SR1 SCB and SR1 SCD
Tips for the Test Scheduled in Week 4
1. What is 'electrochemical series'? [2]
2. Describe how a Galvanic cell can be constructed using magnesium and zinc as electrodes. The electrolytes must be separated in two different tanks. [7]
A magnesium metal (Mg) is immersed in a tank containing its ions such as magnesium nitrate (Mg(NO3)2) [1]. On the other hand, a zinc metal is also immersed in another tank containing its ions such as zinc sulphate (ZnSO4) [1]. Both the metals are connected using a wire [1/2]. A voltmeter is also used to study the potential difference [1/2]. Then, a salt bridge containing concentrated potassium chloride (KCl) is used by putting its ends in both the electrolytes in order to allow for the movement of ions and to complete the electrical circuit [1]. As magnesium is more electropositive than zinc, it will release electrons and form magnesium ions at the negative terminal [1]. The half equation is
Mg --> Mg2+ + 2e-
The electrons are passed from the negative terminal to the zinc metal (the positive terminal) through the wire and the zinc ions in the electrolyte are discharged to form solid zinc [1]. The half equation is
Zn2+ + 2e- --> Zn
The voltmeter pointer deflects to show the magnitude of the voltage at 1.62V [1].
Diagram [2]
Maximum is 7marks
Note 1:
1. Zn and Cu, negative terminal is Zn, electrolytes are zinc sulphate and copper (II) sulphate, 1.10V
2. Zn and Fe, nagative terminal is Zn, electrolytes are zinc sulphate and iron (II) nitrate, 0.32V
3. Mg and Cu, negative terminal is Mg, electrolytes are magnesium nitrate and copper (II) sulphate, 2.72V.
Note 2:
The given electrolytes are theoretically valid and may not be applicable to practical setups.
Note 3:
If the positive terminal is immersed in its ions, make sure the cations are concentrated if 'solution' is mentioned. This is only for those cations (such as Zn2+) that are higher than H+ in the ES, so that Zn2+ is selected to be discharged. If you don't mention whether it is a solution or so, just its ions, then no need to care about the H+ factor because we assume the electrolyte has little or none H+. This is for advanced learning.
For cations lower than H+ in the ES such as Cu2+ in the terminal electrode, dilute solution can be used because it is less electropositive than H+ and Cu2+ will definitely be selected. This is the typical example in the textbook.
3. How can we know whether a chemical reaction between a metal and a salt in a solution will take place? [4]
This can be determined from the electrochemical series [1]. For a chemical reaction to take place, the metal must be more electropositive/reactive than the cation in the salt [1]. For example, in the equation Mg + ZnSO4 --> MgSO4 + Zn, this reaction will take place because Mg is higher than Zn in the electrochemical series and hence more electropositive. So Mg can displace Zn2+ from its salt [1]. In contrast, H2 + 2NaCl --> 2HCl + 2Na will not take place because H is lower than Na in the electrochemical series and less electropositive than Na. So it cannot displace Na+ from the salt [1].
4. In an electrolyte containing magnesium sulphate solution, select and explain the ion that will be discharged at the cathode when the electrical circuit passes through. [2]
H+ will be selectively discharged at the cathode [1] because H+ is lower than Mg2+ in the electrochemical series and hence less electropositive [1], so it H+ ions will accept electrons to form hydrogen gas.
Reminder: Silver is Ag, and its ion is Ag+; gold is Au, and its ion is Au+, both are at the bottom in the ES.
SR1 SCB and SR1 SCD
Tips for the Test Scheduled in Week 17
1. Suggest how alkali metal hydroxide can be prepared in two reactions. State a disadvantage of using this procedure. [5]
4. Explain the trend in the forces of attraction observed in Group 18. Relate your explanation to the melting and boiling points of the elements [4]
The forces of attraction (van der Waals) between the monoatomic molecules of inert gas (Group 18) increase when going down the group from helium to radon [1]. The forces depend on the atomic size [1]. As the atomic radius increases down the group, the van der Waals becomes stronger [1]. As a result, the melting point and boiling point also increase due to the increasing forces of attraction down the group as more heat energy is required to overcome/break the attractive forces [1].
5. Compare and contrast the products of bromine and iodine after they have respectively been placed in the beakers containing water [4]
The reaction of bromine with water produces hydrobromic acid and hypobromous acid, whereas the reaction of iodine with water produces hydroiodic acid and hypoiodous acid. [1] Both the resulting solutions are acidic although the products of iodine only show very weak acidic properties and they turn the blue litmus paper to red [1]. However, the presence of hypobromous acid as a bleaching agent can subsequently turn the red litmus paper to white [1] whereas the hypoiodous acid does not decolourise the blue or red litmus paper because it is not a bleaching agent [1].
