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Topic 9 Oxidation and Reduction Answers
9.1 Exercises
1. Define oxidation and reduction in terms of
a) electron transfer
Oxidation is the loss of electrons from an atom, ion or molecule. Reduction is the gain of electrons by an atom,
ion or molecule.
b) oxidation number
Oxidation is the increase of the oxidation number of an atom. Reduction is the decrease of the oxidation number
of an atom.
2. Why is oxidation and reduction like lending and borrowing?
Oxidation cannot occur without reduction having also occurred, just as you cannot borrow a book without
someone lending it to you. Both involve a transfer.
3. Assign oxidation numbers of the elements for the following groups of species, and give the rule/s
you have applied in assigning oxidation numbers for each group:
a)
0
0
0
0
O2
H2
Cl2
Na
Rule/s: All of the above species have an oxidation number of 0, as they are all in the elemental form (how they
exist naturally, not combined with any other elements).
b)
+1
-1
+
Na
Cl
+2
-
-2
2+
S2 -
Ca
Rule/s: The oxidation number is the same as the charge, as these are monatomic ions.
c)
+1 -2
-1 +1
H2O
HNa
-3 +1
-4 +1
NH3
CH4
Rule/s: Three rules are used here. Firstly, hydrogen always has an oxidation of +1 (except in combination with
reactive metals such as Na when it is -1). Secondly, oxygen always has an oxidation state of –2 (except in H2O2
where it is -1). These known values are used first. Finally, as all these molecules have an overall charge of 0,
the sum of the oxidation numbers must also equal zero.
Conventions for showing oxidation numbers
The oxidation numbers are assigned for individual atoms, not groups of the same
atom within a polyatomic ion. For example the oxidation number of each hydrogen
atom in NH3 is +1 and this is what is written above the hydrogen atom. Because there
are three hydrogen atoms, +1 x 3 = +3, and this total, added to the oxidation number
of nitrogen (–3), gives the overall zero charge for the NH3 molecule.
d)
+1 -1
AgCl
Rule/s: This is an ionic compound. Chlorine is in group 7 and thus forms an ion with a charge of -1, which
means its oxidation number is also –1. Silver (Ag) must have a charge and oxidation number of +1 to balance
out the chloride ion.
+1 +1 -2
HClO
1
Rule/s: The common values for hydrogen (+1) and oxygen (-2) are used. The total charge of the molecule is 0,
so the sum of the oxidation numbers of the individual species must equal 0, therefore chlorine has an oxidation
number of +1 in this case.
Oxidation states of halogens
Normally halogens have negative oxidation states, but on combination with oxygen or
a higher halogen, e.g. fluorine, they have positive oxidation states. This is because
halogens in period 3 (or in a higher period) are able to "expand their octet".
+2 -1
MgBr2
Rule/s: This is an ionic compound. Bromine is in Group 7, and forms an ion with charge –1, so its oxidation
number is also –1. Magnesium is in Group 2, and forms an ion with charge +2, so its oxidation number is +2.
Note that the charge and oxidation number balance out to 0.
+2 -1
CuCl2
Rule/s: This is an ionic compound. Chlorine is in Group 7, and forms an ion with charge –1, so its oxidation
number is also –1. Copper must have a charge and oxidation number of +2 to balance out the two chloride ions.
4. Assign oxidation numbers of all elements in the following species:
a)
+1 +7 -2
KMnO4
Oxygen has the known oxidation number of –2. Potassium is in Group 1, and thus has an oxidation number of
+1. As the molecule has total charge of 0, the sum of the oxidation numbers must equal 0, so manganese has
an oxidation number of +7.
b)
+4 -2
CO2
Oxygen has oxidation number –2, and the molecule has 0 charge. Therefore carbon must have an oxidation
number of +4 making the molecule overall neutral.
c)
-2 +1 -2 +1
C2H5OH
The known values for hydrogen and oxygen are used. As the molecule has a charge of 0, the sum of the
oxidation numbers must be 0, and the two carbon atoms must have a combined oxidation number of –4, so
each carbon atom has an oxidation number of –2.
d)
+7 -2
ClO4
The molecule has a negative charge, so the sum of the oxidation numbers should be -1. Oxygen has the usual 2 value and four of them give a combined oxidation number of –8. The overall sum of oxidation numbers must
add to –1, so Cl must be equal to +7.
e)
+5 -2
3
PO4 
Oxygen has the usual value of –2 and four of them give a combined oxidation number of –8 so the phosphorus
must have an oxidation number of +5 to give the molecule a net charge of –1.
2
5. Oxidation calculations
a) What is the oxidation number for the elements below in their ionic form?
+
Na
Mg
Al
F
+1
2+
3+
-
–1
2-
–2
3-
–3
O
+2
N
+3
b) Calculate the oxidation number of nitrogen in the following:
NH3
-3
N2O5
+5
NH4
-4
N2
0
HNO3
+5
Mg3N2 -3 (Mg is in Group 2, so is +2)
c) Calculate the oxidation number of sulfur in the following:
ZnS
-2
H2SO4 +6
SO2
+4
SO3
+6
SO42-
+6
H2S
-2
6. State the names of the following compounds using oxidation numbers.
It may be helpful to look in the introduction section for the charges and names of the common groups (NO2-)
and (SO42-)
a) Fe2O3
Iron(III) oxide
b) FeO
Iron(II) oxide
c) CuCl2
Copper(II) chloride
d) CoCl2
Cobalt(II) chloride
e) Cu(NO2)2
Copper(II) dioxonitrate(III) (the old name, nitrite, is acceptable).
