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Revision Questions on Energy Changes / Redox / Electrolysis – Answers
Question 1:
Calculate the oxidation state of sulphur in each one of the following compounds.
a)
H2S
(2  (+1)) + S = 0  S = –2
b)
SO2
S + (2  (–2) = 0  S = +4
c)
SO3
S + (3  (–2)) = 0  S = +6
d)
SO32– S + (3  (–2)) = –2  S = +4
e)
K2SO4 (2  (+1)) + S + (3  (–2)) = 0  S = +6
f)
S2O32– (2  S) + (3  (–2)) = –2  S = +2
g)
S4O62– (4  S) + (6  (–2)) = –2  S = +2.5
Question 2:
Chlorine gas can be prepared in the laboratory according to the following chemical reaction.
MnO2(s) + 4HCl(aq)  MnCl2(aq) + 2H2O(l) + Cl2(g)
a)
Complete the table below to show the oxidation states of chlorine and manganese at the
start and at the end of the reaction.
b)
Element
Oxidation State at the Start of
the Reaction
Oxidation State at the End of
the Reaction
Manganese
Mn + (2  (– 2)) = 0  Mn = +4
Mn + (2  (– 1)) = 0  Mn = +2
Chlorine
(+1) + Cl = 0  Cl = –1
Pure element = 0
What is the oxidising agent in this reaction, MnO2 or HCl? Explain your answer.
The manganese(IV) oxide is the oxidising agent, as the manganese has been reduced.
The manganese(IV) oxide has oxidised the chlorine from oxidation state –1 to 0.
c)
What is the reducing agent in this reaction, MnO2 or HCl? Explain your answer.
The hydrochloric acid is the reducing agent as the chlorine has been oxidised.
The hydrochloric acid has reduced the manganese from oxidation state +4 to +2.
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)
d)
State the qualitative test for chlorine gas.
Chlorine turns damp blue litmus paper red and then bleaches it white.
e)
The chlorine gas that is produced by this reaction is damp. Which reagent should be used
to dry the chlorine gas, anhydrous calcium oxide or concentrated sulphuric acid? Explain
your answer.
Concentrated sulphuric acid. Chlorine is an acidic gas and should be dried using an acidic
drying agent. Calcium oxide is a base. If the chlorine were dried using calcium oxide, then
the acidic gas would react with the basic drying agent.
Question 3:
The table below shows some information about two copper ores, tenorite and cuprite. Both contain
copper oxide.
a)
Ore
Formula of Copper
Oxide in Ore
Oxidation Number of
Copper
Percentage of Copper
by Mass
Tenorite
CuO
+2
80.0%
Cuprite
Cu2O
+1
88.9%
i)
What is the formula of the copper compound in tenorite?
CuO
ii)
What is the oxidation number of copper in cuprite, Cu2O?
(2  Cu) + (–2) = 0  Cu = +1
iii)
Calculate the percentage of copper by mass in cuprite, Cu2O.
percentage copper = (mass of copper in 1 mol of Cu2O  mass of 1 mol of Cu2O)  100
percentage copper = ((64 + 64)  (64 + 64 + 16))  100
percentage copper = (128  144)  100
percentage copper = 88.9%
b)
Another ore of copper contains copper(II) sulfide. Complete the dot and cross diagram
below for copper(II) sulphide showing the outer electrons only.
 = Electron of S  = Electron of Cu
Copper Ion
Sulfide Ion
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Question 4:
Dilute hydrogen peroxide decomposes to give oxygen and water when a catalyst is added.
a)
Balance the equation for the reaction and complete the state symbols.
2 H2O2 ( aq )
b)

2 H2O ( l )
+
O2 ( g )
This diagram shows the energy levels of the reactants and products for this reaction.
Complete the diagram by:
 Drawing in the reaction pathway for the reaction.
 Showing the activation energy, Ea, by a single headed arrow.
 Showing the energy change for the reaction, H, by a single headed arrow.
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c)
The overall energy change for the reaction is exothermic.
i)
Explain how the diagram shows this.
The energy of the products is lower than the energy of the reactants.
ii)
Explain, in terms of bond breaking and bond making, why this reaction is
exothermic.
Bond breaking absorbs energy, bond formation releases energy.
In this reaction, the energy released when the bonds in water and oxygen form is
greater than the energy that is absorbed to break the bonds in hydrogen peroxide.
