ANSWERS CHEMISTRY 3.7
1
Answers to Level 3 Chemistry 3.7
Redox Basics (pages 7-8)
Oxidation Number (pages 9-10)
1.
O: The reaction involves the addition of oxygen.
1.
2.
R: The Al ions in the molten alumina (Al2O3) gain electrons.
3.
O: The reaction involves the addition of oxygen.
4.
R: The reaction involves the removal of oxygen or the Fe2+
and Fe3+ ions in the magnetite gain electrons.
5.
O: Electrons are removed from the fluoride ion:
2F–(l)
F2(g) + 2e
6.
O: The reaction involves the addition of oxygen to the
aluminium or the aluminium atoms lose electrons.
a.
d.
g.
j.
m.
p.
s.
v.
y.
7.
R: The silver ions gain electrons to form silver atoms. The
silver ion is an oxidant and is reduced.
z.
MnO(OH): +3
8.
R: The oxidation number/state of manganese decreases
from +4 to +2.
2.
a.
d.
Cr3+:+3
Cr2O72–: +6
9.
O: The oxidation state of carbon increases from -2 to +4 or
the reaction involves the addition of oxygen.
3.
a.
b.
10.
R: The oxidation state of titanium decreases from +4 to zero
or the reaction involves the removal of oxygen.
R: addition of electrons
O: oxidation number of manganese increases from +4
to +7
O: oxidation number of carbon increases from +3 to +4
No change: oxidation number of chromium is +6 in
both ions
O and R: oxidation number of carbon increases from
-2 to -1 and oxidation number of chlorine decreases
from 0 to -1
O: electrons are removed
R: addition of electrons
O and R: removal of electrons from zinc atoms and
addition of electrons to iron(II) ions
R: oxidation number of oxygen decreases from -1
to -2
O: oxidation number of oxygen increases from -1 to 0
Neither: no change in the oxidation number of sulfur
O and R: oxidation number of magnesium increases
from zero to +2 and titanium oxidation number
decreases from +4 to 0
Neither: no change in the oxidation number of any
species
3+
c.
d.
11.
R: The reaction involves the addition of hydrogen.
12.
O and R: The oxidation state of nitrogen increases from -3 to
zero and the oxidation state of chromium reduces from +6 to
+3: (NH4)2Cr2O7(s)
N2(g) + Cr2O3(s) + 4H2O(l)
e.
13.
Neither O nor R: Oxygen and ozone are both forms of
elemental oxygen.
14.
R: The sulfur dioxide removes oxygen from the aqueous
solution therefore denying bacteria the opportunity to
survive.
f.
g.
h.
i.
15.
O and R: The oxidation state of chlorine increases from
+1 (OCl–) to +5 (ClO3–) and the oxidation state of chlorine
reduces from +1 (OCl–) to -1 (Cl–):
3OCl–(aq)
ClO3–(aq) + 2Cl–(aq)
j.
k.
l.
16.
O and R: The oxidation state of carbon increases from zero
(C) to +2 (CO) and decreases from +4 (CO32–) to -1(C22–):
CaCO3(s) + 4C(s)
CaC2(s) + 3CO(g)
m.
17.
O: The ethanol loses hydrogen in the reaction.
NaCl: +1
ClO3–: +5
SO32–: +4
S8: 0
H2O2:-1
BrCl: +1
NO: +2
NH3: -3
K3Fe(CN)6: +3
b. CaCl2: -1
c. OCl–: +1
e. H2S: -2
f. SO2: +4
h. SO42–: +6
i. S2O32–: +2
k. Fe2O3: +3
l. OF2: -1
n. XeO2: +4
o. BrCl3: +3
r. NO3–: +5
q. NO2: +4
t. N2: 0
u. NO2–: +3
w. NH4+: -3
x. NaBiO3 : +5
(the complex ion Fe(CN)63– = 1 x Fe3+
and 6 x CN–)
b.
Cr2O3: +3
c.
CrO42–: +6
Redox Equations (page 14)
1.
Because: • C3H5N3O9 loses oxygen;
• Both the nitrogen and oxygen which are in a combined form in the nitroglycerin become elemental products (oxidation
number of N in compound decreases and oxidation number of O in compound increases).
2.
a.
b.
c.
d.
e.
f.
g.
The hydrogen gains oxygen.
The oxidation number of copper changes from +2 in Cu2+ to +1 in CuI (reduction); The oxidation number of iodine changes from
-1 in I– to 0 in I2 (oxidation).
The nitrogen dioxide loses some oxygen; the ON of nitrogen changes from +4 to +2.
The hydrogen peroxide loses some oxygen; the ON of oxygen changes from -1 to -2.
The hydrogen (in water) loses oxygen; the ON of calcium changes from 0 to +2.
The oxidation number of sulfur changes from -2 in H2S to 0 in S (oxidation); The oxidation number of iron changes from +3 in
Fe3+ to +2 in Fe2+ (reduction).
The aluminium gains oxygen; the Fe2O3 loses oxygen; The oxidation number of aluminium changes from 0 in Al to +3 in Al2O3
(oxidation); The oxidation number of iron changes from +3 in Fe2O3 to 0 in Fe (reduction).
Redox Reactions Revision (pages 15-17)
1.
Observations: The magnesium burns with a brilliant white
light, giving off a white smoke and great heat. A white powder
residue remains on the tongs.
Half equations:
Mg
Mg2+ + 2e
Name of species: magnesium atoms
magnesium ions
Colour of species:
silver
colourless
Half equations:
O2 + 4e
2O2–
Name of species: oxygen molecules
oxide ions
Colour of species:
colourless
white
Species oxidised: magnesium atoms
Species reduced: oxygen molecules
Oxidant:
oxygen gas
Reductant:
magnesium metal
Full equation:
2Mg + O2
2Mg2+ + 2O2–
2.
Observations: Colourless bubbles rise from the surface of
the zinc and the zinc eventually 'disappears'. The solution
remains colourless.
Half equations:
Zn
Zn2+ + 2e
Name of species:
zinc atoms
zinc ions
Colour of species:
silver
colourless
Half equations:
2H+ + 2e
H2
Name of species: hydrogen ions
hydrogen molecules
Colour of species:
colourless
colourless
Species oxidised: zinc atoms
Species reduced: hydrogen ions
Oxidant:
hydrogen ions
Reductant:
zinc metal
Full equation:
Zn + 2H+
Zn2+ + H2
Spectator ions:
The Cl– ions from the hydrochloric acid.
2
CHEMISTRY 3.7 ANSWERS
3.
Observations: A grey solid deposits on the copper metal and
the colourless solution slowly develops a blue colour.
Half equations:
Ag+ + e
Ag
Name of species:
silver ions
silver atoms
Colour of species:
colourless
silver
Half equations:
Cu
Cu2+ + 2e
Name of species: copper atoms
copper ions
Colour of species:
‘coppery’
blue
Species oxidised: copper atoms
Species reduced: silver ions
Oxidant:
silver ions
Reductant:
copper metal
2Ag + Cu2+
Full equation:
2Ag+ + Cu
5.
Observations: The dichromate solution changes colour from
orange to green. When barium chloride solution is added, a
white precipitate is produced.
Half equations: 14H+ + Cr2O72– + 6e
2Cr3+ + 7H2O
Name of species: dichromate ions
chromium(III) ions
Colour of species:
orange
green
Half equations: H2O + SO32–
2H+ + SO42– + 2e
Name of species:
sulfite ions
sulfate ions
Colour of species: colourless
colourless
Species oxidised: sulfite ions
Species reduced: dichromate ions
Oxidant:
acidified dichromate ions
Reductant:
sulfite ions
Full equation: 3SO32– + Cr2O72– + 8H+
2Cr3+ + 3SO42– + 4H2O
A white precipitate was formed when barium chloride solution
was added.
BaSO4(s)
Ba2+(aq) + SO42–(aq)
4.
Observations: The purple permanganate solution is rapidly
decolourised. When the thiocyanate solution is added, the
solution turns dark red.
