Lesson 3.4: Electrolysis and Electroplating PDF

## Lesson 3.4: Electrolysis and Electroplating ### Lesson Summary Electrolysis is a process where electrical energy from an external source is used to drive a nonspontaneous reaction. This is the principle behind the electrolytic cell. Electrolysis has a lot of applications, such as the production...

## Lesson 3.4: Electrolysis and Electroplating ### Lesson Summary Electrolysis is a process where electrical energy from an external source is used to drive a nonspontaneous reaction. This is the principle behind the electrolytic cell. Electrolysis has a lot of applications, such as the production of chemicals, refining metals, extracting metals from its ores, and preventing corrosion. In electrolysis, the amount of products is directly proportional to the amount of current that flows through the electrolytic cell. ### Learning Outcomes 1. Define what is electrolysis and provide an example of its application 2. Discuss the principle of an electrolytic cell 3. Calculate the quantity of charge flowing through the electrolytic cell and the amount of product formed ### Motivation Question Do you know how aluminum metal is produced? How about turning a rusty spoon into a silver one, can you do that? Can you turn a plastic material look like pure gold or silver? If you can, how much material, current, and time do you need to do so? ### Discussion #### Electrolytic Cell and Electrolysis The exact opposite of how a voltaic cell works is the principle of how an electrolytic cell works. An electrolytic cell is an electrochemical cell that uses the process of electrolysis where electrical energy from an external source to drive a nonspontaneous reaction. Let us have an example of a Sn-Cu voltaic cell. From that, let us construct an electrolytic cell. For the voltaic cell, the redox reaction is: - **Sn (s) → Sn2+ (aq) +2é ** [anode; oxidation] - **Cu2+ (aq) + 2e- → Cu (s)** [cathode; reduction] - **Sn (s) + Cu2+ (aq) → Sn2+ (aq) + Cu (s)** $E°cell = 0.48 V$ and $AG° = -93 kJ$ Reversing the cell reaction is then a nonspontaneous reaction, which means that it will not happen on its own as implied by the value of $E°cell$ and $∆G°$. - **Sn2+ (aq) + Cu (s) → Sn (s) + Cu2+ (aq)** $E°cell = - 0.48 V$ and $∆G° = 93 kJ$ But if an external source of electricity with an electric potential greater than 0.48 V such as a battery, we can let this nonspontaneous reaction occur. Thus, the voltaic cell will be converted into an electrolytic cell and reverses the electrode - cathode is now an anode, and an anode is now a cathode. #### Table 1: Comparison of Voltaic and Electrolytic Cell | Cell Type | ∆G | $E°cell$ | Name | Process | Sign | |---|---|---|---|---|---| | Voltaic | < 0 | > 0 | Anode | Oxidation | - | | Voltaic | > 0 | > 0 | Cathode | Reduction | + | | Electrolytic | > 0 | < 0 | Anode | Oxidation | + | | Electrolytic | > 0 | < 0 | Cathode | Reduction | - | #### Application of Electrolysis 1. **Production of Chemicals** - **a) Electrolysis of Molten Sodium Chloride**<br> Molten sodium chloride (NaCl) have Na⁺ and Cl⁻ as ions and it is electrolyzed to form sodium metal and chlorine gas. Downs cell is the version of an electrolytic cell used in industrial large scale electrolysis purposes. In an electrolytic cell, a pair of electrodes are attached to a battery which serves as an electron pump. The battery drives the electrons from the anode to the cathode. The reactions at the electrodes are: - **ANODE (oxidation): 2 CI⁻ (1) → Cl2 (g) + 2e⁻** - **CATHODE (reduction): 2 Na⁺ (1) + 2 e⁻ → 2 Na (1)** - **OVERALL REACTION: 2 Na⁺ (1) + 2 Cl⁻ → Cl2 (g) + 2 Na (1)** This process is a major source of pure sodium and chlorine. The $E°cell$ for the overall reaction is around -4 V which means that it is a nonspontaneous reaction. So to make the reaction proceed, a minimum of 4 V should be supplied by the battery. - **b) Electrolysis