NCEA Level 3 Chemistry (90696) 2011 — page 1 of 4
Assessment Schedule – 2011
Chemistry: Describe oxidation-reduction processes (90696)
Evidence Statement
Q
ONE
Evidence
•
Achievement
Note: reverse diagram is acceptable as long as everything is
reversed.
(a)
Achievement with Merit
Achievement with Excellence
TWO of
•
Diagram correct.
BOTH 1(a) and 1 b (i)(ii)
correct
OR
•
Ni | Ni2+ || Cu2+| Cu
(b)(i)
(ii)
E°cell = E°RHE – E°LHE
(iii)
•
= 0.34 – (–0.23) = +0.57 V
Ni  Ni2+ + 2e
Cu + 2e  Cu
At the anode (Ni electrode) the Ni atoms are oxidised to Ni 2+ ions.
Hence the mass of the Ni electrode will decrease over time and the
solution will appear green due to the formation of Ni2+ ions.
• At the cathode (Cu electrode) the Cu2+ ions are reduced to Cu atoms
and hence the blue solution fades. Cu electrode will increase the
mass/pink/brown deposit forms
• The salt bridge completes the circuit
• The negative ions will flow into the Ni half-cell / anode to balance
the increasing positive charge / decreasing negative charge.
• The positive ions will flow into the copper half-cell / cathode to
balance the decreasing positive charge / increasing negative charge.
•
•
2+
Correct cell diagram AND E°
correct (unit required).
OR
ONE of
• Both equations correct.
• Identifies correct anode and
cathode.
• Notes change in mass at one
electrode.
• Notes change in colour in one
half cell.
• States role of salt bridge.
• States direction of flow of one
ion in salt bridge.
AND
ONE of
• Notes change in mass at both
electrodes with explanation.
OR
• Notes changes in colour in
both half cells linked to
species.
OR
• Complete observation for one
half cell linked to species.
OR
• States role of salt bridge and
fully explains correct flow of
ions.
Full explanation including
colour and mass changes linked
to species.
AND
Role of salt bridge with correct
explanation of ion flow.
NCEA Level 3 Chemistry (90696) 2011 — page 2 of 4
Q
Evidence
Achievement
2MnO4– + 5SO32– + 6H+  2Mn2+ + 3H2O + 5SO42–
+7
+4
+2
+6
TWO
(a)
MnO4– is the oxidant as the oxidation number of the Mn atom
decreases from +7 to +2.
SO32– is the reductant, as the oxidation number of the S atom
increases from +4 to +6.
(b)
H2O2  O2 +2H+ + 2e–
MnO4 + 8H +5e  Mn + 4H2O
–
+
–
2+
THREE of
• Correctly identifies reductant
and oxidant species.
OR
• Correct oxidation numbers for
Mn and S species.
THREE of
• Correctly identifies reductant
and oxidant species and links
to correct oxidation numbers.
OR
• Two correct half-equations
•
2MnO4– + 6H+ + 5H2O2 2 Mn2+ + 8H2O + 5O2
OR
(c)
Reaction B
• colourless H2O2 is oxidised to O2 bubbles /gas.
• oxidation number for O changes from –1 to 0
(OR electrons lost) so is oxidised.
•
Identifies that H2O2 is a
reductant in B / is oxidised
OR
H2O2 acts as a reductant/is oxidised.
Reaction C
• colourless H2O2 is reduced to colourless H2O
• H2O2 oxidation number for O changes from –1 to –2 (OR electrons
gained) so O is reduced
• H2O2 acts as an oxidant/is reduced
Achievement with Merit
•
Observes that colourless O2
gas is produced in B
OR
• Identifies that H2O2 is an
oxidant in C / is reduced
Achievement with Excellence
TWO merit answers plus either:
OR
Correct half and final
equations. (OK if in wrong
boxes)
OR
Identifies that in B H2O2 is a
reductant and links to
oxidation numbers / loss of
electrons and to observation of
O2 bubbles / gas.
OR
Identifies that in C H2O2 is an
oxidant and links oxidation
numbers / gain of electrons
and to lack of observations.
OR
Correctly identifies H2O2 as
reductant in B and oxidant in C
linked to EITHER change in
oxidation numbers/loss gain of
electrons OR observations.
•
Identifies H2O2 as both a
reductant and oxidant and links
to both changes in the
oxidation numbers / loss and
gain of electrons and links
observations to species
involved.
NCEA Level 3 Chemistry (90696) 2011 — page 3 of 4
(d)
H2O2 will react with itself, by being both an oxidant and a reductant.
(H2O2 / H2O) has a more positive reduction potential than (O 2 /
H2O2). Thus the (O2 / H2O2) couple will undergo oxidation forming
O2 and the (H2O2 / H2O) couple will undergo reduction, producing
H2O.
This will result in the concentration of H2O2 decreasing over time.
OR
H2O2 will react with itself, forming water and oxygen, by being both
an oxidant and a reductant.
This is because for the cell H2O2/O2//H2O2/H2O
Ecell= 1.77 – 0.68 = 1.09V
Therefore the reaction is spontaneous/ will occur converting
hydrogen peroxide to water and oxygen
Therefore the concentration of hydrogen peroxide decreases over
time/bottle may burst because of the build up of oxygen/therefore
should be kept cool because the reaction will be slower.
OR
• Recognises that H2O2 can act
as both an oxidant and
reductant
OR
• Calculates E° correctly with
unit.
OR
• Writes both correct half
equations.
H2O2  O2 +2H+ + 2e–
H2O2 +2H+ + 2e– 2H2O
OR
• Writes correct overall
equation.
2H2O2  2H2O + O2
OR
• Recognises products are water
and oxygen.
OR
• Calculate E° correctly AND
links to H2O2 being able to be
both an oxidant and reductant /
oxidises and reduces itself,
AND forms water and oxygen
/ both equations correct.
OR
• Uses reduction potentials to
explain why H2O2 is able to be
both an oxidant and reductant /
oxidises and reduces itself
AND forms water and oxygen
/ both equations correct.
OR
•
Merit plus relates to an issue of
long term storage.
NCEA Level 3 Chemistry (90696) 2011 — page 4 of 4
Q
THREE
Evidence
Achievement
Order: MnO4– / Mn2+ > U4+ / U3+ > U3+ / U
The half-equations for equation (1).
U  U3+ + 3e–
MnO4– + 8H+ + 5e–  Mn2+ + 4H2O
U is losing e–, thus the E°cell of the U3+ / U couple is more negative
than the E°cell MnO4– / Mn2+.
The half-equations for equation (2).
U  U3+ + 3e–
U4+ + e– U3+
U is losing e–, thus the E°cell of the U3+ / U couple is more negative
than the E°cell U4+ / U3+.
The half-equations for equation (3).
U3+  U4+ + e–
MnO4– + 8H+ + 5e–  Mn2+ + 4H2O
U is losing e–, thus the E°cell of the U4+ / U3+ couple is more negative
than the E°cell MnO4– / Mn2+.
Achievement with Merit
Achievement with Excellence
Recognises that MnO4– / Mn2+
has the highest reduction
potential with some valid
reasoning.
Places TWO couples, relative to
each other, correctly, with
correct justification
Places THREE couples, relative
to each other, correctly, with
correct explanation and
appropriate detailed justification.
Minor error acceptable.
OR
Recognises that U3+ / U has the
lowest reduction potential with
some valid reasoning.
OR
Correct order with good
explanation lacking specific
details.
OR
Places all THREE couples,
relative to each other, correctly.
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
2A
3M
OR
1E+1M
2E