Mn2+ + 4H2O
Half equations: 8H+ + MnO4– + 5e
Name of species: permanganate ions manganese(II) ions
Colour of species:
purple
colourless
Fe3+ + e
Half equations:
Fe2+
Name of species:
iron(II) ions
iron(III) ions
Colour of species: pale green
pale yellow
Species oxidised: iron(II) ions
Species reduced: permanganate ions
Oxidant:
acidified permanganate ions
Reductant:
iron(II) ions
Full equation: 5Fe2+ + MnO4– + 8H+
5Fe3+ + Mn2+ + 4H2O
Spectator ions:
K+ and SO42– ions
FeSCN2+
Fe3+ + SCN–
To facilitate the formation of Mn2+ ions. (Note that some acids
such as hydrochloric acid would be unsuitable as they react
with MnO4– ions.)
b.
d.
f.
h.
j.
oxidant:
oxidant:
oxidant:
oxidant:
oxidant:
Redox Equation Exercises (pages 18-22)
1.
a.
c.
e.
g.
i.
k.
oxidant:
oxidant:
oxidant:
oxidant:
oxidant:
oxidant:
MnO4–
H+/MnO4–
Cu2+
HNO3
MnO2
MnO4–
reduced form:
reduced form:
reduced form:
reduced form:
reduced form:
reduced form:
MnO2
Mn2+
Cu
NO2
Mn2+
MnO42–
Cr2O72–
I2
Fe3+
Cl2
HNO3
reduced form:
reduced form:
reduced form:
reduced form:
reduced form:
Cr3+
I–
Fe2+
Cl–
NO
2.
a.
Zn(s)
Zn2+(s) + 2e and Cl2(g) + 2e
2Cl–(s)
Overall: Zn(s) + Cl2(g)
Zn2+(s) + 2Cl–(s)
Observations: A vigorous reaction occurs producing heat and light. The green gas is decolourised and a white solid is formed.
b.
Mg(s)
Mg2+(s) + 2e and I2(g) + 2e
2I–(s)
Overall: Mg(s) + I2(g)
Mg2+(s) + 2I–(s)
Observations: A vigorous reaction occurs producing heat and light. The purple gas is decolourised and a white solid is formed.
c.
Fe(s)
Fe3+(s) + 3e (x2) and Cl2(g) + 2e
2Cl–(s) (x3)
Overall: 2Fe(s) + 3Cl2(g)
2Fe3+(s) + 6Cl–(s)
Observations: The green gas is decolourised as the iron burns in the chlorine and a red-brown smoke is produced.
d.
MnO4–(aq) + 8H+(aq) + 5e
Mn2+(aq) + 4H2O(l) (x2) and SO32–(aq) + H2O(l)
Overall: 2MnO4– + 6H+(aq) + 5SO32–(aq)
2Mn2+(aq) + 3H2O(l) + 5SO42–(aq)
Observations: The purple solution is rapidly decolourised.
e.
Cr2O72–(aq) + 14H+(aq) + 6e
2Cr3+(aq) + 7H2O(l) and SO2(g) + 2H2O(l)
Overall: Cr2O72–(aq) + 2H+(aq) + 3SO2(g)
2Cr3+(aq) + 3SO42–(aq) + H2O(l)
Observations: The orange solution turns green.
f.
2I–(aq)
I2(aq) + 2e and H2O2(aq) + 2H+(aq) + 2e
Observations: The colourless solution turns brown.
g.
I2(aq) + 2e
2I–(aq) and 2S2O32–(aq)
S4O62–(aq) + 2e
Observations: The brown solution turns colourless.
h.
Cu2+(aq) + e
Cu+(aq) (x2) and 2I–(aq)
I2(aq) + 2e
Note: Cu+(aq) is unstable in aqueous solution and precipitates out as CuI(s).
Overall: 2Cu2+(aq) + 4I–(aq)
2CuI(s) + I2(aq)
Observations: The blue solution turns yellow-brown and a white precipitate forms.
i.
Half equations are not appropriate in this example as CO is a molecular substance.
Zn(l) + CO2(g)
Overall equation is: ZnO(s) + CO(g)
Observations: The white solid becomes a silvery liquid and a colourless gas is produced which turns limewater milky.
j.
NO3–(aq) + 2H+(aq) + e
NO2(g) + H2O(l) (x2) and Zn(s)
Zn2+(aq) + 2e
–
+
2+
Overall: Zn(s) + 2NO3 (aq) + 4H (aq)
Zn (aq) + 2NO2(g) + 2H2O(l)
Observations: A vigorous reaction producing a choking brown gas and a colourless solution.
k.
IO3–(aq) + 6H+(aq) + 6e
I–(aq) + 3H2O(l) and 2I–(aq)
Overall: IO3–(aq) + 6H+(aq) + 5I–(aq)
3H2O(l) + 3I2(aq)
l.
Half equations are not appropriate in this example as H2 is a molecular substance.
Overall: Fe2O3(s) + 3H2(g)
2Fe(l) + 3H2O(g)
Observations: The red solid becomes a silvery liquid.
m.
H2O2(aq) + 2H+(aq) + 2e
2H2O(l) and SO32–(aq) + H2O(l)
SO42–(aq) + 2H+(aq) + 2e
Overall: H2O2(aq) + SO32–(aq)
SO42–(aq) + H2O(l)
Observations: No change is observed as both reactants and products are colourless.
n.
Cl2(aq) + 2e
2Cl–(aq) and 2I–(aq)
I2(aq) + 2e
Observations: The colourless solution turns brown.
SO42–(aq) + 2H+(aq) + 2e (x5)
SO42–(aq) + 4H+(aq) + 2e (x3)
2H2O(l) Overall: 2I–(aq) + H2O2(aq) + 2H+(aq)
Overall: I2(aq) + 2S2O32–(aq)
I2(aq) + 2H2O(l)
2I–(aq) + S4O62–(aq)
I2(aq) + 2e (x3)
Observations: The colourless solution turns brown.
Overall: Cl2(aq) + 2I–(aq)
2Cl–(aq) + I2(aq)
ANSWERS CHEMISTRY 3.7
o.
p.
3.
a.
b.
4.
a.
b.
3
Cu(s)
Cu2+(aq) + 2e (x3) and NO3–(aq) + 4H+(aq) + 3e
NO(g) + 2H2O(l) (x2)
3Cu2+(aq) + 2NO(g) + 4H2O(l)
Overall: 3Cu(s) + 2NO3–(aq) + 8H+(aq)
Observations: The copper reacts and a blue solution results. A colourless gas is given off which turns brown on exposure to air.
Mg(s)
Mg2+(aq) + 2e and Cu2+(aq) + 2e
Cu(s)
Overall: Mg(s) + Cu2+(aq)
Mg2+(aq) + Cu(s)
Observations: The blue solution fades and a dark brown/black deposit appears on the magnesium.
Cr3+ ions (solution of chromium(III) sulfate)
Cr2O72–(aq) + 14H+(aq) + 6e
2Cr3+(aq) + 7H2O(l)
Fe2+(aq)
Fe3+(aq) + e (x6)
Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq
2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
sulfur, S
Cl2(g) + 2e
H2S(aq)
Cl2(g) + H2S(aq)
2Cl–(aq)
2H+(aq) + S(s) + 2e
2H+(aq) + S(s) + 2Cl–(aq)
i.
Fe(s)
Fe2+(aq) + 2e (x2)
2Fe2+(aq) + 4OH–(aq)
2Fe(s) + O2(g) + 2H2O(l)
c.
ii.
chlorine, Cl2
2Cl–(aq)
Cl2(g) + 2e
PbO2(s) + 4H+(aq) + 2e
Pb2+(aq) + 2H2O(l)
Cl2(g) + Pb2+(aq) + 2H2O(l)
2Cl–(aq) + PbO2(s) + 4H+(aq)
O2(g) + 2H2O(l) + 4e
4OH–(aq)
Redox Extension Exercises (pages 23-27)
1.
Observations: The purple colour of the permanganate
2.
solution disappears and a brown precipitate forms. When
potassium thiocyanate solution is added to the filtrate, a dark
red solution forms.
Half equations: 2H2O + MnO4– + 3e
MnO2 + 4OH–
Name of species: permanganate ions manganese(IV) oxide
(manganese dioxide)
Colour of species:
purple
brown/black
Half equations:
Fe2+
Fe3+ + e
Name of species:
iron(II) ions
iron(III) ions
Colour of species:
pale green
pale yellow
Species oxidised: iron(II) ions
Species reduced: permanganate ions
Oxidant:
permanganate ions
Reductant:
iron(II) ions
Full equation: MnO4– + 2H2O + 3Fe2+
MnO2 + 3Fe3+ + 4OH–
When thiocyanate ions were added, a red solution was produced.
FeSCN2+
Fe3+ + SCN–
Observations: The purple colour fades to give a colourless
solution and a few small colourless bubbles may be seen.