of Water** <br> Water in your glass at normal condition will not decompose into hydrogen and oxygen gas because the standard free energy for the reaction is so large (∆ G° = 474.4 kJ/mol). The reaction can be induced by using an electrolytic cell made with pairs of the electrode such as Pt immersed in water. The electrodes will be then connected to a battery and the water is added with a diluted H2SO4 to make a 0.1 M solution. The pure water will not be electrolyzed without the acid since there are not enough ions to carry the electric current. After that, gas bubbles will be observed immediately at each electrode. The overall reaction is given by and take note that no H2SO4 is consumed: - **ANODE (oxidation): 2 H2O (1) → O2 (g) + 4 H⁺ (aq) + 4e¯¯** - **CATHODE (reduction): 4 [H⁺ (aq) + e¯¯ → H2(g)]** - **OVERALL REACTION: 2 H2O (1) → 2H2 (g) + O2 (g)** - **c) Electrolysis of Aqueous Sodium Chloride Solution** <br> A bit tricky compared to the previous electrolysis since several species are present that can be oxidized and reduced. In the anode, the following oxidation might occur: - **(1) 2 Cl⁻ (aq) → Cl2 + 2e¯ ** - **(2) 2 H2O (1) → O2 (g) +4 H⁺(aq) + 4e¯** The following reduction reaction might occur: - **(3) 2H⁺ (aq) + 2e → H2 (g)** - **(4) 2H2O (1) + 2e → H2(g) + 2 OH¯(aq)** - **(5) Na⁺ (aq) + e¯¯ → Na (s)** The standard reduction potentials of reaction 1 and 2 are not very different, but the values do suggest that H2O is preferably to be oxidized. But, the gas liberated is not O2 instead it is Cl2 gas. Sometimes the voltage required for a reaction is higher than electrodepotential being indicated resulting in overvoltage. The overvoltage is the difference between the electrode potential and the actual voltage required to cause electrolysis. Since reaction 5 has a very negative standard reduction potential then it is ruled out. Supposedly at standard condition reaction, 3 is preferred overreaction 4; but at pH 7 (that is the pH of NaCl solution) the concentration of H⁺ is very low (about 1 x 10⁻⁷ M) making the reaction 4 a preferred reaction for the cathode. The overall reaction for the electrolysis of aqueous sodium chloride is: - **ANODE (oxidation): 2 Cl⁻ (aq) → Cl2 + 2é** - **CATHODE (reduction): 2H2O (1) + 2e⁻ → H2 (g) + 2 OH¯(aq)** - **OVERALL: 2 Cl⁻(aq) + 2H2O (1) → Cl2 + H2 (g) + 2 OH¯(aq)** We can see that the concentration of Cl⁻ decreases as it is being converted into Cl2 gas while the OH concentration increases. Aside from the formation of H2 and Cl2, we can also have a useful by-product which is the NaOH. The NaOH is obtained by evaporating the aqueous solution after the electrolysis. 2. **Extraction of Metals** - There are two ways of extracting metal from its ore regarding its physical state: - ore in a solid-state is treated with a strong acid to obtain a salt solution and then electrolyzed to liberate the metal; - ore in a molten state and electrolyzed in a furnace. - Let us have an example of the most abundant metal in earth's crust and how it is extracted from its ores: - **Extraction of Aluminum** - The principal ore of aluminum is bauxite which is primarily composed of alumina (aluminum oxide compounds), silica, iron oxides, and titanium oxide. Bauxite is then purified to extract the aluminum oxide compounds. The aluminum oxide compounds have a very high melting point (over 2000 °C) and don't dissolve in water. An easy and cheaper way of dissolving, the alumina is added with molten cryolite an aluminum-containing compound with a much lower melting point than aluminum oxide. - The mixture of aluminum oxide compounds and cryolite is then electrolyzed in a furnace. During the electrolysis, the positively charge aluminum ions gain electrons (reduced) from the cathode resulting in the formation of molten aluminum. The negative charge oxide ions lose electrons (oxidized) forming oxygen molecules at the anode. Taking