This requires some assumed knowledge that the NO2- ion has a –1 charge. Use a bit of common sense here:
you know that the Cu, a metal, must be positive (i.e. most likely +1 or +2) and that the compound is neutral
overall. The fact that there are two NO2- ions for every one copper should give you a clue that the copper ion
has a +2 charge.
f)
Na2SO4
Sodium(I) tetraoxosulfate(VI) (old name: sulfate)
Common oxoanions are given names with the suffixes –ite and –ate. The ion with
the larger number of oxygen atoms is given the suffix ‘-ate’ and that with the smaller
number of oxygen atoms ‘-ite”
Oxoanions of sulfur:
-Sulfates (SO42−), the salts of sulfuric acid (H2SO4), in which sulfur has an
oxidation state of +6
-Sulfites (SO2−), the salts of sulfurous acid (H2SO3) in which sulfur has an
oxidation state of +4
O
O
S
O
OH
S
OH
HO
OH
Sulfuric acid
Sulphurous acid
3
7. Deduce formulas of the inorganic compounds with the following names:
a) sulfuric(VI) acid
H2SO4 (Note that the sulfur has an oxidation number of +6 as shown by the roman numerals and the SO4 group
overall has an oxidation number of –2)
b) sulfurous(IV) acid
(erratum, this should read sulfurous not sulfuric, see previous page for an explanation of the sulfur
oxoanions)
H2SO3 (Note that the sulfur in the SO3 has an oxidation number of +4 as shown by the roman numerals and the
SO3 group overall has an oxidation number of –2)
c) potassium manganate(VII)
KMnO4 (Note that the manganese has an oxidation number of +7)
d) sodium dichromate(VI)
Na2Cr2O7 (Note that chromium is has an oxidation number of +6)
e) copper(I) tetraoxosulfate(VI)/sulphate
Cu2SO4 (Note that the copper has a +1 oxidation state and sulphate group overall has a –2 oxidation state)
f)
nitric(V) acid
HNO3 (Note that the nitrogen has an oxidation state of +5)
8. Indicate which of the following reactions are redox reactions. For the redox reactions, identify the
substance oxidised and the substance reduced by the change in oxidation numbers.
Erratum: in part a) Change Cl-(g) to Cl-(aq) Please note the error in state symbols in the workbook.
a)
+1
-1
+
+1 -1
-
Ag (aq) + Cl (aq) AgCl(s)
This is not a redox reaction, as none of the elements change oxidation number.
The equation written above merely indicates a precipitation reaction (Ag+ and Cl- ions associating to form AgCl
and coming out of solution as a solid precipitate. The transformation step of the elements Ag(s) and Cl2(g) to
ions in solution (Ag+ and Cl-) would have involved redox reactions.
b)
+2 +4 -2
+2 -2
+4 -2
CaCO3(aq) CaO(s) + CO2(g)
This is not a redox reaction as none of the elements change oxidation number.
c)
-2 +1 -1
-2 +1 -
2 +1 -2 +1
-1
–
CH3Br(aq)+ OH (aq) CH3OH(aq) + Br –(aq)
This is not a redox reaction as none of the elements change oxidation number. This is a substitution reaction.
You will learn about these reactions in topic 10.
d)
+5 -2
-1
–
+1
-
0
+1 -2
+
BrO3 (aq) + 5Br (aq) + 6H (aq) 3Br2(l) + 3H2O(l)
This is a redox reaction, as the bromine in BrO3 is reduced (decreases oxidation number), and the bromide ion
is oxidised (increases oxidation number.)
e)
0
+1 -2
+1 -1
0
2F2(g) + 2H2O(l) 4HF(aq) + O2(g)
This is a redox reaction. The fluorine is reduced as its oxidation number decreases from 0 to -1. The oxygen is
oxidised as it increases in oxidation from -2 to 0.
4
f)
0
-1
-1
–
0
–
Cl2(g) + 2Br (aq) 2Cl (aq) + Br2(g)
This is a redox reaction. The chlorine is reduced as it decreases in oxidation number from 0 to -1. The bromine
is oxidised as it increases in oxidation number from -1 to 0.
9. For the following reactions, identify the substance oxidised and the substance reduced by the
change in oxidation numbers.
a) reaction that destroys ozone in the stratosphere:
+2 -2
0
+4 -2
0
NO(g) + O3(g) NO2(g) + O2
Nitrogen is oxidised as the oxidation number increases from +2 to +4. One oxygen atom is reduced from ozone
(O3) as the oxidation number decreases from 0 to -2 in the product NO2.
b) formation of rust (simplified equation):
0
0
+3 -2
Fe(s) + O2(g) Fe2O3(s)
The iron is oxidised as the oxidation number increases from 0 to +3. Oxygen is reduced as the oxidation
number decreases from 0 to -2.
c) the reaction of hydrazine with nitrogen tetroxide is utilised in rocket propellant:
-2 +1
+4 -2
0
+1 -2
2N2H4(g) + N2O4(g) 3N2(g) + 4H2O(g)
Nitrogen from N2H4 is oxidised as it increases in oxidation number from -2 to 0. Nitrogen from N2O4 is reduced
as it decreases in oxidation number from +4 to 0.
d) the reaction that occurs when a car battery is discharging:
0
+4 -2
+1 +6 -2
+2 +6 -2
+1 -2
Pb(s) + PbO2(aq) + H2SO4(aq)
2PbSO4(aq) + 2H2O(l)
The solid lead is oxidised as the oxidation number increases from 0 to +6. The lead in PbO2 is reduced as the
oxidation number decreases from +4 to +2.