Question 5:
Car manufacturers are developing fuel cells for use in cars.
Fuel cells produce electrical energy from the reaction between a fuel and oxygen. Two possible
fuels for use in fuel cells are hydrogen and methanol. The table below gives some data about
these two fuels.
a)
Fuel
Melting Point / C
Boiling Point / C
Energy Change of
Combustion kJ / mol
Hydrogen
–259
–252
–256
Methanol
–97.7
64.5
–715
The table gives values for the energy change of combustion for each fuel in kJ / mol.
i)
Calculate the energy output for 1 g of each fuel.
1 mol of H2 weighs 1 + 1 = 2 g
energy output by 1 g of H2 = –256  2 = –128 kJ / g
1 mol of CH3OH weighs 12 + 16 + 4 = 32 g
energy output by 1 g of methanol = –715  32 = –22.3 kJ / g
ii)
Use the values that you have calculated, and information in the table, to discuss the
advantages and disadvantages of using each fuel in cars.
When burned, 1 g of hydrogen produces more energy than 1 g of methanol.
Methanol is a liquid at room temperature and pressure, while hydrogen is a gas.
Therefore, 1 g of methanol will occupy a smaller volume than 1 g of hydrogen.
Methanol is more easily stored and transported compared to hydrogen.
iii)
Hydrogen and methanol have different effects on the environment when used as
fuels. Outline two environmental differences between the two fuels.
Combustion of hydrogen produces water as the only chemical product. Water is not
a pollutant. Combustion of methanol produces water and carbon dioxide. Carbon
dioxide contributes to climate change (global warming). Incomplete combustion of
methanol may produce carbon monoxide, which is a toxic gas.
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b)
The energy output of a fuel cell can be shown using an energy profile diagram. Draw an
energy profile diagram for the combustion of methanol. Your diagram should include labels
for the reaction enthalpy change and activation energy.
Question 6:
When copper is extracted from its ore, it contains carbon impurities.
a)
Suggest how the carbon impurities get into the copper.
Copper can be extracted from its oxide using carbon (the more reactive carbon displaces
the less reactive copper from its oxide): 2CuO(s) + C(s)  2Cu(s) + CO2(g)
Unreacted carbon will remain as an impurity in the copper.
b)
Copper for electrical wiring needs to have a very high purity.
Pure copper is made by electrolysis, using aqueous copper(II) sulfate as an electrolyte.
i)
Write equations, including state symbols, for the reactions at each electrode.
Anode (+ve): Copper is oxidised, Cu(s)  Cu2+(aq) + 2e–
Cathode (–ve): Copper(II) ions are reduced, Cu2+(aq) + 2e–  Cu(s)
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ii)
What happens to the concentration of the aqueous copper(II) sulfate during the
electrolysis? Explain your reasoning.
The concentration of aqueous copper(II) sulfate remains constant during the
electrolysis. As copper(II) ions are reduced to copper atoms at the cathode, they are
replaced in solution by copper atoms being oxidised to copper(II) ions at the anode.
Sulfate ions are not discharged and are not removed from the solution.
Question 7:
Chlorine is made by electrolysis of concentrated aqueous sodium chloride.
The diagram below shows apparatus that can be used to electrolyse concentrated aqueous
sodium chloride.
Chlorine
Hydrogen
Dilute
Sodium
Chloride
a)
Label the products on the diagram.
b)
Write ionic equations, including state symbols, for the reactions taking place at the anode
and the cathode.
Reaction at the anode:
Reaction at the cathode:
2Cl–(aq)  Cl2(g) + 2e–
2H+(aq) + 2e–  H2(g)
Question 8:
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Sodium metal is extracted from molten sodium chloride by electrolysis.
a)
i)
Write an ionic half-equation, with state symbols, to show the reaction that happens
at the anode.
2Cl–(l)  Cl2(g) + 2e–
ii)
Describe a simple test, and its result, that would identify the gas given off at the
anode.
Chlorine gas will turn moist blue litmus paper red, and will then bleach it white.
b)
Calcium chloride is added to the sodium chloride to lower the melting point of the mixture.
i)
Explain why lowering the melting point of the mixture makes the process cheaper to
run.
Because the melting point of the electrolyte is lowered, less energy is required to
heat it to its melting point. Less money is spent on fuel, therefore the process is
cheaper.
ii)
The molten sodium contains metallic impurities. Name the main metallic impurity
that you would expect to find and explain how it forms.