Half equations: 8H+ + MnO4– + 5e
Mn2+ + 4H2O
Name of species: permanganate ions manganese(II) ions
Colour of species:
purple
colourless
Half equations:
C2O42–
2CO2 + 2e
Name of species:
oxalate ions carbon dioxide molecules
Colour of species: colourless
colourless
Species oxidised: oxalate ions
Species reduced: permanganate ions
Oxidant:
acidified permanganate ions
Reductant:
oxalate ions
Full equation:
2MnO4– + 16H+ + 5C2O42–
2Mn2+ + 8H2O + 10CO2
3.
Observations: A pale green,pungent gas is given off which
4.
turns starch-iodide paper blue-black. Black suspension
decolourises with time.
Half equations:
4H+ + MnO2 + 2e
Mn2+ + 2H2O
Name of species: manganese(IV) ions manganese(II) ions
Colour of species:
brown/black
colourless
Half equations:
2Cl–
Cl2 + 2e
Name of species:
chloride ions
chlorine ions
Colour of species:
colourless
pale green
Species oxidised: chloride ions
Species reduced: manganese(IV) oxide
Oxidant:
manganese(IV) oxide
Reductant:
chloride ions
Full equation: MnO2 + 4H+ + 2Cl–
Mn2+ + Cl2 + 2H2O
The starch-iodide paper turned blue-black.
Cl2 + 2I–
2Cl– + I2 then I2 + starch
blue-black reaction product.
Observations: A brown solution and possibly a grey
precipitate forms.
Half equations:
2I–
I2 + 2e
Name of species:
iodide ions
iodine molecules
Colour of species:
colourless
brown (in solution)
Half equations:
2H+ + H2O2 + 2e
2H2O
Name of species: hydrogen peroxide water molecules
molecules
Colour of species: colourless
colourless
Species oxidised: iodide ions
Species reduced: hydrogen peroxide molecules
Oxidant:
hydrogen peroxide molecules
Reductant:
iodide ions
2H2O + I2
Full equation:
H2O2 + 2I– + 2H+
The colourless iodide solution turned brown.
5.
Observations: The purple permanganate solution fades until
6.
colourless and bubbles of a colourless gas are given off.
Half equations:
8H+ + MnO4– + 5e
Mn2+ + 4H2O
Name of species: permanganate ions manganese(II) ions
Colour of species:
purple
colourless
Half equations:
H2O2
2H+ + O2 + 2e
Name of species: hydrogen peroxide
oxygen molecules
molecules
Colour of species:
colourless
colourless
Species oxidised: hydrogen peroxide molecules
Species reduced: permanganate ions
Oxidant:
acidified permanganate ions
Reductant:
hydrogen peroxide molecules
Full equation: 2MnO4– + 6H+ + 5H2O2
2Mn2+ + 8H2O + 5O2
7.
Observations: A grey precipitate forms and the colourless
solution turns brown.
Half equations:
6H+ + IO3– + 6e
I– + 3H2O
Name of species:
iodate ions
iodide ions
Colour of species: colourless
colourless
Half equations:
2I–
I2 + 2e
Name of species:
iodide ions
iodine molecules
Colour of species:
colourless
brown
Species oxidised: iodide ions
Species reduced: iodate ions
Oxidant:
iodate ions
Reductant:
iodide ions
–
Full equation: IO3 + 6H+ + 5I–
3I2 + 3H2O
The solution turned brown and a grey precipitate may be
observed.
Observations: The brown colour of the iodine solution
disappears. The resulting solution is colourless.
Half equations:
2S2O32–
S4O62– + 6e
Name of species: thiosulfate ions
tetrathionate ions
Colour of species:
colourless
colourless
Half equations:
I2 + 2e
2I–
Name of species: iodine molecules
iodide ions
Colour of species: brown (in solution)
colourless
Species oxidised: thiosulfate ions
Species reduced: iodine molecules
Oxidant:
iodine molecules
Reductant:
thiosulfate ions
Full equation: 2S2O32– + I2
S4O62– + 2I–
The brown colour of the iodine solution disappeared.
Observations: The blue solution turns brown and a white
precipitate is produced.
Half equations:
2I–
I2 + 2e
Name of species: iodate ions
iodine molecules
Colour of species: colourless
brown (in solution)
Half equations:
I– + Cu2+ + e
CuI
Name of species: copper(II) ions
copper(I) iodide
(an ionic solid)
+ iodide ions
Colour of species: blue and colourless
white
Species oxidised: iodide ions
Species reduced: copper(II) ions
Oxidant:
copper(II) ions
Reductant:
iodide ions
Full equation: 2Cu2+ + 4I–
2CuI + I2
8.
4
CHEMISTRY 3.7 ANSWERS
9.
Observations: The copper metal reacts and ‘disappears’,
a brown gas with a pungent smell is given off and a green
solution is formed.
Half equations:
Cu
Cu2+ + 2e
Name of species: copper atoms
copper(II) ions
Colour of species: 'coppery'
blue
Half equations: 2H+ + NO3– + e
NO2 + H2O
Name of species: nitrate ions
nitrogen dioxide molecules
Colour of species: colourless
brown
Species oxidised: copper atoms
Species reduced: nitrate ions
Oxidant:
acidified nitrate ions (nitric acid)
Reductant:
copper atoms
Full equation: Cu + 2NO3– + 4H+
Cu2+ + 2NO2 + 2H2O
10.
Spontaneous Reactions (page 28)
1.
2.
3.
Mg(s)
Fe(s)
Sn(s)
Mg2+(aq) + 2e
Fe2+(aq) + 2e
Sn2+(aq) + 2e
and
and
and
Ag+(aq) + e
Cu2+(aq) + 2e
Pb2+(aq) + 2e
Ag(s)
Cu(s)
Pb(s)
Observations: A pale green pungent gas is given off which
turns starch-iodide paper blue-black. A yellow residue is left
in the bottom of the test tube.
Cl2 + 2e
Half equations:
2Cl–
Name of species:
chloride ions
chlorine molecules
Colour of species:
colourless
pale green
Half equations:
4H+ + PbO2 + 2e
Pb2+ + H2O
Name of species: lead(IV) oxide
lead(II) ions
(ionic solid)
Colour of species:
brown
colourless
Species oxidised: chloride ions
Species reduced: lead(IV) oxide
Oxidant:
lead(IV) oxide
Reductant:
chloride ions
Full equation: PbO2 + 4H+ + 2Cl–
Pb2+ + Cl2 + 2H2O
Mg(s) + 2Ag+(aq
Fe(s) + Cu2+(aq)
Sn(s) + Pb2+(aq
Mg2+(aq) + 2Ag(s)
Fe2+(aq) + Cu(s)
Sn2+(aq) + Pb(s)
Electron Transfer (pages 30-31)
Results/Observations
• A reading of about 2 mA appears on the milliammeter.
• When thiocyanate solution is added to the arm containing the iron(II) solution, a pink colour is produced.
• On running the cell for 5 minutes, the purple permanganate colour gets weaker and the pink colour intensifies (can be left set up overnight
and observed next day).
Discussion Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
To prevent general mixing of the solutions but still allow
electrical contact.
2 mA
From the negative (iron(II)) electrode to the positive (MnO4–)
electrode.
Because it contains more ions and conducts better than
water.
Iron(II) ammonium sulfate: Fe2+, NH4+, SO42–
potassium permanganate: K+, MnO4–
dilute sulfuric acid:
H+, HSO4–, SO42–
Negative electrode: oxidation (Fe2+
Fe3+)
Positive electrode: reduction (MnO4–
MnO2)
Fe2+
Fe3+ + e
MnO4– + 2H2O + 3e
MnO2 + 4OH– *
–
* (If the MnO4 is in acid solution, Mn2+ is produced).
The Fe2+ ions were oxidised to Fe3+ ions at the negative
electrode and the SCN– ions reacted with these to form a red
complex: Fe3+ + SCN–
FeSCN2+
Because the purple MnO4– ions were reduced at the positive
electrode to brown MnO2.*
* (if left overnight to react, brown MnO2 can be observed on
the cotton wool plug.)
As the circuit is allowed to run, the concentration of purple
MnO4– ions decreases as more Mn2+ ions are produced
and the concentration of Fe2+ ions decreases as more Fe3+
ions are produced – therefore, at this electrode [FeSCN2+]
increases.
Electrons from the oxidation of Fe2+ ions at the negative
electrode travel through the wire and are used in the reduction
of MnO4– ions at the positive electrode.