Q 10-11 Theory of Knowledge questions to be discussed with your TOK teacher.
9.2 Exercises
1. What is a redox equation?
A redox equation is a balanced equation describing a redox reaction in which one species is oxidised and one
species is reduced.
2. What is a half equation?
A half equation demonstrates only one half of the overall redox reaction; either the oxidation or the reduction
half. The two half equations (oxidation and reduction) are added together to produce the overall redox reaction.
3. List three different ways to describe a reducing agent.
1. A reducing agent supplies electrons to another species; it reduces the other species.
2. A reducing agent is oxidised during a redox reaction.
3. A reducing agent contains an element whose oxidation number increases during the reaction.
5
4. How can we tell from a chemical equation whether a reaction is a redox equation?
Assign oxidation numbers to all elements. If you see a compound that has an element that has either an
increase or decrease in oxidation number from one side of the equation to the other, then you know it is a redox
reaction.
5. Write half-equations for
a) the reduction of copper(II) ions in aqueous solution to copper metal
2+
Cu (aq) + 2e Cu(s)
Balance the charge by adding electrons. In this case the copper is reduced as it gains electrons and the
oxidation number decreases from +2 to 0.
b) aluminium metal is oxidised to Al3+ in aqueous solution
Al(s) Al3+(aq) + 3e
Note from question 5 a) and b):
You can see that when a species is reduced the electrons appear on the reactant side
of the half equation and that when a species is oxidised the electrons appear on the
product side. This highlights the definition of reduction and oxidation with respect to
electrons being either lost or gained by a species.
6. The following species are common oxidising agents. Deduce their reduction half-equations.
a) O2
O2 + 4e 2O2- Remember, if a species is an oxidising agent, it must itself be reduced. So look for a decrease
in oxidation number in the element/species. Reduction is gain of electrons. Look at the element which is most
likely to gain electrons in polyatomic species.
b) Cl2
Cl2 + 2e 2Cl
c) H2O2/H+
H2O2 + 2H+ + 2e 2H2O
Notice that the oxidation number of oxygen decreases from –1 in the peroxide (H2O2)
to –2 in water. Oxygen is reduced, i.e. it gains electrons. If you think logically about
this, it is much more likely that the oxygen acts as the oxidant rather than the
hydrogen ions acting as the oxidant and being reduced, i.e. hydrogen does not gain
electrons easily at all!
-
+
d) MnO4 /H
+
2+
-
MnO4 + 8H + 5e Mn + 4H2O Oxygen has a –2 oxidation state in MnO4 , ie oxygen cannot be reduced
-
further, so it must be Manganese, Mn that gains the electrons. Mn has +7 oxidation state in MnO4 , and +2 in
the manganese ions, it has gained electrons and has been reduced.
e) Br2
Br2 + 2e 2Br + 4H2O
f)
-
+
ClO /H
ClO + 2H+ + 2e 2Cl + H2O Chlorine has an oxidation state of +1 in ClO- and this is reduced to a –1
oxidation state in the chlorine Cl- ions.
6
g) HNO3
HNO3 + H+ + e NO2 + H2O Nitrogen has an oxidation state of +5 in HNO3. This is the highest oxidation state
for nitrogen, so it can only be reduced. Nitrogen in HNO3 is reduced to +4 in NO2
7. The following species are common reducing agents. Deduce oxidation half-equations.
a) Na
Na Na+ + e
b) Zn
2+
Zn Zn
+ 2e
c) Mg
Mg Mg2+ + 2e
d) Fe
Fe Fe2+ + 2e
e) Fe2+
Fe2+ Fe3+ + e
f)
H2O2
+
H2O2 O2 + 2H + 2e
g) SO2
H2O + SO2 SO3 + 2H+ + 2e
h) SO32H2O + SO3 SO4 2 + 2H+ + 2e
8. Complete and balance the following equations, by giving balanced half equations followed by the
overall redox equation if applicable. For the equations that are redox reactions identify the oxidizing
agent. For reactions that are not redox explain why.
a) H2(aq)+ Br2(g) Oxidation:
H2 2H+ + 2e
Reduction:
Br2 + 2e 2Br
Overall:
Br2 + H2 2Br + 2H+
H2 is the reducing agent and Br2 is the oxidising agent.
It is easy to tell which element will be reduced and which will be oxidised by simply
comparing their electronegativities. Br is very electronegative, so likely to gain
electrons.
2+
b) Fe (s) + Cl2(g) 2+
3+
-
Oxidation:
Fe (s)+ Fe
+e
Reduction:
Cl2(g) + 2e 2Cl
Overall:
2Fe2+(s) + Cl2(g) 2Fe3+ + 2Cl- Iron in a +2 oxidation state, Fe(II), is oxidised to Fe(III),
-
-
i.e. the +3 oxidation state.
