358 ANSWERS
6 ▶ a X is between iron and hydrogen in the reactivity series.
It displaces hydrogen from dilute hydrochloric acid,
and copper from copper(II) sulfate, and so must be
above hydrogen and copper. It won’t displace iron
from iron(II) sulfate, and so must be below iron.
b i yes: X nitrate and silver
ii no reaction
iii no reaction (at least, not in the short term)
iv yes: X chloride and copper
v yes: X sulfate and hydrogen
7
▶ a hydrogen
b aluminium chloride
c 2Al(s) + 6HCl(aq) → 2AlCl
3
(aq) + 3H
2
(g)
d Aluminium is covered by a very thin, but very strong,
layer of aluminium oxide which prevents the acid
getting at the aluminium underneath. On heating,
the acid reacts with the oxide and removes it. The
aluminium then shows its true reactivity, and produces
a vigorous reaction.
8
▶ Drop a very small piece into cold water. If it reacts, judge
its reactivity relative to K, Na, Ca or Mg.
If it doesn’t react, add a small piece to dilute hydrochloric
acid and warm if necessary. Rapid reaction in the cold
would place it as ‘similar to magnesium’. A few bubbles
of hydrogen in the cold, but more on heating would place
it as ‘similar to iron or zinc’.
If there is no reaction, then it is ‘below hydrogen’.
9
▶ a The iron/steel must be exposed to oxygen (air) and water.
b Painting prevents the oxygen and water from coming
into contact with the iron/steel.
c It is iron/steel coated with zinc.
d It would take much longer before the car went rusty.
Zinc is more reactive than iron and so corrodes in
preference to the iron. The zinc prevents iron rusting
even when the coating is scratched. When paint is
scratched the steel underneath will rust – this will not
happen when the steel is galvanised, therefore the life
of the car is extended.
10
▶ a
Pearsonium
chloride
solution
Mollium
chloride
solution
Rosium
chloride
solution
Amelium
chloride
solution
pearsonium
7 7 7 7
mollium
3 7 3 3
rosium
3 7 7 7
amelium
3 7 3 7
b mollium > amelium > rosium > pearsonium
c mollium + pearsonium chloride → mollium chloride +
pearsonium
CHAPTER 15
1 ▶ Iron is below carbon in the reactivity series and so can be
extracted by heating with carbon.
From the first experiment:
1 cm
3
of gas has a mass of
0.16
_____
89
= 0.00180 g
100 cm
3
has a mass of 100 × 0.00180 = 0.180 g
h The student did not heat the carbonate for long
enough so not all the carbonate decomposes.
Some of the gas escaped/leaked from the apparatus.
4
▶ Carbon dioxide is one of the gases responsible for the
greenhouse eect. These gases absorb IR (infrared)
radiation that has been emitted from the Earth’s surface.
They then re-radiate it in all directions, warming the
atmosphere. Heating of the atmosphere could lead to
climate change.
CHAPTER 14
1 ▶ a sodium, aluminum, iron, copper
b i magnesium oxide, copper
ii Mg(s) + CuO(s) → MgO(s) + Cu(s) (Include state
symbols in all equations for preference.)
iii Copper(II) oxide (CuO) has been reduced to
copper (Cu) because it has lost oxygen - reduction
is the loss of oxygen.
iv Copper(II) oxide (CuO) is the oxidizing agent
because it oxidizes the magnesium (gives oxygen
to it) and is, in the process, reduced.
c i Zinc is higher in the reactivity series because it
takes the oxygen from the cobalt(II) oxide
ii A reducing agent is a substance which reduces
something else. Zinc removes oxygen from the
cobalt(II) oxide. Removal of oxygen is reduction.
iii Zinc because it gains oxygen - oxidation is gain of
oxygen.
d aluminium, manganese, chromium (Statement 1: Al is
above Cr. Statement 2: Mn is below Al. Statement 3:
Mn is above Cr. Putting this together gives the final
list.)
2
▶ a oxidised; gain of oxygen
b reduced; loss of oxygen
c oxidised; loss of electrons
d reduced; gain of electrons
3
▶ Magnesium is above lead because it removes the oxygen
from the lead(II) oxide.
4
▶ a Either: grey iron filings become coated with brown
solid. Or: solution fades from blue to colourless (very
pale green).
b iron (Fe) has been oxidised (to Fe
2
+) because it has
lost electrons - oxidation is loss of electrons.
c Fe(s) + CuSO
4
(aq) → FeSO
4
(aq) + Cu(s)
5
▶ a nickel, copper, silver
b i Either: colour of solution changes from blue to
green. Or: nickel becomes coated with brown
solid.
ii Ni(s) + CuSO
4
(aq) → NiSO
4
(aq) + Cu(s)
iii Ni(s) + Cu
2+
(aq) → Ni
2+
(aq) + Cu(s)
Nickel has been oxidised by loss of electrons.