NCEA Level 2 Chemistry (90311) 2008 — page 1 of 3
Assessment Schedule – 2008
Chemistry: Describe oxidation-reduction reactions (90311)
Q
ONE
(a)
(b)
(c)
(d)
TWO
(a)
(b)
Evidence
Achievement
Achievement with
Merit
THREE correct.
3+
6+
0
3–
The oxidation number of iron in iron (III)
oxide decreases from 3+ to 0, hence is
reduced,
and oxidation number of carbon in carbon
monoxide increases from 2+ to 4+, hence is
oxidised.
One set of oxidation
numbers (ON)
correct and labelled
as oxidation /
reduction
OR
Both sets of ON
correct (may not be
labelled as oxidation
/ reduction or may
be labelled
incorrectly)
OR
Iron labelled as a
decrease in
oxidation number,
thus reduction and
carbon labelled as
an increase in
oxidation number,
thus oxidation (if
ON’s evident on
equation).
Explains the
oxidation number of
iron decreases from
3+ to 0, hence is
reduced
AND
Carbon increases
from 2+ to 4+ hence
is oxidised.
The magnesium / Mg ‘loses’ electrons,
hence is oxidised
and the hydrogen ion / H+ / H from H2SO4
‘gains’ an electron, hence is reduced.
Describes electron
loss / gain for one
reactant and labelled
as oxidation /
reduction
Explains electron
loss for Mg /
magnesium and links
to oxidation
OR
describes electron
loss and gain for
both reactants (may
not be labelled as
oxidation / reduction
or may be
incorrectly labelled)
OR
Describes electron
transfer from Mg to
H+.
AND
electron gain for H+ /
hydrogen ions / H
in H2SO4 and links
to reduction.
Achievement with
Excellence
NCEA Level 2 Chemistry (90311) 2008 — page 2 of 3
(c)
Copper identified as species oxidised and
reductant (as the oxidation number of
copper has increased from zero to 2+).
Cu  Cu2+ + 2e–
Nitrate ion / nitric acid identified as the
species reduced and oxidant (as the
oxidation number of nitrogen in the nitrate
ion is decreased from 5+ to 4+).
NO3– + 2H+ + e–  NO2 + H2O
Observations
Brown / pink metal / Cu disappears /
smaller(Cu a blue (or green) solution forms
(Cu2+)).
Bubbles / gas / brown gas is formed (NO2).
Colourless solution changes to blue (Cu2+).
THREE
(a)
MnO4– + 8H+ + 5e-  Mn2+ + 4H2O
(b)
2HClCl2 + 2H+ + 2e- OR 2Cl–Cl2 + 2e-
(c)
2MnO4–+ 6H++10HCl2Mn2++ 8H2O+5Cl2
OR
Describes TWO of
 Copper identified
as the species
being oxidised
 Nitrate / N
identified
as the species
being reduced
 Oxidant and
reductant
identified
 An observation is
described
 Half equations.
Explains THREE of
 copper identified
as the species
being oxidised
 nitrate / N
identified
as the sp being
reduced (reason
correct if used)
 oxidant and
reductant
identified
 an observation is
linked to species
 half equations.
ONE half-equation
correct
OR
final equation
correct.
Both half equations
correct and used to
produce full
equation with
simplest ratio of
species.
A full discussion
with copper
identified as the
species being
oxidised and HNO3/
nitrate identified as
the species being
reduced (reason
correct if used)
AND
oxidant and
reductant identified
AND
an observation is
linked to species
AND
half equations.
2MnO4–+16H++10Cl– 2Mn2++8H2O +5Cl2
FOUR
(a)
(i) purple to colourless / pale pink.
(ii) pale green / colourless / to
yellow / orange / yellow-brown /
red-brown.
ONE observation
correct.
(b)
(i) orange to green / blue
(ii) orange sp: Cr2O72– and green/blue: Cr3+
Observation correct.
BOTH ions correctly
linked to their colour
FIVE
Oxidation and reduction processes
Chlorine is reduced to chloride / will oxidise
bromide ions to form bromine.
Bromide ions are oxidised to bromine / will
reduce chlorine to form chloride.
The reaction of
chlorine to chloride
is described
OR
the reaction of
bromide ions to
bromine is
described
OR
a full balanced
equation is given
OR
iodine does not
react with bromide
ions
OR
initial and final
observation is
described.
The reaction of
chlorine with
bromide is described,
AND
an observation linked
to a species
AND
a full balanced
equation is used.
Observations
Both solutions are initially colourless (or
pale green / yellow / yellow-green) for
chlorine / chlorine water)
A yellow / orange / brown solution of
bromine is formed.
Equations
Cl2 + 2Br  2Cl + Br2
–
–
OR
Cl2 + 2KBr  2KCl + Br2
Iodine does not act as an oxidant in this
reaction, as it will not oxidise bromide ions.
OR
Two of the above
three points
AND
explanation includes
iodine not reacting
with bromide ions.
A full discussion of
the reaction of
chlorine with
bromide
AND
an observation
linked to a species
AND
a full balanced
equation is used
AND
the discussion
includes iodine not
reacting with
bromide ions.
NCEA Level 2 Chemistry (90311) 2008 — page 3 of 3
SIX
Right Hand Side: The cathode increases in
mass as Cu2+ ions gain electrons from the
cathode and form Cu. Cu2+ + 2e–  Cu
Left Hand Side: The anode “dissolves”
slowly as Cu atoms lose electrons to the
anode and become Cu2+ hence decreases in
mass. Cu  Cu2+ + 2e–
Intensity of blue colour doesn’t change as
the rate of dissolving of Cu  Cu2+ at one
electrode (anode) is equaled by the rate of
copper sulfate forming Cu2+  Cu at other
electrode (cathode).
RHS: Identifies Cu
forms OR mass
increase at negative
electrode / cathode
AND half equation.
OR
LHS:Identifies Cu2+
form OR mass
decrease at positive
electrode/anode
AND half equation.
RHS: explains Cu is
formed at cathode,
and mass increase,
links decrease in
blue colour of
electrolyte to less
Cu2+, and half
equation.
RHS and LHS:
Discusses mass
increase of cathode,
mass decrease of
anode, and intensity
of blue colour.
Discussion is
supported by
equations.
OR
RHS and LHS:
explains Cu2+ and Cu
are formed, mass
increase at cathode /
negative electrode
and mass decrease at
anode / positive
electrode
AND
half equations
OR
links to unchanging
colour of electrolyte
OR
LHS:
Explains Cu2+ is
formed at anode, and
mass decrease, links
increase in blue
colour of electrolyte
to more Cu2+, and
half equation.
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
Total of FIVE opportunities answered at
Achievement level or higher.
Total of at least SIX opportunities
answered with FOUR at Merit level or
higher.
Total of at least SIX opportunities
answered with TWO at Excellence level
and TWO at Merit level or higher.
5A
4M+2A
2E+2M+2A