The Cl2 reduced, therefore the Cl2 is the oxidising agent.
c) N2(g) + H2(g) Oxiation:
3H2 6H+ + 6e
Reduction:
N2 + 6e 2N3
7
Overall:
N2 + 3H2 2NH3
N2 is the oxidising agent and H2 is the reducing agent.
d) HCl(aq)+ CaCO3(s) This is a reaction of the form acid + carbonate salt + CO2 + water
Balanced equation with oxidation states of elements assigned:
+1 -1
+2 +4 -2
+2 -1
+4 -2
+1 -2
2HCl(aq) +CaCO3(s) CaCl2(s)+ CO2(g) + H2O(l)
This is not a redox reaction as none of the elements change oxidation number.
e) C3H8(g) + O2(g) This is a combustion reaction of the form hydrocarbon + oxygen water + CO2
Balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
For a propane fuel cell under acidic conditions the half equations are as follows:
Oxidation:
C3H8 + H2O 3CO2 + 20H+ + 20e
Reduction:
O2 + 4e + 4H + 2H2O (x 5 to balance electrons)
Overall:
C3H8 + 5O2 3CO2 + 4H2O
+
Carbon in C3H8 is the reducing agent and O2 is the oxidising agent.
You have done well if you have worked this out. In an examination situation you would be asked only to balance
the equation for this combustion equation not provide half equations. You can tell which species is being
oxidised by looking at which has oxygen being added to it, here it is the propane, C3H8 oxygen is being added to
it. Working out the numbers of electrons transferred is a little tricky if you only look how the oxidation number of
carbon (it seems that each carbon in C3H8 has an oxidation number of 8/3 or 2 2/3) so look at the oxygen
instead. The oxidation number of oxygen changes from 0 in O2 on the left of the equation to –2 in CO2 and H2O
on the right. There are 10 oxygen’s on each side, so for 10 oxygen’s to be reduced from a 0 to a –2 oxidation
state, 20 electrons are required.
f)
CH3OH(l)+ O2(g) Oxidation:
CH3OH + H2O CO2 + 6H+ + 6e (x 2)
Reduction:
4e + 4H+ + O2 2H2O (x 3)
Overall:
2CH3OH + 3O2 2CO2 + 6H2O
Carbon in CH3OH is the reducing agent and O2 is the oxidising agent.
g)
+1 +6 -2
+1 -1
+1 -2
0
H2SO4(aq) + HI(g) H2S(g) + I2 (g) +
Oxidation:
2HI I2 + 2H+ + 2e (x 4)
Reduction:
H2SO4 + 8e+ 8H+ H2S + 4H2O
Overall:
H2SO4 + 8HI H2S + 4I2 + 4H2O
Sulfur in the H2SO4 is the oxidising agent and iodine in HI is the reducing agent.
h) CuO(s) + HCl(aq) This is a reaction of the form acid + base salt + water
Balanced equation: CuO(s) + 2HCl(aq) CuCl2 + H2O
This is not a redox reaction as none of the elements change oxidation number.
8
i)
Mg(s) + Cu2+ Oxidation:
Mg(s) Mg2+ + 2e
Reduction:
2e + Cu Cu(s)
Overall:
Mg (s) + Cu2+ Mg2+ + Cu (s)
2+
Mg(s) is the reducing agent and Cu2+ is the oxidising agent.
j)
MnO4-(aq) + H+ (aq) + Fe2+(aq)
Oxidation:
Fe2+ Fe3+ + e (x 5)
Reduction:
8H+ + 5e+ MnO4 Mn2+ + 4H2O
Overall:
8H + 5Fe
+
2+
2+
3+
+ MnO4 Mn + 4H2O + 5Fe
Fe2+ is the reducing agent and manganese in MnO4− is the oxidising agent.
It may be difficult at first to see what the products are going to be, but remember to
look at which elements/species are most likely to be oxidised or reduced as a clue to
working it out. Iron, Fe2+ could either be oxidised to Fe3+ or reduced to Fe(s). Oxygen
can only either remain in the same oxidation state or be oxidised as it is in its lowest
oxidation state of –2 and can now only lose electrons (be oxidised). Mn is in a high
oxidation state of +7 in MnO4- so is most likely reduced. From other examples you
have completed, you may have noticed that MnO4- ions are often reduced to Mn2+ ions
2+
and water, H2O. To reduce Mn with oxidation state +7 to Mn ions, with oxidation
state +2, 5 electrons are required.
9. Why are group 1 and 2 metals such good reducing agents?
Group 1 and 2 metals are highly electropositive and readily lose electrons to become positively charged ions
(cations) to achieve a full outer shell. Oxidation is a loss of electrons (remember OIL RIG), and if the species is
oxidised it must therefore be acting as a reducing agent.
10. Magnesium burns in air, giving a very bright light.
a) write the equation for this reaction.
2Mg(s) + O2(g) 2MgO(s)
b) write the half reaction for the oxygen.
Reduction:
O2 + 4e 2O2
c) Describe the transfer of electrons in the reaction and hence identify the oxidising agent and
reducing agent.
The oxygen element gains electrons, so it is reduced and is itself an oxidising agent. The magnesium atom
loses two electrons and is oxidised to Mg2+. It is the reducing agent as it reduces oxygen atoms to O2-.
11. Use the change in oxidation numbers to identify which of these reactions are redox reactions. If the
reaction is a redox reaction, name the element reduced, the element oxidized, the reducing agent, and
the oxidizing agent.
a)
0
+1 -2
+1 -2 +1
0
2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
This is a redox reaction as the oxidation numbers change. Na is oxidized as the oxidation number increases
from 0 to +1, and thus is the reducing agent. Hydrogen is reduced and is the oxidizing agent.
