cl- ions. To get an idea of how this is done, let’s examine the Eo values for the production of Na and Cl2.

o Clearly Na+ must gain electrons to be reduced and Cl- must lose electrons and be oxidized. The half-reactions are:

Na+ + e- à Na E = -2.71 V 2 Cl- à Cl2 + 2 e- E = -1.36 V

• When we balance the 2 half-reactions and check our E values we get the following: 2 Na+ + 2 Cl- à 2 Na + Cl2 E = -4.07 V • The negative value of E tells us that the reaction is not spontaneous and favours the reactants. So, how do we

produce the elements then? We can see that there is 4.07 V that opposes the reaction. So if we add 4.07 V or greater to the system the reaction will have enough energy to occur. This is the essence of electrolysis.

• ELECTROLYSIS: the practice of applying an external electrical potential in order to bring about a non-spontaneous

redox reaction.

• In order for electrolysis to occur, we use an electrolytic cell. o This is a container holding the substance to be electrolyzed. o Electrodes, connected to the voltage source, are immersed in the substance. o Since there must be a complete circuit, the substance being electrolyzed must be in a state which enables a

current to flow. (usually a liquid or an aqueous solution) Types of Electrolytic cells

• There are three different types of electrolytic cells, simply called: Type 1, Type 2, and Type 3. Type 1 Cell

• In a Type 1 cell, the substance being electrolyzed is in LIQUID state. An example of this is if we electrolyzed pure barium bromide in its liquid state.

• Note: the accepted way of representing a direct current is: The negative terminal is the shorter line; the positive

terminal is the longer line. o Electrons leave from the negative end of a battery

• Note: The INERT ELECTRODES means that the anode and cathode are conductors that do not react with anything in the cell. Platinum works for this, but it is expensive. Carbon (graphite) is a good choice.

• At the Cathode: (electrons gained) Ba+2 (l) + 2 e- à Ba (s) E = -2.90 V

• At the Anode: (electrons lost) 2 Br- à Br (l) + 2 e- E = -1.06 V • This means the minimum voltage needed for this cell to operate is: -2.90 + (-1.06) = -3.96 V

Type 2 Cell

• If a substance is dissolved in water to become an aqueous solution, this is a Type 2 cell. • This is usually preferable over a type 1 cell because high temperatures can be avoided as compared to using a

molten salt. • In a type 2 cell, it must be kept in mind that at each electrode there are TWO possible reactions that can occur. So at

the anode there are two possible reactions, AND at the cathode there are two possible reactions. The reason why is because, aside from the substance that is dissolved, the water may also be oxidized or reduced.

• Example: What are the two possible half reactions at each electrode, and which of these two half-reactions will

actually occur when aqueous calcium iodide is electrolyzed using inert electrodes? o In this cell, the CaI2 in solution means that Ca+2 and I- are present. As well, H20 is present. o If we look at the ANODE (where oxidation occurs): Here I- may be oxidized or H20 may be oxidized. The ½

reactions are: 2 I- à I2+ 2e- E = -0.53 V H20 à ½ O2 + 2H+ (10-7M) + 2e- E = -0.82 V

§ When we compare these two, we see the iodine ion has a greater potential so that reaction will occur at the anode.

o If we look at the CATHODE (where reduction occurs): Ca+2 may be reduced or H20 may be reduced. The ½ reactions are:

Ca+2 + 2e- à Ca (s) E = -2.76 V 2 H20 + 2e- à H2 (g) + 2OH- (10-7M) E = -0.41 V

§ When we compare these two, we see the water has a greater potential so that reaction will occur at the cathode.

• As this cell operates CaI2 is consumed, while I2, H2 and Ca(OH) 2 are produced. • The minimum amount of energy needed for this reaction to happen is: -0.53 V + (-0.41 V) = -0.94 V

Type 3 Cell

• A type 3 cell is a type of cell used in electroplating. • Electroplating is the plating one metal on to another (applying a thin coat over top of another coat). This is a form of

electrolysis. • All electroplating cells have the following features:

o An external low-voltage source is used. o The item to be plated is the cathode. o The electrolyte contains in solution the ion that will plate out o The anode is a piece of the metal that is plating out onto the cathode. o The anode ½ reaction is the reverse of the cathode ½ reaction.

• Electroplating is not a simple as this. Some experimenting with the voltage, current, the electrolyte components, and the preparation of the cathode is frequently necessary to obtain a smooth even coat of the plated metal.

• Example: Plating tin onto a piece of nickel.