12.
The standard reduction potentials for the cobalt and
iron(II/III) half cells are –0.28 V and 0.77 V respectively.
⇒ The tendency is for oxidation to occur at the cobalt half
cell: Co(s)
Co2+(aq) + 2e
and reduction to occur at the iron half cell:
Fe3+(aq) + e
Fe2+(aq)
The overall cell reaction is:
2Fe3+(aq) + Co(s)
2Fe2+(aq) + Co2+(aq)
The cobalt strip is the negative electrode in the cell and
the platinum wire is the positive electrode. When the cell
is operating, electrons flow from the cobalt electrode to
the platinum electrode. Under standard conditions, the cell
voltage will be 1.05 V.
As the cell runs, the concentration of pink Co2+ ions will
increase and the concentration of yellow-orange Fe3+ ions
will decrease as they are replaced by pale green Fe2+ ions.
It may be noticed that the cobalt cell gets pinker and the iron
cell turns greener.
Electrochemical Construction (page 32)
Voltmeter to
measure energy
of electrons
H
iron strip
( + electrode)
9
magnesium strip
( – electrode)
Salt bridge (filter
paper soaked in
potassium sulfate
solution)
iron(II) sulfate
solution
(1 mol L–1)
magnesium
sulfate solution
(1 mol L–1)
Making Electrochemical Cells (pages 34-37)
Results (cell voltages for 1 mol L–1 solutions are indicated)
Cell
Cell Voltage/V
a
Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
1.10*
b
Cu(s)/Cu2+(aq)//I2(aq),I–(aq)/C(s)
0.20
c
C(s)/I (aq),I2(aq)//Fe (aq),Fe (aq)/C(s)
0.23
d
1.30
e
Zn(s)/Zn2+(aq)//I2(aq),I–(aq)/C(s)
Zn(s)/Zn2+(aq)//Fe3+(aq),Fe2+(aq)/C(s)
1.53
f
Cu(s)/Cu2+(aq)//Fe3+(aq),Fe2+(aq)/C(s)
0.43
–
3+
2+
* For Zn2+ and Cu2+ ion concentrations of 0.1 mol L–1, the cell voltage
will be about 1.13 V at 25ºC.
Task: Draw Your Cells
Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
e–
9
wire
–
zinc
metal
salt bridge
electron flow
+
copper
metal
cotton wool
plugs
Zn2+ ions
0.1 mol L–1
Cu2+ ions
0.1 mol L–1
ANSWERS CHEMISTRY 3.7
C(s)/I–(aq),I2(aq)//Fe3+(aq),Fe2+(aq)/C(s)
wire
e–
9
electron flow
–
+
salt bridge
graphite
electrode
graphite
electrode
cotton wool
plugs
Fe3+(aq)
and Fe2+(aq)
both at
0.1 mol L–1
I2(aq) and
I–(aq) both at
0.1 mol L–1
Discussion
• A salt bridge contains ions which are free to move into either half
cell without interfering with the reactions occurring in the half cells.
When the cell is operating, ions move within the internal circuit
through the salt bridge. Consequently, the salt bridge will become
contaminated with cations and anions from each half cell. A salt
bridge so contaminated could interfere with the reactions in other
half cells if it were not replaced.
• Cell voltages calculated using Eºcell = EºRHE – EºLHE (or Eºred – Eºox):
(a)
Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
Eºcell = +0.34 – (-0.76) = 1.10 V
(b)
Cu(s)/Cu2+(aq)//I2(aq),I–(aq)/C(s)
Eºcell = +0.54 – (+0.34) = 0.20 V
(c)
C(s)/I–(aq),I2(aq)//Fe3+(aq),Fe2+(aq)/C(s)
Eºcell = +0.77 – (+0.54) = 0.23 V
(d)
Zn(s)/Zn2+(aq)//I2(aq),I–(aq)/C(s)
Eºcell = +0.54 – (-0.76) = 1.30 V
(e)
Zn(s)/Zn2+(aq)//Fe3+(aq),Fe2+(aq)/C(s)
Eºcell = +0.77 – (-0.76) = 1.53 V
(f)
Cu(s)/Cu2+(aq)//Fe3+(aq),Fe2+(aq)/C(s)
Eºcell = +0.77 – (+0.34) = 0.43 V
Representing Cells (page 40)
1.
4.
Pt/H2(g), H+(aq) // Cu2+(aq)/Cu(s)
Mg(s)/Mg2+(aq) // Ag+(aq)/Ag(s)
2.
5.
• Cell reactions:
(a)
Cu2+(aq) + 2e
Zn(s)
Cu2+(aq) + Zn(s)
(b)
I2(aq) + 2e
Cu(s)
I2(aq) + Cu(s)
(c)
Fe3+(aq) + e
2I–(aq)
2Fe3+(aq) + 2I–(aq)
(d)
I2(aq) + 2e
Zn(s)
I2(aq) + Zn(s)
(e)
Fe3+(aq) + e
Zn(s)
2Fe3+(aq) + Zn(s)
(f)
Fe3+(aq) + e
Cu(s)
2Fe3+(aq) + Cu(s)
Cu(s)
Zn2+(aq) + 2e
Cu(s) + Zn2+(aq)
2I–(aq)
Cu2+(aq) + 2e
2I–(aq) + Cu2+(aq)
Fe2+(aq)
(x2)
I2(aq) + 2e
2Fe2+(aq) + I2(aq)
2I–(aq)
Zn2+(aq) + 2e
2I–(aq) + Zn2+(aq)
Fe2+(aq)
(x2)
Zn2+(aq) + 2e
2Fe2+(aq) + Zn2+(aq)
Fe2+(aq)
(x2)
Cu2+(aq) + 2e
2Fe2+(aq) + Cu2+(aq)
• Chart: Strongest oxidant
Fe3+(aq) + e
I2(aq) + 2e
Cu2+(aq) + 2e
Zn2+(aq) + 2e
Weakest oxidant
Weakest reductant
Fe2+(aq)
2I–(aq)
Cu(s)
Zn(s)
Strongest reductant
5
• Predictions:
Cu(s) + I2(aq)
(a)
Would not expect Cu2+(aq) + 2I–(aq)
– However, in practice, I– reduces Cu2+ to Cu+ and Cu+
reacts with I– to form a white precipitate of CuI in a red-brown
solution of iodine: 2Cu2+(aq) + 4I–(aq)
2CuI(s) + I2(aq)
(b)
Would expect a reaction where the yellow colour of the
Fe3+ solution fades to almost colourless/pale green:
2Fe3+(aq) + Zn(s)
2Fe2+(aq) + Zn2+(aq)
(c)
Would expect a reaction where the red-brown colour of the
iodine solution fades to colourless:
2I–(aq) + Zn2+(aq)
I2(aq) + Zn(s)
(d)
Would predict no reaction between Fe2+ ions and Cu2+ ions.
Zn(s)/Zn2+(aq) // Pb2+(aq)/Pb(s)
Fe(s)/Fe2+(aq) // I2(aq), I–(aq)/C(s)
3.
6.
Al(s)/Al3+(aq) // Ni2+(aq)/Ni(s)
Pt(s)/H2(g), H+(aq) // Fe3+(aq),Fe2+(aq)/Pt(s)
Cells and Reduction Potentials (page 41)
1.
a.
b.
c.
E°cell = E°RHE – E°LHE = -0.44 – (-0.76) = +0.32 V
E°cell = E°RHE – E°LHE = 0.77 – 0.54 = +0.23 V
E°cell = E°RHE – E°LHE = 1.49 – 0.08 = +1.41 V
Zn(s) + Fe2+(aq)
Zn2+(aq) + Fe(s)
–
3+
2I (aq) + 2Fe (aq)
I2(aq) + 2Fe2+(aq)
–
+
2MnO4 (aq) + 16H (aq) + 10S2O32–(aq)
2Mn2+(aq) + 8H2O(l) + 5S4O62–(aq)
Testing Your Predictions! (pages 42-43)
Predictions
Calculations
1.
2.
3.
4.
5.
6.