9
b)
+4 -2
+4 -2
N2O4(g) 2NO2(g)
This is not a redox reaction, as the oxidation numbers do not change.
c)
0
+1 -1
0
+1 -1
Cl2(g) + 2NaBr(aq) Br2(g) + 2NaCl(aq)
This is a redox reaction as the oxidation numbers change. Br is oxidized as the oxidation number increases
from -1 to 0, and thus is the reducing agent. Cl is reduced and is the oxidizing agent.
2
O2 + 4e 2O 
d)
+1 +4 -2
+1 +6 -2
+1 +6 -2
+4 -2
+1 -2
Na2CO3(s) + H2SO4(aq)
Na2SO4(aq) + CO2(g) + H2O(l)
This is not a redox reaction, as the oxidation numbers do not change.
12. Balance these redox equations:
a) HNO3 +H2S S + NO + H2O
Oxidation:
H2S S + 2H+ + 2e (x 5)
Reduction:
HNO3 + 5e + 5H+ NO + 3H2O (x 2)
Overall:
2HNO3 + 5H2S 2 NO + 6H2O + S
b) Cl2(g) + KOH(aq) KClO3(aq) + KCl (aq) + H2O
Oxidation:
Cl2 + 6H2O 2ClO3- + 10e- + 12H+
Reduction:
Cl2 + 2e Cl (x 5 to get lower equation)
Overall:
6Cl2(g) + 12KOH(aq) 2KClO3(aq) + 10KCl (aq) + 6H2O
-
Remember, you can use the change in oxidation state of the chlorine to work out how
to balance this equation, it is often easier than using the half equations.
13. In question 12, name the compounds that have the atom with the highest oxidation number.
The atoms with the highest oxidation number are chlorine in the compound KClO3 (potassium chlorate) with a
+5 oxidation state and nitrogen in nitric acid, HNO3, with a +5 state. (ClO3- = chlorate ion)
9.3 Exercise
1. What is a reactivity series?
A list of elements in order of their reactivity. Those that are higher on the list are more reactive than the
elements lower. They can also displace elements that are lower on the list from a solution of their ions.
2. Why may hydrogen be included in a metal activity series?
Like metals, it can form positive ions, e.g. H+. Comparison to hydrogen allows us to define an element as
electropositive or electronegative. Elements higher in the series than H2 are electropositive and elements lower
than H2 are electronegative.
10
3. Use the reactivity series below to complete the following equations. For any reactions that will not
occur write “no reaction”.
K > Na > Al > Fe > H2 > Cu and F > Cl > I
A displacement reaction can only occur if the metal of the elemental species is higher on the reactivity series
than the metal/hydrogen ion of the ionic species.
a) 2Al + 3H2SO4 Al2(SO4)3 + 3H2
b) Cl2 + 2KI 2KCl + I2
c) Cu + FeSO4 No reaction, as copper is the elemental species and it is lower on the reactivity series than iron.
d) 2Na + 2H2O 2NaOH + H2
4. Answer the following questions and for each “yes” response, write a balanced reaction equation:
a) Can H2 reduce Cu2+ ions in an aqueous solution to copper metal?
Yes. H2(g) + Cu2+ (aq) Cu(s) + 2H+ (aq)
b) calcium metal reduce Pb2+ ions in an aqueous solution to lead metal?
2+
2+
Yes. Ca(s) + Pb (aq) Pb(s) + Ca (aq)
c) Can lead metal reduce Fe3+ ions in an aqueous solution to Fe2+?
Yes. Pb(s) + 2Fe3+ (aq) Pb2+ (aq) + 2Fe2+ (aq)
See the IB Chemistry Data Booklet for a standard electrode potential table.
Pb is above Fe3+ for the reaction Fe3+ Fe2+ + e so this reaction occurs.
Would Pb reduce Fe2+ to Fe(s)?
5. Group l metals react violently with water.
a) Write an equation for the reaction of sodium with water.
2Na(s) + H2O(l) 2NaOH(aq) + H2(g) A vigorous reaction!
b) Why does reactivity with water increase down group 1?
All group 1 metals react violently with water. The reactivity of the metals increase down the group as the size
and number of shells of electrons of the metal increases. The more shells an atom has, the less tightly held the
outer electrons are and the more reactive they become.
c) Why are group 1 metals such good reducing agents?
Group 1 metals only have one outer shell electron which they lose readily in order to obtain a full outer shell of
electrons. As they easy lose electrons (are oxidised) this makes them excellent reducing agents.
6. Zinc reacts with acids to produce a salt and hydrogen.
a) Write an equation for this reaction
Zn(s) + HCl(aq) ZnCl2(s) + H2(g)
7. Balance equations for reactions
a) Write balanced chemical equations, with state symbols, for these reactions. If a reaction does
not occur write “no reaction”
2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g)
Sn + NaNO3 No reaction
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Cu + ZnSO4 No reaction
2K(s) + 2H2O(l) 2KOH(aq) + H2(g)
b) Now explain why you have written no reaction where you have.
Sn is the elemental species and cannot displace Na from the NaNO3 salt as Sn is lower on the reactivity series
than Na.
Cu is the elemental species and cannot displace Zn from the ZnSO4 salt as Cu is lower on the reactivity series
than Zn.
