Answer :
Certainly! Let's break down the steps to address the questions provided:
### Question 3:
Calculate the current required to produce 20 dm³ of chlorine gas by electrolysis at STP for one hour.
1. Molar Volume of Gas at STP:
The molar volume of any ideal gas at standard temperature and pressure (STP) is 22.4 liters per mole.
2. Calculate the moles of chlorine gas:
[tex]\[ \text{Moles of chlorine gas} = \frac{\text{Volume of chlorine gas}}{\text{Molar volume at STP}} = \frac{20 \text{ liters}}{22.4 \text{ liters/mol}} \approx 0.89286 \text{ moles} \][/tex]
3. Faraday Constant:
The Faraday constant is the amount of electric charge per mole of electrons, approximately 96500 Coulombs per mole (C/mol).
4. Chloride ion charge:
The production of one molecule of chlorine gas (Cl₂) requires 2 moles of Cl⁻ ions (each needing 1 electron to be removed). Therefore the charge involved is:
[tex]\[ \text{Charge required} = \text{moles of Cl₂} \times \text{number of electrons per mole} \times \text{Faraday constant} \][/tex]
[tex]\[ = 0.89286 \text{ moles} \times 2 \text{ moles of electrons} \times 96500 \text{ C/mol} \approx 172321.43 \text{ C} \][/tex]
5. Current Required:
Current (I) is the rate of flow of charge (Q) over time (t).
[tex]\[ I = \frac{Q}{t} \][/tex]
Here, the time [tex]\( t = 1 \text{ hour} = 3600 \text{ seconds} \)[/tex].
[tex]\[ I = \frac{172321.43 \text{ C}}{3600 \text{ s}} \approx 47.87 \text{ A} \][/tex]
Therefore, the current required is approximately 47.87 Amperes.
### Question 4:
An electric current was passed in series through a solution of calcium chloride and copper (II) sulphate using carbon electrodes. If 2.5 liters of chlorine gas measured at STP were produced:
#### Part (a):
What volume of oxygen would also be produced?
1. Moles of chlorine gas produced:
[tex]\[ \text{Moles of chlorine gas} = \frac{2.5 \text{ liters}}{22.4 \text{ liters/mol}} \approx 0.11161 \text{ moles} \][/tex]
2. Ratio of chlorine to oxygen gas:
Chlorine and oxygen gases produced during electrolysis have a specific stoichiometric relation. For example, during electrolysis of water and chloride ions at the anode:
[tex]\(2Cl^- \rightarrow Cl_2 + 2e^-\)[/tex] and
[tex]\(4OH^- \rightarrow O_2 + 2H_2O + 4e^-\)[/tex]
From the stoichiometry of the reactions, the moles of oxygen gas produced when we have certain moles of chlorine gas will be half:
[tex]\[ \text{Moles of oxygen} = \frac{\text{Moles of chlorine}}{2} = \frac{0.11161 \text{ moles}}{2} \approx 0.0558 \text{ moles} \][/tex]
3. Volume of oxygen gas produced at STP:
[tex]\[ \text{Volume of oxygen} = \text{Moles of oxygen} \times \text{Molar volume at STP} \][/tex]
[tex]\[ = 0.0558 \text{ moles} \times 22.4 \text{ liters/mol} = 1.25 \text{ liters} \][/tex]
Therefore, the volume of oxygen produced is 1.25 liters.
#### Part (b):
What was the mass of copper produced?
1. Electrochemical equivalence:
From Faraday's laws of electrolysis, the number of moles of a substance produced is proportional to the quantity of electricity (charge). Copper(II) sulfate produces copper by:
[tex]\[ Cu^{2+} + 2e^- \rightarrow Cu \][/tex]
2. Moles of copper produced:
The number of moles of copper is half the moles of chlorine, given that the same quantity of charge is passed through:
[tex]\[ \text{Moles of copper} = \frac{\text{Moles of chlorine}}{2} = \frac{0.11161 \text{ moles}}{2} \approx 0.0558 \text{ moles} \][/tex]
3. Mass of copper produced:
The atomic mass of copper (Cu) is approximately 63.55 grams/mole.
[tex]\[ \text{Mass of copper} = \text{Moles of copper} \times \text{Atomic mass of copper} \][/tex]
[tex]\[ = 0.0558 \text{ moles} \times 63.55 \text{ g/mol} \approx 3.55 \text{ grams} \][/tex]
Therefore, the mass of copper produced is approximately 3.55 grams.