Eºcell = Eºred – Eºox = +0.34 – (-0.76) = +1.10 V
Eºcell = +0.77 – (+0.34) = +0.43 V
Eºcell = +0.77 – (+0.54) = +0.23 V
Eºcell = +0.54 – (+0.77) = -0.23 V
Eºcell = +0.34 – (+0.54) = -0.20 V
Eºcell = -0.76 – (+0.77) = -1.53 V
Reaction Prediction:
spontaneous
spontaneous
spontaneous
non-spontaneous
non-spontaneous
non-spontaneous
Spontaneous Reactions
Non-Spontaneous
Reactions
Zn(s) + Cu2+(aq)
Cu(s) + Fe3+(aq)
I–(aq) + Fe3+(aq)
Fe2+(aq) + I2(aq)
Cu2+(aq) + I–(aq)
Fe2+(aq) + Zn2+(aq)
Results
Reactants
Cu(s), Fe3+(aq)
I–(aq), Fe3+(aq)
Fe2+(aq), I2(aq)
Observations
A brown colouration appears on the surface of the zinc and the blue colour of
the copper sulfate starts to fade.
The yellow-orange iron(III) solution fades and a blue solution forms.
The yellow/orange iron(III) solution becomes a red-brown solution.
No visible change is observed. The iodine solution remains brown.
Cu2+(aq), I–(aq)
The blue copper sulfate solution turns brown and a white precipitate is formed.
Fe2+(aq), Zn2+(aq)
No reaction is observed.
Zn(s), Cu2+(aq)
Spontaneous or Not?
spontaneous
spontaneous
spontaneous
non-spontaneous
The reaction indicated is
non-spontaneous (see discussion below)
non-spontaneous
Discussion
All reactions except the Cu2+/I– reaction can be successfully predicted. In this reaction, Cu2+ ions are reduced to Cu+ ions and are precipitated
as white copper(I) iodide, CuI. The I– ions are oxidised to I2 which accounts for the brown solution. The overall reaction can be expressed as:
2Cu2+(aq) + 4I–(aq)
2CuI(s) + I2(aq)
6
CHEMISTRY 3.7 ANSWERS
More Questions Using Eº (pages 44-52)
1.
a.
d.
e.
magnesium, Mg
b.
thiosulfate ion, S2O32–(aq) c.
iron(III) ion, Fe3+(aq)
2+
–
Mg(s)/Mg (aq), Pt/Cl (aq),Cl2(aq)
Using E° values, bromine is a better oxidant than iron(III) ions and iron(II) ions are a better reductant than bromide ions.
Alternatively consider the cell: Pt(s)/Fe2+(aq), Fe3+(aq) // Br2(aq), Br–(aq)/Pt
Apply the relationship E°cell = E°RHE – E°LHE ⇒ E°cell = +1.09 – (+0.77) = +0.32 V
Since E°cell is positive, the spontaneous reaction is between the oxidant (Br2) at the RHE and the reductant (Fe2+) at the LHE
⇒ the reaction Br2(aq) + 2Fe2+(aq)
2Fe3+(aq) + 2Br–(aq) is spontaneous.
2.
a.
Co3+(aq)
3.
Half Equation
S(s) + 2e
Fe2+(aq) + 2e
Pb2+(aq) + 2e
2H+(aq) + 2e
I2(aq) + 2e
MnO2(s) + 4H+(aq) + 2e
Cl2(aq) + 2e
Co3+(aq) + e
4.
5.
6.
7.
8.
9.
10.
11.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
a.
b.
Ca(s)
S2–(aq)
Fe(s)
Pb(s)
H2(g)
2I–(aq)
Mn2+(aq) + 2H2O(l)
2Cl–(aq)
Co2+(aq)
c.
Ca2+(aq)
d.
Co2+(aq)
Eº/volt
-0.48
-0.44
-0.13
0.00
+0.54
+1.23
+1.36
+1.81
Note: Some of these reactions
Yes – spontaneous. Fe is a stronger reductant than Pb: Fe(s) + Pb2+(aq)
Fe2+(aq) + Pb(s)
be very slow but they
Yes – spontaneous. Cl2 is a stronger oxidant than Pb2+ ions: Pb(s) + Cl2(g)
Pb2+(aq) + 2Cl–(s) may
are possible from energy
considerations. The reaction
No – not spontaneous. Cl2 is a stronger oxidant than MnO2.
between lead and chlorine
No – not spontaneous. The Pb2+ ion is a weaker oxidant than the Co3+ ion.
is likely to stop because of
2–
–
the coating of insoluble lead
Yes – spontaneous. I2 is a stronger oxidant than S: I2(aq) + S (aq)
2I (aq) + S(s)
chloride on the unreacted lead.
No – not spontaneous. Both H+ and Ca2+ are oxidants.
Ca2+(aq) + H2(g)
Yes – spontaneous. The H+ ion is a stronger oxidant than the Ca2+ ion: Ca(s) + 2H+(aq)
No – not spontaneous. I2 is a weaker oxidant than MnO2.
Yes – spontaneous. Fe is a stronger reductant than H2: Fe(s) + 2H+(aq)
Fe2+(aq) + H2(g)
Yes – spontaneous. The H+ ion is a stronger oxidant than the Pb2+ ion: Pb(s) + 2H+(aq)
Pb2+(aq) + H2(g)
2–
No – not spontaneous. The S ion is a weaker reductant than Ca.
Yes – spontaneous. The Co3+ ion is a stronger oxidant than Cl2: 2Co3+(aq) + 2Cl–(s)
2Co2+(aq) + Cl2(g)
Since E°cell is negative, the spontaneous reaction will be the reductant (Cr) on the RHS reacting with the oxidant (Sn2+) on the
LHS: 3Sn2+(aq) + 2Cr(s)
3Sn(s) + 2Cr3+(aq).
b.
Use the relationship:
E°cell = E°RHE – E°LHE ⇒ E°RHE = E°cell + E°LHE = -0.60 + (-0.14) = -0.74 V
Consider the cell Pt/I–(aq), I2(aq) // Fe3+(aq), Fe2+(aq)/Pt.
As written, E°cell = E°RHE – E°LHE = +0.77 – (+0.54) = + 0.23 V. Since the value of E°cell is positive, the spontaneous reaction is between
the reductant on the LHS and the oxidant on the RHS: 2I–(aq) + 2Fe3+(aq)
I2(aq) + 2Fe2+(aq).
The E°cell value is negative. Therefore the spontaneous reaction is between SO2(aq) and MnO4–(aq).
Half equations:
SO2(g) + 2H2O(l)
SO42–(aq) + 4H+(aq) + 2e
MnO4–(aq) + 8H+(aq) + 5e
Mn2+(aq) + 4H2O(l)
Overall: 2MnO4–(aq) + 5SO2(g) + 2H2O(l)
2Mn2+(aq) + 5SO42–(aq) + 4H+(aq)
a.
i.
Since E°cell is positive, the spontaneous reaction is the reductant on the LHS reacting with the oxidant on the RHS:
Ni(s) + Cu2+(aq)
Ni2+(aq) + Cu(s)
ii.
Copper electrode – where reduction is occurring.
iii.
E°cell = E°RHE – E°LHE ⇒ E°LHE = E°RHE – E°cell ⇒ E°LHE = +0.34 – (+0.57) = -0.23 V, i.e. ENi/Ni2+ = -0.23 V
b.
Lead is not as strong a reductant as nickel (less negative E° value), therefore no reaction would be observed when the lead
rod was placed in the solution of nickel nitrate. Lead is a stronger reductant than copper, therefore copper would be displaced
from copper(II) nitrate solution by the lead. The blue colour of the copper(II) ion would fade and a pink deposit of metallic copper
would precipitate out.
Cell: Li(s)/Li+(aq) // F2(g), HF(aq)/Pt; E°cell = E°RHE – E°LHE ⇒ E°cell = +3.06 – (-3.03) = +6.09 V
2Li+ + 2HF
Since E°cell is positive the spontaneous reaction is: 2Li + F2 + 2H+
Fluorine is an extremely reactive and poisonous gas. A licence is required for anyone to produce the gas. Lithium metal reacts
vigorously with water or any aqueous medium. In practical terms, it is not possible to assemble the chemicals required for this reaction.
a.
i.
To provide a medium that will allow charge (ions) to move from one half cell to the other half cell.
ii.
Material X(s) is platinum/graphite. A good conductor of electricity and chemically inert in the conditions that the cell
operates under.
Fe2+(aq) + 2e.
iii.
Fe(s) is the reductant because electrons are released at this electrode by the process: Fe(s)
Therefore the process at the other electrode must be: Fe3+(aq) + e
Fe2+(aq).
The overall cell reaction is therefore: Fe(s) + 2Fe3+(aq)
3Fe2+(aq).
2+
+
+
b.
i.