8. A student studying the reactivity of four metals in displacement reactions, listed the results as these
ionic equations:
2+
2+
Fe(s) + Cu (aq) Fe (aq) + Cu(s)
Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)
Mg(s) + Fe2+(aq) Mg2+(aq) + Fe(s)
a) Deduce a reactivity series based on the behaviour of the metals in the above equations
Iron can displace (or reduce) copper, so it is higher on the reactivity list. Copper can reduce silver. Magnesium
can reduce iron. Therefore in order of most reactive we have the series:
Mg > Fe > Cu > Ag
b) Define oxidation in terms of electron transfer, using one of the equations above as an example
Oxidation is defined as the loss of electrons. In the first equation Fe(s) is being oxidised to Fe2+(aq) as it is
losing electrons.
c) Define reduction in terms of electron transfer, using one of the equations above as an example
Reduction is defined as the gain of electrons. In the first equation Cu2+ is being reduced to Cu(s) as it is gaining
electrons.
d) Which is the strongest reducing agent and why?
Magnesium is the strongest reducing agent as it is highest on the reactivity series. It is most easily oxidised
metal in the this table, i.e. magnesium most easily loses electrons to form Mg2+(aq).
e) Which is the strongest oxidising agent and why?
Silver is the strongest oxidising agent as it is lowest on the reactivity series. It is most easily reduced ion, i.e.
silver most easily gains electrons to form Ag(s).
f)
Deduce whether a silver coin will react with aqueous magnesium chloride.
No, as the silver is the solid elemental species and is lower on the reactivity series than magnesium, and thus
could not displace it.
9.4 Exercises
1. Define the following terms:
a) electrode
A conductor that collects electrons as they are produced by oxidation reactions in a half cell and then supplies
them to the other half cell where they are used for a reduction reaction.
b) anode
The electrode at which oxidation occurs. In the voltaic cell it is the negatively charged electrode.
c) cathode
The electrode at which reduction occurs. In the voltaic cell it is the positively charged electrode.
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d) electrolyte
The liquid that conducts electricity as a result of the presence of positive or negative ions. Electrolytes are
molten ionic compounds or solutions containing ions, i.e. solutions of ionic salts or compounds that ionise in
solution.
e) half cell
Half of a voltaic cell. It is comprised of an electrode in a solution of ions.
f)
salt bridge
An electrical connection between two half cells that completes the electrical circuit and keeps the cells
electrically neutral.
g) voltaic cell
Comprised of two half cells, a voltaic cell produces electrical energy from spontaneous redox reactions.
h) fuel cell
An electrochemical energy conversion device, for example the hydrogen cell. Fuel cells differ from batteries.
They produce electricity from external sources of fuel, which needs to be continually replenished. The hydrogen
cell uses hydrogen as a fuel and oxygen as the oxidant.
2. Primary and secondary cells are both types of voltaic cells. What is the difference between a
primary cell and a secondary cell?
A primary cell is a voltaic cell in which the chemical reaction producing the electricity is not reversible and the
cell cannot be recharged by application of a charging current. For example, non-rechargeable batteries are
primary fuel cells. A secondary cell is a voltaic cell in which the chemical reaction producing the electricity is
reversible and the cell can be recharged. Rechargeable car batteries are made from secondary cells.
3. How does the salt bridge maintain charge balance in the solutions of the half cells?
The salt in the salt bridge contains positive and negative ions in solution. The negative ions move to the anode
to balance the charge from the positive ions produced by oxidation. Similarly the positive ions from the salt
move to the cathode to balance the charge due to the consumption of positive ions which are reduced to the
solid metal.
4. Draw a diagram for a voltaic cell with zinc and copper as the two electrodes and dilute sulfuric acid
as the electrolyte. This is known as the Daniell Cell.
Magnesium is more reactive (is higher in the reactivity series) than iron. This means that magnesium is a
stronger reducing agent, and thus is oxidised to produce Mg2+ cations and electrons. As electrons are produced
by the oxidation of magnesium, the magnesium half-cell forms the anode. Therefore the iron forms on the
cathode, and is reduced from the aqueous cations into the solid metal.
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a) Write balanced half-equations for the reactions occurring at each electrode.
Anode:
Zn(s) Zn2+(aq) + 2e
Cathode:
Cu (aq) + 2e Cu(s)
2+
b) On your diagram:
i)
label each electrode as either the anode or the cathode
ii) show the movement of the ions in the salt bridge
iii) assign a charge to each electrode
c) State which of the two metals is more reactive, give a reason for your choice.
Zinc is higher on the reactivity series than copper and therefore more reactive. It undergoes oxidation from Zn(s)
to Zn2+.
d) What would you expect to happen if
i)
A piece of zinc metal was placed in an aqueous solution of Cu2+ ions
Since zinc is more reactive zinc would undergo oxidation: Zn(s) Zn2+(aq) + 2e and the copper ions would be
2+
2+
2+
reduced to copper metal: Cu (aq) + 2e Cu(s). Overall: Zn(s) + Cu (aq) Zn (aq) + Cu(s)
ii) A piece of copper metal was placed in an aqueous solution of Zn2+ ions
2+
The copper is not reactive enough to reduce the Zn ions from the solution, so there would be no reaction.
e) What is the main energy conversion in the Daniell cell?
The spontaneous reactions between the electrodes and electrolytes produces a potential difference between
the two electrodes and this causes the flow of electrons between them. This is electricity! The energy
conversion is chemical energy to electrical energy.
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5. Consider the following diagram of a voltaic cell:
electron flow
cathode
anode
a) On the diagram, label the anode and cathode, indicate the direction of electron flow through the
wire and the movement of ions in the salt bridge
b) Write half equations for the anode and cathode and an overall cell reaction equation.
