Unit 6: Electrolysis - Calculations & Applications
Applying the principles of electrolysis to electroplating and quantitative calculations.
6.22 Electroplating
Electroplating is a practical application of electrolysis used to coat an object with a thin layer of a metal. This is done for decorative purposes (e.g., silver-plating jewellery) or for protection against corrosion (e.g., zinc-plating steel).
The Setup:
- Cathode (-): The object to be plated.
- Anode (+): A bar of the pure plating metal.
- Electrolyte: An aqueous solution containing ions of the plating metal (e.g., silver nitrate solution for silver plating).
The Process (e.g., Silver-plating a spoon):
- The spoon is made the cathode and a bar of pure silver is the anode.
- At the anode, the silver metal is oxidised, losing electrons and dissolving into the solution as silver ions.
$Ag(s) \rightarrow Ag^+(aq) + e^-$ - At the cathode, silver ions from the solution are attracted to the negatively charged spoon, where they gain electrons and are reduced to a thin, even layer of solid silver metal.
$Ag^+(aq) + e^- \rightarrow Ag(s)$
The overall effect is the transfer of metal from the anode to the cathode, keeping the concentration of the electrolyte constant.
Solved Examples:
- What is the main purpose of electroplating?
Solution: To coat an object with a layer of another metal for decoration or corrosion protection. - In electroplating, is the object to be plated the anode or the cathode?
Solution: The cathode (the negative electrode). - What should the anode be made of when plating an object with copper?
Solution: A pure bar of copper. - What would be a suitable electrolyte for plating an iron nail with zinc?
Solution: A solution containing zinc ions, such as zinc sulphate ($ZnSO_4$) or zinc nitrate ($Zn(NO_3)_2$). - Write the half-equation for the reaction occurring at the cathode when silver-plating a ring.
Solution: $Ag^+(aq) + e^- \rightarrow Ag(s)$. - What happens to the mass of the anode during electroplating?
Solution: It decreases as the metal is oxidised and dissolves into the electrolyte. - Why is it important that the anode is made of the plating metal?
Solution: To replenish the metal ions in the electrolyte as they are deposited onto the cathode, keeping the concentration constant and ensuring a smooth, continuous plating process. - What energy conversion takes place during electroplating?
Solution: Electrical energy is converted to chemical energy. - Is electroplating a spontaneous process?
Solution: No, it is a non-spontaneous process that requires an external power supply, which is the definition of electrolysis. - Chromium plating is used on car bumpers for protection. What is the cathode in this process?
Solution: The car bumper.
6.23 Using Electrolysis to Prove Water Composition
Electrolysis provides a classic experimental method to determine the chemical composition of water. Pure water is a very poor conductor of electricity, so a small amount of an electrolyte, typically dilute sulphuric acid ($H_2SO_4$), is added to allow a current to flow.
The Reactions:
- At the Cathode (-): Hydrogen ions (from the acid and water) are reduced to hydrogen gas.
$2H^+(aq) + 2e^- \rightarrow H_2(g)$ - At the Anode (+): Hydroxide ions (from water) are oxidised to oxygen gas.
$4OH^-(aq) \rightarrow O_2(g) + 2H_2O(l) + 4e^-$
Overall Reaction: $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$
Observation and Conclusion:
When the gases produced at each electrode are collected, the volume of hydrogen gas produced at the cathode is found to be twice the volume of oxygen gas produced at the anode.
According to Avogadro's Law, at the same temperature and pressure, the ratio of volumes of gases is equal to the ratio of their moles. Therefore, the 2:1 volume ratio proves that water is composed of hydrogen and oxygen atoms in a 2:1 ratio, confirming the chemical formula $H_2O$.
Solved Examples:
- Why is an electrolyte like sulphuric acid added to water before electrolysis?
Solution: To provide mobile ions so that the water can conduct electricity. Pure water has too few ions. - What gas is collected at the negative electrode (cathode) during the electrolysis of water?
Solution: Hydrogen gas ($H_2$). - What gas is collected at the positive electrode (anode)?
Solution: Oxygen gas ($O_2$). - What is the observed volume ratio of hydrogen to oxygen produced?
Solution: 2 volumes of hydrogen to 1 volume of oxygen (2:1). - How does this experiment confirm the formula $H_2O$?
Solution: The 2:1 volume ratio of gases corresponds to a 2:1 mole ratio, indicating there are two hydrogen atoms for every one oxygen atom in a water molecule. - Write the overall balanced equation for the electrolysis of water.
Solution: $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$. - How could you test the identity of the gas produced at the cathode?
Solution: Collect the gas and test with a lit splint; it will extinguish with a "squeaky pop," confirming hydrogen. - How could you test the identity of the gas produced at the anode?
Solution: Collect the gas and test with a glowing splint; it will relight, confirming oxygen. - If 10 cm³ of oxygen is collected, what volume of hydrogen is expected?
Solution: 20 cm³. - Which scientific law links the volume of a gas to its number of moles?
Solution: Avogadro's Law.
6.24 Faraday's Constant & Quantitative Calculations
The amount of substance produced during electrolysis is directly proportional to the amount of electricity that passes through the cell. This relationship is quantified by Faraday's constant (F).
Faraday's constant is the total electric charge contained in one mole of electrons.
$F = 96,500$ Coulombs per mole ($C \cdot mol^{-1}$)
The total charge (Q) passed can be calculated from the current (I) and time (t):
$Q = I \times t$ (where Q is in Coulombs, I is in Amperes, and t is in seconds).
Steps for Electrolysis Calculations:
- Calculate the total charge passed: $Q = I \times t$.
- Calculate the moles of electrons passed: Moles of $e^-$ = Q / F.
- Write the relevant half-equation for the process.
- Use the mole ratio from the half-equation to find the moles of substance produced or consumed.
- Convert moles of substance to the desired quantity (e.g., mass, volume of gas).
Solved Examples:
- What is Faraday's constant?
Solution: The charge of one mole of electrons, approximately 96,500 C/mol. - A current of 2.0 A is passed through a solution for 500 seconds. How much charge has passed?
Solution: $Q = I \times t = 2.0 A \times 500 s = 1000 C$. - How many moles of electrons are in 19,300 C of charge?
Solution: Moles of $e^-$ = Q / F = 19,300 C / 96,500 C/mol = 0.2 mol. - What mass of copper is deposited at the cathode when a current of 1.5 A flows through a $CuSO_4$ solution for 30 minutes? (Ar of Cu = 63.5)
Solution:
1. $Q = 1.5 A \times (30 \times 60 s) = 2700 C$.
2. Moles of $e^-$ = 2700 / 96500 = 0.028 mol.
3. Half-equation: $Cu^{2+} + 2e^- \rightarrow Cu$.
4. Moles of Cu = Moles of $e^-$ / 2 = 0.028 / 2 = 0.014 mol.
5. Mass of Cu = 0.014 mol × 63.5 g/mol = 0.89 g. - What volume of hydrogen gas (at STP) is produced by the electrolysis of water using a current of 5.0 A for 1 hour? (Molar volume at STP = 22.4 L/mol)
Solution:
1. $Q = 5.0 A \times 3600 s = 18000 C$.
2. Moles of $e^-$ = 18000 / 96500 = 0.1865 mol.
3. Half-equation: $2H^+ + 2e^- \rightarrow H_2$.
4. Moles of $H_2$ = Moles of $e^-$ / 2 = 0.1865 / 2 = 0.09325 mol.
5. Volume of $H_2$ = 0.09325 mol × 22.4 L/mol = 2.09 L.