The main metallic component will be calcium from the calcium chloride. This is
formed when calcium ions are reduced at the cathode: Ca2+(l) + 2e–  Ca(l)
c)
Sodium chloride can also be electrolysed in aqueous solution. Describe the differences in
the products of the electrolysis of concentrated aqueous sodium chloride compared to
molten sodium chloride.
During electrolysis of concentrated aqueous sodium chloride, chlorine gas will still be
produced at the anode, but hydrogen gas (not sodium) will be produced at the cathode.
Question 9:
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Aqueous copper(II) sulfate is electrolysed using carbon electrodes.
a)
Give the formulae of all the ions present in the solution.
Anions:
Cations:
b)
SO42 – and OH–
Cu2+ and H+
A copper coating forms on the cathode, and a gas is evolved at the anode.
i)
Write a half-equation for the formation of copper at the cathode.
Cu2+(aq) + 2e–  Cu(s)
ii)
Name the gas formed at the anode and describe a test for this gas.
Oxygen gas is produced at the anode.
Oxygen gas will relight a glowing splint.
c)
After some time, the blue colour of the aqueous copper(II) sulfate fades, and the pH of the
solution decreases. Explain why these changes take place.
The solution is blue due to the presence of copper(II) ions dissolved in it. The blue colour of
the aqueous copper(II) sulfate fades as the concentration of the copper(II) ions in solution
decreases. Copper(II) ions are removed from solution as they are reduced to copper metal
at the cathode. As copper(II) ions and hydroxide ions are removed from solution during the
electrolysis, hydrogen ions and sulfate ions (i.e. sulfuric acid) remain, causing the pH of the
solution to decrease.
d)
A student investigated the relationship between the mass of copper formed and the total
charge passed through the solution. This is the graph of the results.
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i)
What mass of copper is formed when a charge of 600 coulombs is passed
through the solution?
0.20 g
ii)
Use the graph to predict the charge needed to form 1 g of copper, and hence
predict the charge needed to deposit 1 mole of copper.
0.20 g requires 600 coulombs to deposit it
 1.00 g of copper will require (1  0.20)  600 = 3000 coulombs to deposit it
1.00 g of copper is deposited by 3000 coulombs
1 mole of copper weighs 64 g
 1 mole of copper will require 64  3000 = 192000 coulombs to deposit it
Question 10:
A new type of electroplating is known as “brush electroplating”. It is used to electroplate zinc onto
very large iron supports to be used in buildings. The iron supports are too big to be plated in a
normal electrolysis tank. During the process, a metal brush spreads a layer of aqueous zinc sulfate
over the iron surface. A battery gives the metal brush a positive charge and gives the iron support
a negative charge. A layer of zinc forms on the surface of the iron support.
a)
The surface of the iron acts as a cathode. Zinc ions from the solution form zinc on the
surface of the iron. Write an ionic half-equation, with state symbols, for this reaction.
Zn2+ + 2e–  Zn(s)
b)
Two different designs of metal brush are available. One type of brush is made from zinc,
one type is made from platinum. As the electrolysis takes place, each brush has a different
effect on the concentration of zinc ions in the solution.
i)
What will happen to the concentration of the zinc ions during the electrolysis if the
brush is made from platinum?
The concentration of zinc ions in aqueous solution will decrease.
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ii)
What will happen to the concentration of the zinc ions during the electrolysis if the
brush is made from zinc?
The concentration of zinc ions in aqueous solution will remain constant.
iii)
Platinum brushes are much more expensive than zinc brushes. However, zinc
brushes need replacing regularly, but platinum brushes do not. Explain why.
If the brush is made of zinc, then the bristles of the brush will be oxidised to zinc
ions: Zn(s)  Zn2+(aq) + 2e–. Platinum is an inert electrode. It is not oxidised.
c)
During the process, a worker needs to hold the brush. Which of the following materials
would be a good choice for the handle of the brush? Give a reason for your answer.
chromium
d)
copper
graphite
iron
poly(ethene)
Material:
Poly(ethene)
Reason:
Poly(ethene) is an electrical insulator. It does not conduct electricity.
Explain why iron supports coated with zinc do not rust, even if the zinc coating is damaged.
Zinc prevents the iron from rusting through sacrificial protection. Because zinc is more
reactive than iron, zinc will preferentially oxidise / corrode in place of the iron.
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