2+
-
Anode: Fe (s) Fe (aq) + 2e
Cathode: Ag2+(aq) + 2e- Ag(s)
Overall: Fe(s) + Ag2+(aq) Fe2+(aq) + Ag(s)
6. Draw a diagram to show an electrochemical cell using a strip of magnesium in a solution of
magnesium nitrate, and an iron rod in a solution of iron(ll) nitrate.
electron flow
magnesium
electrode
iron electrode
anode
cathode
magnesium
nitrate
solution
iron(ll)
nitrate
solution
a) Write half equations for the anode and cathode and an overall cell reaction equation.
Anode: Mg(s) Mg2+(aq) + 2eCathode: Fe2+(aq) + 2e- Fe(s)
2+
2+
Overall: Mg(s) + Fe (aq) Mg (aq) + Fe(s)
b) Label the anode, the cathode, the positive electrode, the negative electrode and the salt bridge.
c) Give the direction of flow for:
i)
electrons in external connecting wire.
in a voltaic cell electrons always flow from anode to cathode.
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ii) nitrate ions in the magnesium nitrate solution.
as more Mg2+(aq) form, the NO3- will remain in the LHS half cell.
iii) positive ions in the salt bridge if this contains KCl(aq)
2+
move to the cathode the replenish the loss of Fe
ions as they are reduced to Fe(s).
iv) iron(ll) ions in the iron(ll) nitrate solution.
2+
Fe
are converted to Fe(s) by reduction. The Fe(s) is deposited on the cathode.
d) Where do the electrons flowing in the external wire come from? Where are they going?
2+
The electrons come from the Mg(s) anode as Mg(s) is oxidised to Mg . The electrons move to the cathode
where they are used in the reduction of Fe2+ to Fe(s).
2+
7. The following cell was constructed using two metals X and Y and a solution of their salts, X
and
2+
Y . The electron flow when the cell was operating is shown on the diagram.
anode (-)
cathode (+)
Write a half equation for the reaction occurring at each electrode.
Anode: Y(s) Y2+(aq) + 2eCathode: X2+(aq) + 2e- X(s)
a) On the diagram:
i)
Show the movement of ions in the salt bridge
ii) Label each electrode as the anode or the cathode and assign a charge
b) Which of the two metals is more reactive?
2+
Y is the more reactive metal as it can reduce the other metal, X to X(s)
c) What would you expect to observe if
i)
2+
ions No reaction
2+
ions
A piece of metal X was placed in an aqueous solution of Y
ii) A piece of metal Y was placed in an aqueous solution of X
2+
The Y(s) metal will reduce X ions to X(s) and would be reduced in size as X(s) deposited.
8. Write the appropriate half-equations for the anode and cathode reactions for the following cells
given the overall reaction:
a) nickel-cadmium cell: Ni2O3(s) + Cd(s) + 3H2O Cd(OH)2(s) + 2Ni(OH)2(s)
2+
Anode: Cd(s) Cd
Cathode: Ni
3+
-
-
+ 2e
1e Ni
2+
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b) car battery (during discharge): PbO2(s) + Pb(s) + 4H+(aq) + 2SO2-4 (aq) 2PbSO4(s) + 2H2O(l)
Anode: Pb(s) Pb2+ + 2e4+
Cathode: Pb
-
2+
+ 2e Pb
9. Which of the following statements is correct?
A: In a voltaic cell the anode is the positive electrode.
B: Increasing the size of the electrode increases the cell voltage.
C: In a voltaic cell the electrons travel in the external wire to the cathode from the anode
D: Within a voltaic cell the anions migrate towards the cathode
Answer: D The positive ions are attracted to the electrons on the cathode.
A: the anode is negative, the electrons are moving out. B: size only increases the cell’s operating time, not its
voltage. C: the electrons travel from anode to cathode.
9.5 Exercises
1. Voltaic and electrolytic cells are both types of electrochemical cells. What is the difference between
a voltaic cell and an electrolytic cell?
In a voltaic cell spontaneous redox reaction generate electricity between two half-cells that contain ions which
are allowed to flow between the two solutions through the salt bridge. Two different metals must be used and
they must be separated from each other. It is the different reaction potentials of the two metals which generates
the electricity.
An electrolytic cell uses electricity from an outside source to generate a non-spontaneous redox reaction.
Another difference is that in the one cell the anode and the cathode are both suspended in the same solution
and they may be made from the same metal.
The suffix –lysis comes from the Greek word meaning to loosen or split up.
2. In a voltaic cell, the anode is negatively charged, it is the negative electrode. In an electrolytic cell,
the anode is positively charged, it is the positive electrode. Explain. (Hint, what types of ions always
flow to the anode/cathode in both a voltaic and electrolytic cells and what is their origin)
Oxidation occurs at the anode and reduction at the cathode in both cases. In the voltaic cell the oxidation
process leaves an excess of electrons at the anode, making it negatively charged. In an electrolytic cell the
anode is positive and the cathode negative, due to the external power supply attached. The cations are
attracted to the cathode (negative electrode) and the power supply provides enough potential for these ions to
be reduced. The anions are attracted to the anode (positively charged) and these are oxidised (gain electrons)
3. Define the terms:
a) Electrolysis
The process by which a current from an external source generates a non-spontaneous redox reaction.
b) Electroplating
The coating of a metal onto the surface of the cathode as the metal ions are reduced at the cathode into the
solid metal.