Tips for the Test Scheduled in Week 10
1. Find the mass of aluminium oxide (Al2O3) that is equivalent to 6.13 X 1024 ions. [RAM Al: 27; O: 16; Avogadro constant: 6 X 1023]. [5]
2. Find the mass of aluminium nitrate [Al(NO3)3] that is equivalent to 5.41 X 1024 ions. [RAM Al: 27; N: 14; O: 16; Avogadro constant: 6 X 1023]. [5]
Answer: 480 g (use the above steps to get this answer) [5]
3. Write and balance this equation: The solid of aluminium reacts with dilute sulfuric acid, yielding aluminium sulfate and hydrogen gas. [2]
[2]
Carbon-12 exists as a solid at room temperature and thus can be handled easily [1]. It can combine with many elements [1]. Besides, carbon-12 is also commonly used as a reference standard in mass spectrometers [1].
Tips for the Test Scheduled in Week 9
1. What information could be obtained from this symbol to tell about this element? [3]
[1]
Tips for the Test Scheduled in Week 4
1. What is 'electrochemical series'? [2]
Electrochemical series is an arrangement of metals based on the tendency of each metal atom to donate electrons [1]. The greater the tendency to donate electrons, the more electropositive is the metal and the higher it is in the electrochemical series [1]
2. Describe how a Galvanic cell can be constructed using magnesium and zinc as electrodes. The electrolytes must be separated in two different tanks. [7]
A magnesium metal (Mg) is immersed in a tank containing its ions such as magnesium nitrate (Mg(NO3)2) [1]. On the other hand, a zinc metal is also immersed in another tank containing its ions such as zinc sulphate (ZnSO4) [1]. Both the metals are connected using a wire [1/2]. A voltmeter is also used to study the potential difference [1/2]. Then, a salt bridge containing concentrated potassium chloride (KCl) is used by putting its ends in both the electrolytes in order to allow for the movement of ions and to complete the electrical circuit [1]. As magnesium is more electropositive than zinc, it will release electrons and form magnesium ions at the negative terminal [1]. The half equation is
Mg --> Mg2+ + 2e-
The electrons are passed from the negative terminal to the zinc metal (the positive terminal) through the wire and the zinc ions in the electrolyte are discharged to form solid zinc [1]. The half equation is
Zn2+ + 2e- --> Zn
The voltmeter pointer deflects to show the magnitude of the voltage at 1.62V [1].
Diagram [2]
Maximum is 7marks
Note 1:
1. Zn and Cu, negative terminal is Zn, electrolytes are zinc sulphate and copper (II) sulphate, 1.10V
2. Zn and Fe, nagative terminal is Zn, electrolytes are zinc sulphate and iron (II) nitrate, 0.32V
3. Mg and Cu, negative terminal is Mg, electrolytes are magnesium nitrate and copper (II) sulphate, 2.72V.
Note 2:
The given electrolytes are theoretically valid and may not be applicable to practical setups.
Note 3:
If the positive terminal is immersed in its ions, make sure the cations are concentrated if 'solution' is mentioned. This is only for those cations (such as Zn2+) that are higher than H+ in the ES, so that Zn2+ is selected to be discharged. If you don't mention whether it is a solution or so, just its ions, then no need to care about the H+ factor because we assume the electrolyte has little or none H+. This is for advanced learning.
For cations lower than H+ in the ES such as Cu2+ in the terminal electrode, dilute solution can be used because it is less electropositive than H+ and Cu2+ will definitely be selected. This is the typical example in the textbook.
3. How can we know whether a chemical reaction between a metal and a salt in a solution will take place? [4]
This can be determined from the electrochemical series [1]. For a chemical reaction to take place, the metal must be more electropositive/reactive than the cation in the salt [1]. For example, in the equation Mg + ZnSO4 --> MgSO4 + Zn, this reaction will take place because Mg is higher than Zn in the electrochemical series and hence more electropositive. So Mg can displace Zn2+ from its salt [1]. In contrast, H2 + 2NaCl --> 2HCl + 2Na will not take place because H is lower than Na in the electrochemical series and less electropositive than Na. So it cannot displace Na+ from the salt [1].
4. In an electrolyte containing magnesium sulphate solution, select and explain the ion that will be discharged at the cathode when the electrical circuit passes through. [2]
H+ will be selectively discharged at the cathode [1] because H+ is lower than Mg2+ in the electrochemical series and hence less electropositive [1], so it H+ ions will accept electrons to form hydrogen gas.
Reminder: Silver is Ag, and its ion is Ag+; gold is Au, and its ion is Au+, both are at the bottom in the ES.