the Al2O3 as one of the aluminum compounds, the following is one of the redox reaction that occurs: - **CATHODE (reduction): 2Al³⁺ (1) + 6e⁻ → 2AI (1)** - **ANODE (oxidation): 3O²⁻ (1) → 3O2 (g) + 6e⁻** - **OVERALL REACTION: 2AI³⁺ (1) + 3O²⁻ (1)→2Al (1) + 3O2 (g)** 3. **Refining of Metals** - Electrorefining is a process in which electrolysis is used to increase the purity of metal extracted from the ore. The electrorefining of copper is the world's largest industry for the refining of metals. In copper's refining, the impure Cu from the ore is oxidized in the anode while being submerged in an electrolyte solution. The Cu²⁺ ions formed from the oxidation of impure Cu will migrate to the cathode and the reduced to Cu. The Cu will deposit in the cathode and its purity is higher compared to those in the anode. The overall reaction for this is: - **ANODE (oxidation): Cu (s) → Cu²⁺ (aq) + 2e⁻** - **CATHODE (reduction): Cu²⁺(aq) + 2e⁻ → Cu (s)** - **OVERALL REACTION: Cu (s) + Cu²⁺(aq) → Cu²⁺(aq) + Cu (s)** - Take note that the concentration of Cu²⁺ ions from the electrolyte solution doesn't change. For every Cu from the impure will give an equivalent Cu²⁺ ion and thus forming the same amount of Cu deposits but with higher purity. 4. **Electroplating** - Electroplating is the process where a metal is being plated or coated by other metal and commonly for aesthetic purposes and prevention of corrosion. Creating an electrolytic cell where the anode is the metal that is to be used for plating while the cathode is the material (metal or even plastic) that is to be plated. The electrolyte solution contains an ion of that metal to be used for plating. - Looking at the figure, the iron spoon plated with Ag. The spoon serves as the cathode while the Ag metal is the anode. The Ag in the anode will be oxidized to Ag⁺ ions. The formed Ag⁺ ions will then migrate to the spoon in which it will be reduced to Ag. The Ag will deposit and coat the spoon giving it a "silver" finished. #### Stoichiometry of Electrolysis - So far if you've observed, th current flowing through the electrolytic cell yields a product. The stoichiometry of electrolysis is the relationship between the amount of current and product and this is discovered by Michael Faradays. Faradays' law of electrolysis states that the quantity of substance produces at each electrode is directly proportional to the amount of current flowing through the cell. - Using the balanced half reaction equation, we will know how much product is formed with a given amount of current and how much current is needed to produce with the given amount of products. Let us have a sample problem: - **SAMPLE PROBLEM:** An engineer is tasked to plate a faucet with 0.86 g of chromium from an electrolytic bath containing aqueous Cr2(SO4)3. If 12.5 minutes is allowed for the plating, how much current is needed? - **ANSWER:** - **a) Write a balanced half-reaction:** - **Cr³⁺ (aq) + 3e⁻ → Cr (s)** - **b) Calculate the amount (mol) of e⁻ transferred for the mass of Cr needed:** - **mol of e⁻ transferred = 0.86 g × (1 mol Cr/52 g Cr) × (3 mol e⁻/1 mol Cr)** = **0.050 mol e⁻** - **c) Calculate the charge using the Faradays constant:** - **Charge (C) = mol of e⁻ × Faradays constant** - **Charge (C) = 0.050 mol e⁻ x (9.65 x10⁴C/1 mol e⁻)** = **4.8 x 10³ C** - ** d) Calculate the current:** - **Current (A) = (charge (C))/(time(s))** - **Current (A) = (4.8 x 10³ C)/(12.5 min x 60 s)** = **6.4 A** - Below is a summary diagram that can help you in solving problems about stoichiometry of electrolysis: | MASS (g) of substance oxidized or reduced| AMOUNT (mol) of substance oxidized or reduced| AMOUNT (mol) of electrons transferred | balanced half-reaction | Faradays constant | CHARGE (C) | time (s) | CURRENT (A) | |---|---|---|---|---|---|---|---| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | **Figure 38. Stoichiometry of Electrolysis**