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4. How is the current conducted during electrolysis?
The current is produced by an external power supply and the electron flow is from anode to cathode.
5. For the electrolysis of molten sodium chloride, draw a fully labelled diagram to show how the ions
move when the circuit is complete, and give the electrode reactions.
The previous page shows a diagram for a general purpose electrolytic cell. For
electrolysis of molten NaCl the cell must be heated, the electrodes inert (usually
graphite) and the products Cl(g) and liquid Na(l) protected from air, water and
themselves.
In industry the molten sodium floats up from the cathode and the chlorine gas is kept
separate by the shield as shown in the diagram below.
cathode: Na+(l) + 1e- Na(l)
anode: 2Cl-(l) Cl2(g) + 2eb) Write half equations to show the reactions at each electrode when fused magnesium chloride is
electrolysed.
-
-
Anode: 2Cl (l) Cl2(g) + 2e
Cathode: Mg2+(l) + 2e- Mg(l)
Fused in this context means heated until molten. No electrolysis will occur until the
salt is molten, i.e. melted until liquid.
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6. A metallic object, to be plated with copper, is one electrode in an electrolytic cell: a bar of pure
copper is the other electrode.
a) State whether the object to be plated is the anode or cathode during the plating process, and
give your reason.
The object to be plated must be the cathode, as reduction always occurs at the cathode so the copper ions will
be reduced at the cathode to solid copper.
b) Name a solution that could be used as the electrolyte.
Copper sulfate, CuSO4(aq).
CuSO4(aq) provides mobile copper ions for transport of current and also for plating.
The SO42-(aq) are stable and the pure copper anode will react by forming Cu2+ ions
because it is above water, the other species present that could react, in an activity
series.
c) Give half equations for the electrode reactions.
In this case the chlorine anion will be oxidised to chlorine gas at the anode. Reduction of the copper ions into
solid copper will occur at the cathode.
anode: Cu(s) Cu2+(aq) + 2e (the electrode itself reacts)
cathode: Cu2+(aq) + 2e Cu(s)
7. The Edison cell is a storage battery that may receive hard treatment and yet gives good service for
years. It may be left uncharged indefinitely and still be recharged. It provides 1.3 V. Its electrolyte is
21% potassium hydroxide, KOH(aq), to which a small amount of lithium hydroxide, LiOH, is added.
The reaction on discharge is:
Fe(s) + 2Ni(OH)3 Fe(OH)2 + Ni(OH)2
a) Give the electrode reactions during discharge and charging.
discharge
3+
-
2+
cathode: Ni (aq) + e Ni (aq)
2+
-
anode: Fe(s) Fe (aq) + 2e
3+
Fe is higher on the reactivity series than Ni (aq) so is able to reduce Ni
3+
ions to Ni
2+
charging
cathode: Fe2+(aq) + 2e- Fe(s)
anode: Ni2+(aq) Ni3+ + e2+
3+
The external power supply forces the iron ions to be reduced back to Fe(s) and the Ni oxidised to Ni .
The Edison cell
The Edison cell is a rechargeable battery. When it is discharging (i.e. being used) it
acts as a voltaic cell. The spontaneous redox reactions that occur as a result of the
different reactivities of the metals and their ions generate electricity. When the battery
is being charged however, it acts as an electrolytic cell. The external power supply
forces the redox reactions to happen in the opposite direction so that it is ready for use
again.
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b) For discharge, which electrode is the anode? Is this the same as for the charging? Explain.
The iron electrode is the anode. It is being oxidised. No. During charging the nickel ions, Ni2+(aq), are being
oxidised so the electrode in their solution is the anode, not the Fe.
Oxidation occurs at the anode and reduction at the cathode, for both discharge AND
for charging (i.e. when acting as voltaic or electrolytic cells). The difference is which
metal or ion is being reduced. When discharging, the most reactive metal is oxidised
at the anode in a spontaneous redox reaction. When charging the most reactive metal
ion is reduced at the cathode by the external power supply.
8. Electrolysis occurs during the recharging of a car battery.
During discharge the overall reaction in a car battery is:
+
2-
PbO2(s) + Pb(s) + 4H (aq) + 2SO4 (aq) 2PbSO4(s) + 2H2O(l)
a) Rewrite this equation to show the overall reaction during charging.
2PbSO4(s) + 2H2O(l) PbO2(s) + Pb(s) + 4H+(aq) + 2SO42- (aq)
b) Write half-equations for the anode and cathode reactions occurring during charging.
anode: Pb(s) Pb2+ + 2ecathode: PbO2 + 4H+(aq) + 2e- Pb2+(aq) + 2H2O
9. Why do solid salts not conduct electricity? What are two methods by which salt can be used to
conduct electricity?
Solid salts do not conduct electricity, as the ions are not able to physically move when they are held in the solid
state in the crystal lattice, and there can be no movement of electrons. Aqueous salts and molten salts are both
able to conduct electricity because the ions and electrons are able to move freely.
10. What products are discharged at the anode and cathode during the electrolysis of the following
molten salts:
a) KBr
Cathode: K(l)
Anode: Br2(g)
b) NaCl
Cathode: Na(l)
Anode: Cl2(g)
c) MgCl2
Cathode: Mg(l)
Anode: Cl2(g)
The metals are molten because of the high temperatures required to keep the salts in
the liquid state. These active metals must not be allowed to come into contact with air
or moisture or with the elemental halogens, so they are stored under paraffin oil.
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