Thus, we've thoroughly answered each part of the question step-by-step.
### Question 3:
Calculate the current required to produce 20 dm³ of chlorine gas by electrolysis at STP for one hour.
1. Molar Volume of Gas at STP:
The molar volume of any ideal gas at standard temperature and pressure (STP) is 22.4 liters per mole.
2. Calculate the moles of chlorine gas:
[tex]\[ \text{Moles of chlorine gas} = \frac{\text{Volume of chlorine gas}}{\text{Molar volume at STP}} = \frac{20 \text{ liters}}{22.4 \text{ liters/mol}} \approx 0.89286 \text{ moles} \][/tex]
3. Faraday Constant:
The Faraday constant is the amount of electric charge per mole of electrons, approximately 96500 Coulombs per mole (C/mol).
4. Chloride ion charge:
The production of one molecule of chlorine gas (Cl₂) requires 2 moles of Cl⁻ ions (each needing 1 electron to be removed). Therefore the charge involved is:
[tex]\[ \text{Charge required} = \text{moles of Cl₂} \times \text{number of electrons per mole} \times \text{Faraday constant} \][/tex]
[tex]\[ = 0.89286 \text{ moles} \times 2 \text{ moles of electrons} \times 96500 \text{ C/mol} \approx 172321.43 \text{ C} \][/tex]
5. Current Required:
Current (I) is the rate of flow of charge (Q) over time (t).
[tex]\[ I = \frac{Q}{t} \][/tex]
Here, the time [tex]\( t = 1 \text{ hour} = 3600 \text{ seconds} \)[/tex].
[tex]\[ I = \frac{172321.43 \text{ C}}{3600 \text{ s}} \approx 47.87 \text{ A} \][/tex]
Therefore, the current required is approximately 47.87 Amperes.
### Question 4:
An electric current was passed in series through a solution of calcium chloride and copper (II) sulphate using carbon electrodes. If 2.5 liters of chlorine gas measured at STP were produced:
#### Part (a):
What volume of oxygen would also be produced?
1. Moles of chlorine gas produced:
[tex]\[ \text{Moles of chlorine gas} = \frac{2.5 \text{ liters}}{22.4 \text{ liters/mol}} \approx 0.11161 \text{ moles} \][/tex]
2. Ratio of chlorine to oxygen gas:
Chlorine and oxygen gases produced during electrolysis have a specific stoichiometric relation. For example, during electrolysis of water and chloride ions at the anode:
[tex]\(2Cl^- \rightarrow Cl_2 + 2e^-\)[/tex] and
[tex]\(4OH^- \rightarrow O_2 + 2H_2O + 4e^-\)[/tex]
From the stoichiometry of the reactions, the moles of oxygen gas produced when we have certain moles of chlorine gas will be half:
[tex]\[ \text{Moles of oxygen} = \frac{\text{Moles of chlorine}}{2} = \frac{0.11161 \text{ moles}}{2} \approx 0.0558 \text{ moles} \][/tex]
3. Volume of oxygen gas produced at STP:
[tex]\[ \text{Volume of oxygen} = \text{Moles of oxygen} \times \text{Molar volume at STP} \][/tex]
[tex]\[ = 0.0558 \text{ moles} \times 22.4 \text{ liters/mol} = 1.25 \text{ liters} \][/tex]
Therefore, the volume of oxygen produced is 1.25 liters.
#### Part (b):
What was the mass of copper produced?
1. Electrochemical equivalence:
From Faraday's laws of electrolysis, the number of moles of a substance produced is proportional to the quantity of electricity (charge). Copper(II) sulfate produces copper by:
[tex]\[ Cu^{2+} + 2e^- \rightarrow Cu \][/tex]
2. Moles of copper produced:
The number of moles of copper is half the moles of chlorine, given that the same quantity of charge is passed through:
[tex]\[ \text{Moles of copper} = \frac{\text{Moles of chlorine}}{2} = \frac{0.11161 \text{ moles}}{2} \approx 0.0558 \text{ moles} \][/tex]
3. Mass of copper produced:
The atomic mass of copper (Cu) is approximately 63.55 grams/mole.
[tex]\[ \text{Mass of copper} = \text{Moles of copper} \times \text{Atomic mass of copper} \][/tex]
[tex]\[ = 0.0558 \text{ moles} \times 63.55 \text{ g/mol} \approx 3.55 \text{ grams} \][/tex]
Therefore, the mass of copper produced is approximately 3.55 grams.
Thus, we've thoroughly answered each part of the question step-by-step.