• To do this a direct current is used (ex. battery) and the nickel is connected to the negative terminal of the battery. This

will make the nickel the cathode. • On the positive terminal of the battery, the pure tin is connected. We want the tin in solution to gain electrons and

plate onto the nickel. We find on the Table of Standard Reduction Potentials the half-reaction showing the reduction of tin: Sn+2 (aq) + 2 e- à Sn (s)

• This means that an electrolyte containing Sn+2 is needed. We use Sn(NO3) 2 since we know it will be soluble. (Remember that NO3- is soluble with all cations.)

• So at the cathode we get: Sn+2 + 2e- à Sn (s) • And at the anode we get: Sn (s) à Sn+2 + 2e-

Industrial Application of Electrolysis – Electrorefining Metals

• These electrolysis processes are used in the processing and refining of metals and metal ores. The process of refining lead, zinc, and aluminum are extensions of the electroplating cell.

Corrosion

• Corrosion is an example of a reaction that occurs spontaneously because of a positive E. • Corrosion is the oxidation of a metal – most often we associate this with rusting. • Rusting most often occurs with iron. Here’s how it works:

o If iron is being attacked by an oxidizing agent, then it will lose electrons: Fe (s) à Fe+2 + 2e- E = 0.41 V o The E for this half-reaction is 0.41 V. If oxygen from the air is acting as the oxidizing agent (in the presence of

moisture) the half-reaction is: ½ O2 (g) + 2 H+ (10-7M) + 2 e- à H20 E = 0.82 V o Since 2 moles of electrons have been lost by Fe, and 2 moles of electrons have been gained by oxygen, we

can add them together to give: Fe (s) + ½ O2 (g) + 2 H+ à Fe+2 + H20 E = 1.23 V o The positive E tells us the reaction is spontaneous. Iron will certainly rust under these conditions. Eventually

compounds like FeO, Fe2O3, Fe(OH)2, and Fe(OH)3 form to give the crumbly reddish-brown substance we call rust.

Preventing Corrosion

• There are two basic methods to try to prevent corrosion: o Altering the conditions that are necessary for corrosion o Preventing the oxidizing agent from coming into contact with the metal

Cathodic Protection • Essentially the method involves placing a more easily-oxidized metal in contact with the metal to be protected. For

example, oil and gas pipeline are made of steel (which is mostly iron). Commonly, to protect these pipelines from being oxidized, blocks of magnesium are placed in contact with the pipeline every few hundred meters.

• This works because the E for the oxidation of Mg is greater than that for Fe (i.e. Mg is more easily oxidized than Fe). Mg à Mg+2 + 2e- E = 2.37 V

Fe (s) à Fe+2 + 2e- E = 0.41 V • When oxygen, or any other oxidizing agent comes in contact with the pipeline, the Mg releases electrons to the

surface of the steel where the oxidizing agent picks them up. In other words, a little electrochemical cell is set up with Mg acting as the anode, and Fe as the cathode. This is why the method is called cathodic protection. The result: the magnesium corrodes, not the iron.

Coating with Another Metal

• Many times, to prevent corrosion of steel or iron, they are coated with another metal. The two most used metals for coating are tin and zinc.

Tin-coating • Cans used for food are coated in tin. If we look at the Table of Standard Reduction Potentials we see that Sn is less

easily oxidized then iron. So it will be more resistant than Fe to oxidation. But tin also (like many other metals – Ni, Al, Cr, Zn) when it is exposed to oxygen, it forms a thin layer of SnO on its surface which is resistant to further oxidation. This method is effective by has a drawback – if the tin oxide layer is cracked, then the oxidation of the iron is accelerated because Fe is more easily oxidized then Sn.

Zinc-coating (also called galvanizing) • Involves dipping the iron into molten zinc. This is effective in two ways. Zinc loses electron even more readily than Fe,

but when this happens the Zn forms a tough layer of oxide that is very effective in preventing further corrosion. Secondly, even if the zinc coating is broken, the zinc will provide cathodic protection to the iron (because it grabs electrons better than Fe).

The Breathalyzer

• The breathalyzer makes use of an oxidation-reduction and an electronic analysis to give a precise measure of the breath alcohol content (BAC) of an individual.

• The suspected driver blows into a balloon. This sample of air is then put into a solution of acidified potassium dichromate solution. The dichromate ion is a strong oxidizing agent which is orange in color. If alcohol is present in the breath sample, the dichromate oxides the alcohol. As a result dichromate is reduced and the blue-green Cr+3 is

produced. The amount of Cr+3 produced can be determined electronically by a device called a spectrophotometer. This measures how much light of the specific frequency of the blue-green color is absorbed by the solution. The amount of light absorbed is directly proportional to the [Cr+3] which in turn is determined by the amount of alcohol that was in the breath sample.