Cu(s)/Cu (aq) // Ag (aq)/Ag(s) or the reverse, Ag(s)/Ag (aq) // Cu2+(aq)/Cu(s)
ii.
Apply the relationship: E°cell = E°RHE – E°LHE ⇒ E°cell = (+0.80) – (+0.34) = +0.46 V
Note: If the alternative cell is used then E°cell = -0.46 V
At this electrode electrons are gained by silver ions converting them to silver atoms.
iii.
Ag+(aq)/Ag(s)
iv.
The silver ion, Ag+(aq).
v.
Blue, due to the copper(II) ion, Cu2+(aq), being formed.
a.
Tin, Sn
b.
The amalgam filling (the tin)
Sn2+(aq) + 2e
c.
Half equations: Sn(s)
and
O2(g) + 4H+(aq) + 4e
2H2O(l)
Overall: 2Sn(s) + O2(g) + 4H+(aq)
2Sn2+(aq) + 2H2O(l)
d.
E°cell = E°RHE – E°LHE ⇒ E°cell = +0.82 – (-0.14) = +0.96 V
e.
It is an electrolyte and acts as the salt bridge, allowing ions to
move through the cell.
ANSWERS CHEMISTRY 3.7
12.
13.
a.
b.
Magnesium, Mg
Voltmeter to measure
e–
energy of electrons
i.
See diagram opposite.
2+
2+
ii.
Net ionic equation: Mg(s) + Cu (aq)
Mg (aq) + Cu(s)
iii.
E°cell = E°RHE – E°LHE ⇒ E°cell = +0.34 – (-2.37) = +2.71 V
magnesium
copper
strip
strip
iv.
For the reaction of magnesium with dilute sulfuric acid, set up
Salt bridge
the following cell:
(potassium
Mg(s)/Mg2+(aq) // H+(aq), H2(g)/Pt(s)
sulfate solution)
and E°cell = 0.00 – (-2.37) = +2.37 V
cotton wool
plugs
Since E°cell is positive
⇒ the reaction Mg(s) + 2H+(aq)
Mg2+(aq) + H2(g) is
magnesium sulfate
copper(II) sulfate
spontaneous.
solution (1 mol L–1)
solution (1 mol L–1)
For the reaction of copper with dilute sulfuric acid, set up the
following cell: Cu(s)/Cu2+(aq) // H+(aq), H2(g)/Pt(s) and E°cell = 0.00 – (+0.34) = -0.34 V
Since E°cell is negative ⇒ the reaction Cu(s) + 2H+(aq)
Cu2+(aq) + H2(g) is not spontaneous.
For the reaction of copper with dilute nitric acid, set up the following cell: Cu(s)/Cu2+(aq) // NO3–(aq), H+(aq), NO(g)/Pt(s)
and E°cell = 0.96 – (+0.34) = +0.62 V
Since E°cell is positive ⇒ the spontaneous reaction is: 3Cu(s) + 8H+(aq) + 2NO3–(aq)
3Cu2+(aq) + 2NO(g) + H2O(l)
a.
b.
c.
See diagram opposite.
Electron flow through the wire is from the nickel to the silver electrode.
Ions move in and out of the salt bridge to maintain an electrical
balance.
sodium, potassium or ammonium nitrate
For the cell: Ni(s)/Ni2+(aq) // Ag+(aq)/Ag(s),
E°cell = 0.80 – (-0.23) = +1.03 V
d.
e.
14.
15.
16.
7
a.
V
Half equations:
Fe(s)
and O2(g) + 2H2O(l) + 4e
Overall: 2Fe(s) + O2(g) + 2H2O(l)
and 2Fe2+(aq) + 4OH–(aq)
Fe2+(aq) + 2e
4OH–(aq)
2Fe2+(aq) + 4OH–(aq)
2Fe(OH)2
9
e– flow
e– flow
voltmeter
salt bridge
nickel
metal
nickel
nitrate
solution
silver
metal
anode
cathode
silver
nitrate
solution
b.
For the cell: Fe(s)/Fe2+(aq) // O2(g), OH–(aq)/Pt(s), E°cell = +0.40 – (-0.44) = +0.84 V
Since E°cell is positive ⇒ the spontaneous reaction is that of the overall reaction shown above for the rusting of iron.
a.
b.
Hg22+
i.
Hg2Cl2(s)/Hg(l)
ii.
For the cell: Sn(s)/Sn2+(aq) // Hg2Cl2(s)/Hg(l), E°cell = +0.24 – (-0.15) = +0.39 V
iii.
Sn(s) and Hg2Cl2(s)
-0.39 V
Zn2+(aq) + 2e
b.
HgO(s) + 2H+(aq) + 2e
Hg(l) + H2O(l)
Zn(s)
Since oxidation takes place at the negative electrode in a cell, the zinc electrode is the negative electrode.
Liquid mercury (a cumulative poison) is produced as the cell is used. The cell should be disposed of so that mercury is not
released to the atmosphere. Continuous inhalation of small quantities of mercury vapour can eventually be fatal.
c.
a.
c.
d.
Reaction Predictions (pages 53-54)
1.
2.
a.
Cell Diagram: Fe(s)/Fe2+(aq)//Ni2+(aq)/Ni(s)
Eºcell = EºRHE – EºLHE = -0.23 – (-0.44) = +0.21 V
b.
Cell Diagram: Pb(s)/Pb (aq)//Zn (aq)/Zn(s)(s)
Eºcell = EºRHE – EºLHE = -0.76 – (-0.13) = -0.63 V
direction written.
2+
⇒
Since Eºcell is positive, reaction is spontaneous as written.
2+
⇒
Since Eºcell is negative, reaction will not proceed spontaneously in the
c.
Cell Diagram: Sn(s)/Sn2+(aq)//S(s),H+(aq),H2S(g)/Pt
Eºcell = EºRHE – EºLHE = +0.14 – (-0.14) = +0.28 V ⇒ Since Eºcell is positive, reaction will proceed spontaneously in the direction
written.
a.
Cell Diagram: Pt/Cl–(aq),Cl2(g)//Br2(aq),Br–(aq)/Pt
Eºcell = EºRHE – EºLHE = 1.09 – (+1.36) = -0.27 V
⇒ Since Eºcell is negative, the reaction Br2(aq) + 2Cl–(aq)
2Br–(aq) + Cl2(g) would not proceed spontaneously.
b.
Cell diagram would be: Pb(s)/Pb2+(aq)//Cu2+(aq)/Cu(s)
Eºcell = +0.34 – (-0.13) = +0.47 V ⇒ The reaction Pb(s) + Cu2+(aq)
Pb2+(aq) + Cu(s) could occur.
Observation: As the reaction proceeded, the blue colour of the copper(II) sulfate solution would fade and a dark brown/black
deposit would appear on the lead rod.
c.
Possible reactions would be:
For (1):
⇒
For (2):
⇒
For (3):
⇒
d.
2Ag(s) + Pb2+(aq)
2Ag+(aq) + Pb(s)
... (1)
2+
Zn(s) + Pb (aq)
Zn2+(aq) + Pb(s)
... (2)
Mg(s) + Pb2+(aq)
Mg2+(aq) + Pb(s)
... (3)
Eºcell = EºRHE – EºLHE = -0.13 – (+0.80) = -0.93 V
Since Eºcell is negative, reaction (1) is not spontaneous/will not occur.
Eºcell = EºRHE – EºLHE = -0.13 – (-0.76) = +0.63 V
Since Eºcell is positive, reaction (2) is spontaneous/could occur.
Eºcell = EºRHE – EºLHE = -0.13 – (-2.38) = +2.25 V
Since Eºcell is positive, reaction (3) is spontaneous/could occur.
Cu(s) + Fe2+(aq)
Eºcell = +0.34 – (-0.44) = +0.78 V
Reaction would be: Cu2+(aq) + Fe(s)
⇒ Since Eºcell is positive, reaction could proceed spontaneously as written.
⇒ You cannot store a copper(II) sulfate solution in an iron container since reaction would occur.
8
CHEMISTRY 3.7 ANSWERS
e.
Solutions containing Fe2+(aq) are oxidised in air due to the spontaneous reaction:
4Fe3+(aq) + 2H2O(l)
(Eºcell = +1.23 – (+0.77) = +0.46 V)
4Fe2+(aq) + O2(g) + 4H+(aq)
2Fe2+(aq) + 2H2O(l) is also spontaneous.