SR1 SCB and SR1 SCD
Tips for the Test Scheduled in Week 17
1. Suggest how alkali metal hydroxide can be prepared in two reactions. State a disadvantage of using this procedure. [5]
The first reaction should allow the alkali metal to react with oxygen gas to produce alkali metal oxide [1]. The equation is 4M + O2 --> 2M2O [1]. The second reaction is to allow the alkali metal oxide to react with water to produce akali metal hydroxide [1]. The equation is M2O + H2O --> 2MOH [1]. This procedure is tedious as compared to the direct reaction of alkali metals with water, which also produces alkali metal hydroxide. [1]
2. State the specific characteristics of iron [4]
1. Iron can form ions in two oxidation states, Fe (II) and Fe (III)
2. Iron can form complex ions
3. Iron can form coloured compounds
4. Iron can act as a catalyst in the Haber process for the manufacture of ammonia
3. State the application of the following elements [3]
Helium: It is used as an artificial atmosphere for divers [1]
Argon: It is used to fill electric bulbs to prevent the oxidation of the tungsten filament [1]
Krypton: It is used to fill high speed photographic flash lamps [1]
4. Explain the trend in the forces of attraction observed in Group 18. Relate your explanation to the melting and boiling points of the elements [4]
The forces of attraction (van der Waals) between the monoatomic molecules of inert gas (Group 18) increase when going down the group from helium to radon [1]. The forces depend on the atomic size [1]. As the atomic radius increases down the group, the van der Waals becomes stronger [1]. As a result, the melting point and boiling point also increase due to the increasing forces of attraction down the group as more heat energy is required to overcome/break the attractive forces [1].
5. Compare and contrast the products of bromine and iodine after they have respectively been placed in the beakers containing water [4]
The reaction of bromine with water produces hydrobromic acid and hypobromous acid, whereas the reaction of iodine with water produces hydroiodic acid and hypoiodous acid. [1] Both the resulting solutions are acidic although the products of iodine only show very weak acidic properties and they turn the blue litmus paper to red [1]. However, the presence of hypobromous acid as a bleaching agent can subsequently turn the red litmus paper to white [1] whereas the hypoiodous acid does not decolourise the blue or red litmus paper because it is not a bleaching agent [1].
Tips for the Test Scheduled in Week 10
1. Find the mass of aluminium oxide (Al2O3) that is equivalent to 6.13 X 1024 ions. [RAM Al: 27; O: 16; Avogadro constant: 6 X 1023]. [5]
*Please be informed that you can use the term 'molar mass' to represent RAM, RMM and RFM where applicable. Practically, if units such as g/mol are involved in your calculation, it is more logic to use 'molar mass' as RAM, RMM and RFM do not have a unit.
2. Find the mass of aluminium nitrate [Al(NO3)3] that is equivalent to 5.41 X 1024 ions. [RAM Al: 27; N: 14; O: 16; Avogadro constant: 6 X 1023]. [5]
Answer: 480 g (use the above steps to get this answer) [5]
3. Write and balance this equation: The solid of aluminium reacts with dilute sulfuric acid, yielding aluminium sulfate and hydrogen gas. [2]
[2]
4. Use the chemical equation in (3) to answer this question: Find the number of atoms and the volume of hydrogen gas in STP conditions if 45 g aluminium is involved in this reaction. [RAM Al: 27; H: 1; Avogadro constant: 6 X 1023; molar volume: 22.4 dm3/mol at STP]. [6]
Answer: Number of atoms = 3 X 1024 atoms [3]. Volume of hydrogen gas = 56 dm3 [3] You must show the steps involved in your calculation.
5. Why is carbon-12 chosen as the standard to measure the relative atomic mass. [3]
Tips for the Test Scheduled in Week 9
This element is called Sodium, which has a mass number/neucleon number of 23 and the proton number/atomic number of 11 [1]. Besides, it has 23-11 = 12 neutrons in its neucleus [1]. In its neutral state, it also possesses 11 electrons [1].
2. If this element tends to form an ion, how would you draw the new electron configuration? Predict whether it will form a positively charged ion or a negatively charged ion. [2]
The following is the electron configuration of the element after it has become an ion. The element will form a positively charged ion (cation). [1]
[1]
3. What is the use of a melting point in chemistry? [4]
A melting point of a chemical substance can be used to assess the purity of the substance [1]. If the substance shows a melting point that is different from that usually obtained from its pure substance, it means the substance is no longer pure [1]. It also indicates the stength of attractive forces between the particles [1]. A high melting point is associated with high attractive forces (and the reverse is the low melting point) [1].
Electrochemical series is an arrangement of metals based on the tendency of each metal atom to donate electrons [1] The greater the tendency to donate electrons, the more electropositive is the metal and the higher it is in the electrochemical series [1]
Electrochemical series is an arrangement of metals based on the tendency of each metal atom to donate electrons [1]. The greater the tendency to donate electrons, the more electropositive is the metal and the higher it is in the electrochemical series [1]
This comment has been removed by the author.
ReplyDeleteWrite down your formula here. Let me see first.
Deleteno of mol= 5.41 times 10^24 divide Avogadro constants ? without using the no of compund formula
ReplyDeletewhat is 5.41? show the unit....as long as the equation makes sense, and is understandable...then go ahead.
DeleteDominick, sorry for the inconvenience caused. I didn't realize its disappearance after adding Exercise 4 to the blog.
ReplyDelete