The reaction: 2Fe(s) + O2(g) + 4H+(aq)
(Eºcell = +1.23 – (-0.44) = +1.67 V) ⇒ Higher Eºcell value.
⇒ The tendency of Fe(s) to lose electrons is greater than Fe2+(aq), so the presence of Fe(s) will inhibit the tendency of
Fe3+(aq)
Fe2+(aq)
The Dry Cell (Leclanché Cell) (page 55)
1.
2.
3.
Zn(s) + 2MnO2(s) + 2NH4+(aq)
Zn2+(aq) + 2MnO(OH)(s) + 2NH3(aq)
Graphite; a good conductor of electricity.
It remains in solution in the cell which is sealed off from the atmosphere.
The Dry Cell (pages 57-59)
Diagrams
Observations
Cross-Section View:
metal cap
graphite
electrode
metal exterior casing
graphite electrode
black solid (powder)
plastic cap
(insulator)
insulation
metal case
(zinc)
graphite electrode
black powder
metal cap
All of the above held together with a
cardboard wrapper down the side.
zinc casing
cardboard
insulation at base
Results
Observing Dry Cell – Steps 4 And 5:
The black powder does not dissolve in water but when filtered,
a portion of the filtrate produces a white, crystalline solid on
evaporation. Note that if this solid is heated with aqueous sodium
hydroxide, the characteristic pungent smell of ammonia gas is
detected, indicating that the solid must contain NH4+ ions.
Making A Dry Cell:
As the cell operates, the universal indicator in the solution in the
vicinity of the carbon rod turns green and then blue.
Construction of Cell
V
V = 1.5 V
graphite rod
Solution containing
25 mL 1 mol L–1 ZnCl2,
25 mL saturated NH4Cl and
10 drops of universal indicator.
zinc metal strip
100 mL beaker
Discussion
• The Leclanché cell was invented in 1866 by Georges Leclanché
(1839-1892).
• The cell is called a ‘dry cell’ because there is no evidence of any
liquids in the cell as the chemical interior is compactly enclosed.
The expression ‘dry cell’ is somewhat of a misnomer because the
cell contains a moist paste and in fact would not work if its contents
was completely dry.
• A commercially made cell consists of a graphite electrode
surrounded by a moist paste of ammonium chloride, zinc
chloride, carbon black, starch and manganese dioxide (to prevent
polarisation). The carbon black decreases the electrical resistance
of the cell, enabling it to deliver a relatively high current.
• Electrode Reactions:
Anode ( - electrode):
Zn(s)
Zn2+(aq) + 2e
Cathode ( + electrode): NH4+(aq) + 2e
NH3(aq) + H2(g)
Note that the manganese dioxide ‘polariser’ prevents accumulation
of hydrogen gas which would otherwise stop the operation of the
cell. The overall process can be represented as:
MnO2(s) + NH4+(aq) + e
MnO(OH)(s) + NH3(aq)
• At the top of the cell, a plastic insulator prevents the negatively
charged zinc case from coming into contact with the positively
charged carbon rod and its metal cap. At the base of the cell,
cardboard insulation is used to prevent contact between the zinc
case and the carbon rod.
• As a cell discharges and goes ‘flat’, the zinc casing corrodes
forming a white compound containing Zn2+ ions.
• In each of the cells constructed in the practical ‘Making
Electrochemical cells’ (page 33), at least one of the reactants was
in the form of ions in solution and to prevent reactants coming into
contact, the two half cells were physically separated. In the dry
cell, both reactants are essentially solids placed in one container,
which also contains a moist paste of electrolyte separated only by
porous paper.
• In commercial cells, a paste of ammonium chloride and zinc
chloride is used. The paste minimises the water content of the
cell. The paste conducts electricity (like an aqueous solution)
but takes up a smaller volume, allowing a greater space for the
reactant in the cell. Also, the cell is less likely to leak if there is no
liquid electrolyte.
• Observations:
(a) Ammonium chloride is formed when the filtrate is evaporated.
(b) The colour change around the carbon rod is due to the
generation of an alkaline solution caused by the production of
2NH3(aq) + H2(g)
ammonia in solution: i.e. NH4+(aq) + 2e
• Advantages of dry cell: Compact, not messy like wet cells,
its portability, can be conveniently enclosed within electrical
appliances.
Disadvantage of dry cell: Cannot be recharged – when the cell
goes ‘flat’ it has to be discarded (waste/expense).
The Lead-Acid Battery (pages 60-61)
1.
2.
3.
4.
a.
Pb(s)
Pb2+(aq) + 2e
b.
PbO2(s) + 4H+(aq) + 2e
Pb2+(aq) + 2H2O(l)
2+
2+
Pb(s)
d.
Pb (aq) + 2H2O(l)
PbO2(s) + 4H+(aq) + 2e
c.
Pb (aq) + 2e
The concentration is reduced. Hydrogen ions are removed as shown in equation (1.b.) and the sulfate ion is removed in equivalent
amounts in the formation of insoluble lead sulfate.
Any car suffers from vibration. Over a period of time the powdered lead dioxide forming one plate (the anode) is dislodged from its
packing and settles at the bottom of the battery case as a powder.
Pb2+(aq) + 2e (oxidation)
and
PbO2(s) + 4H+(aq) + 2e
Pb2+(aq) + 2H2O(l) (reduction).
a.
Pb(s)
2Pb2+(aq) + 2H2O(l)
Overall reaction: Pb(s) + PbO2(s) + 4H+(aq)
b.
lead sulfate, PbSO4
c.
The cell can be re-charged.
ANSWERS CHEMISTRY 3.7
9
The Lead-Acid Cell (pages 62-63)
Results
• A red-brown colouration appears at the + electrode (cathode).
• A reading is gained on the voltmeter (about 2 V).
• The light bulb glows.
–
Discussion Questions
+
1.
As a result of the formation of lead(IV) oxide (PbO2), the
dilute sulfuric acid
cathode develops a red-brown colouration.
PbSO4(s) + 2e
2.
At anode: Pb(s) + SO42–(aq)
At cathode:
lead(IV) oxide plates
(lead dioxide)
lead plates
PbO2(s) + 4H+(aq) + SO42–(aq) + 2e
PbSO4(s) + 2H2O(l)
3.
See diagram opposite.
4.
Unlike the cells in car batteries, the laboratory cell did not initially contain any lead(IV) oxide which had to be produced at the positive
electrode by charging the cell. The laboratory cell has only one anode and one cathode, whereas cells used in car batteries each have
three cathodes sandwiched between four anodes.
Electrolysis (pages 65-67)
1.
2.
3.
4.
5.
a.
a.
b.
c.
d.
C
b.
A
c.
C
d.
C
e.
Electrical
At the anode (positive electrode)
The negative electrode (cathode).
NaCl(l) is molten sodium chloride (i.e. solid sodium
chloride that has been heated until it melted at 804ºC).
NaCl(aq) is sodium chloride that has been dissolved in
water to produce an aqueous solution.
a.
See diagram opposite.
b.
Cathode: A pink coloured solid is deposited on the
cathode and the mass of the cathode increases.
Anode: The copper anode shrinks (Cu atoms move into
solution as Cu2+ ions) and the mass of the anode
decreases.
c.
Cathode: Cu2+(aq) + 2e
Cu(s)
Anode:
Cu(s)
Cu2+(aq) + 2e
d.
The purification of copper produced thermochemically.
a.
Mg2+ and Cl–
b.
There is no change in oxidation number as the
precipitate contains the same ions.
c.
neutralisation
Mg(OH)2 + 2HCl
MgCl2 + 2H2O
d.
i.
2Cl–
Cl2 + 2e
ii.
Mg2+ + 2e
Mg
e.
See diagram opposite.
Some examples are provided in the following table:
A
f.
A
g.
C
h.
+
electron flow
C
–
i.
C
j.
A
electron flow
D.C. electrical
supply
copper
electrodes
Cathode
(reduction)
Anode
(oxidation)
Cu
ions
OH
ions
–
anode (oxidation)
2+
+
copper sulfate
solution
–
cathode (reduction)
Cl–
Mg2+
Industrial Application
Electrolyte
Electrodes
Conditions
Uses
Production of
molten aluminium
1000ºC with a flux
Transport (planes, boats), Containers (cans),
carbon
Aluminium
oxide
of cryolite
Leisure equipment (tennis rackets, etc.)
copper sulfate with
Electrical wiring, Roofing, Coinage metals,
Purification of Copper
copper
65ºC
dilute sulfuric acid
Alloys (e.g. brass, bronze)
Production of
molten magnesium
steel cathode,
650ºC plus small amounts
Alloyed with aluminium to make strong,
Magnesium
chloride
graphite anode
of sodium chloride
light metals. Production of titanium.
Production of Chlorine,
carbon anode, mercury
brine (sodium
Hydrogen and Sodium
cathode or titanium
room temperature
*
chloride solution)
Hydroxide*
anode, nickel cathode
* The mercury cathode cell produces purer sodium hydroxide but has the environmental hazard of using toxic mercury. The process
can be summarised by the equation: 2Na+(aq) + 2Cl–(aq) + 2H+(aq) + 2OH–(aq)
2Na+(aq) + 2OH–(aq) + H2(g) + Cl2(g)
brine
sodium hydroxide
Uses: Sodium hydroxide – making soap, purifying aluminium ore, removal of oil and grease.
Hydrogen – making margarine, rocket fuel, manufacture of ammonia.
Chlorine – purifying water, bleaching wood pulp, manufacture of PVC.
Electrolysis of Copper Chloride Solution (pages 68-69)
Method
1.
Cu2+ and Cl–
• Good conductor of electricity.
• Does not react with species produced in the reaction.
2.
There is a build-up of a pink deposit at the negative electrode
(cathode) and bubbles of a pale green, pungent gas are
produced at the positive electrode (anode).
3.
Moist litmus paper is bleached and starch/iodide paper turns
black.
The gas produced is chlorine, Cl2.
4.
There is a pink deposit on the negative electrode. There is no
change to the positive electrode.
The copper-plated electrode was negative.
5.
The solution was blue due to copper(II) ions, Cu2+
Oxidation, because the copper atoms lose electrons and
become copper(II) ions, Cu2+.
Nitric acid is acting as an oxidant.
Discussion Exercises
Cu
(reduction)
1.
Cathode reaction: Cu2+ + 2e
Anode reaction:
2Cl–
Cl2 + 2e (oxidation)
2.
Sodium is made by electrolysis of a molten mixture of sodium
chloride and calcium chloride. Chlorine is a by-product.
The electrode reactions can be represented as follows:
At graphite anode:
2Cl–
Cl2 + 2e
At steel cathode: Na+ + e
Na
10
CHEMISTRY 3.7 ANSWERS
Electroplating – An Application of Electrolysis (pages 70-71)
Results
Initial mass of brass electrode
Final mass of brass electrode
Change in mass of brass electrode
= 8.89 g
= 9.05 g
= 0.16 g
Initial mass of copper electrode
=
Final mass of copper electrode
=
Change in mass of copper electrode =
Discussion Questions
1.
Because grease from the hand forms an insulating layer
preventing Cu2+ ions from being reduced in that region – this
gives rise to uneven plating.
2.
The difference in colour of the brass and copper enables the
plating to be seen. The negative charge attracts Cu2+ ions
and they are reduced.
3.
No! Virtually zero current would flow through the cell and
there would be no Cu2+ ions in the cell solution to be reduced
and therefore form a ‘plating’.
4.
5.
6.
7.
8.
Cathode: (-ve electrode): Cu2+ + 2e
Cu
(reduction)
Anode: (+ve electrode): Cu
Cu2+ + 2e (oxidation)
Loss of mass at the anode should equal gain of mass at the
cathode. If the copper anode contains impurities, it will lose
more mass than the cathode gains.
6.15 g
5.94 g
0.21 g
No! The number of Cu2+ ions produced at the anode will
equal the number of Cu2+ ions reduced at the cathode.
The impure (95%) copper sample is attached to the positive
terminal in an electrolytic cell containing a CuSO4/H2SO4
solution and a pure copper cathode. As electrolysis proceeds,
Cu2+ ions migrate from the impure anode to the pure cathode
leaving impurities as a ‘mud’ at the bottom of the cell.
• Electrorefining of metals such as copper.
• Electroplating – e.g. iron objects are plated with chromium
or nickel to prevent rusting.
• Electrical reduction of oxides to produce active metals
(e.g. Al2O3
Al).
Oxidation-Reduction Reactions Summary (pages 72-77)
1.
a.
b.
c.
d.
2.
3.
a.
The black copper(II) oxide turns orange-brown or pink
and condensation appears on the walls of the reaction
container.
The metal reacts vigorously giving off a brown gas and
producing a green solution.
Heat is evolved during the reaction and the orange
dichromate solution turns green. The characteristic
pungent smell of sulfur dioxide fades.
The purple permanganate colour disappears and is
replaced by a faint brown solution.
Half equations:
Fe
Fe3+ + 3e
and
Cl2 + 2e
2Cl–
Overall equation:
2Fe + 3Cl2
2Fe3++ 6Cl–
b.
Half equations:
IO3– + 6H+ + 6e
and
2I–
Overall equation: IO3– + 6H+ + 5I–
I– + 3H2O
I2 + 2e
3I2 + 3H2O
c.
Half equations:
2Cu2+ + 2I– + 2e
and
2I–
2+
Overall equation:
2Cu + 4I–
2CuI
I2 + 2e
2CuI + I2
d.
Half equations:
Cl– + H2O
Cl2 + 2e
Cl2 + H2O
OCl– + 2H+ + 2e
and
2Cl–
Overall equation: OCl– + 2H+ + Cl–
internal circuit
(salt bridge)
external circuit
(wire)
9
–
+
(oxidation)
anode
electrolyte
4.
5.
(reduction)
cathode
X–
anions
cations
half cells
X+
electrolyte
In an electrochemical cell the electrode at which oxidation
occurs is called the anode or negative electrode.
The electrode at which reduction occurs is called the cathode
or positive electrode.
In a Daniell cell, the zinc electrode is the anode and the
copper electrode is the cathode.
A salt bridge contains ions that are free to move so that
they can balance charges formed in the half cells. Cations
(positive ions) migrate to the cathode and anions (negative
ions) migrate to the anode . This part of the cell is called the
internal circuit.
a.
i.
-2
ii.
+4
iii.
+4
b.
2H2S + O2
2H2O + 2S
2H2O + 3S
c.
2H2S + SO2
d.
2H2S
2H2O + 3S,
e.
For the reaction 2H2S + SO2
E°cell = E°red – E°ox = +0.45 – (+0.17) = +0.28 V
⇒ since E°cell is positive, the reaction as written is
spontaneous.
6.
Part A: Standard Cell
a.
b.
c.
d.
e.
f.
g.
Reduction (Fe3+ + e
Fe2+).
The orange/brown colour fades to colourless/pale
green.
I2 + 2e ).
Oxidation (2I–
A brown colour (due to I2(aq)) would appear in the
solution.
Pt(s)/I–(aq), I2(aq) // Fe3+(aq), Fe2+(aq)/Pt(s)
E°cell = E°RHE – E°LHE = +0.77 – (+0.54) = +0.23 V
i.
The solution on the right hand side would
change from colourless/pale green to orange.
ii.
The solution on the left hand side will fade from
pale green (due to Cl2(aq)) to colourless.
Chlorine, Cl2, is a more powerful oxidant that Fe3+(aq),
so will oxidise the Fe2+(aq) ions to Fe3+(aq), hence the
colour changes described above.
Overall cell reaction:
Cl2(aq) + 2Fe2+(aq)
2Cl–(aq) + 2Fe3+(aq)
Part B: Button Cells
Zn2+ + 2Ag + H2O
a.
Ag2O + 2H+ + Zn
b.
Zn(s)/Zn2+(aq)//Ag2O(s), H+(aq)/Ag(s)
c.
E°cell = E°RHE – E°LHE = +1.17 – (-0.76) = 1.93 V
d.
It acts as a salt bridge – it separates the reacting
species but allows ions to move from one half cell to
the other to maintain electrical neutrality.
e.
It acts as an electrolyte, allowing the current to flow.
It also complexes the Zn2+ ions formed at the cathode:
Zn2+(aq) + 4OH–(aq)
Zn(OH)42–(aq)
f.
The concentrations of the species reacting are not
standard concentrations of 1 mol L–1.
There are no products present – there is no
accumulation of Zn2+ ions as they are complexed as
Zn(OH)42–ions.
g.
Nickel is more resistant to corrosion than zinc.
The cell is not prone to leaking since the outer nickel
casing is not oxidised. By comparison, the zinc casing
in a dry cell is converted to Zn2+ ions and is slowly used
up resulting in holes forming.