NCEA Level 3 Chemistry (90696) 2009 — page 1 of 4
Assessment Schedule – 2009
Chemistry: Describe oxidation-reduction processes (90696)
Evidence Statement
Note: Codes e.g. (x2, a2, m2, e2) indicate evidence towards grade for question.
Q
ONE
(a)
Evidence
• 2I–
Achievement
 I2 + 2e–
• MnO4– + 8H+ + 5e–  Mn2+ + 4H2O
• 2MnO4– + 16H+ + 10I–  2Mn2+ + 8H2O + 5I2
(b)
TWO OF

I–  I2
MnO4–  Mn2+
(a1)
Achievement with Merit
THREE OF
Achievement with Excellence
THREE OF
• Correctly balanced half
equations (m1)
• Full equation correct (e1)
• Observations linked to species
for one oxidation or reduction
change (m2)
• Observations linked to species
for the oxidation and both
reduction changes. (e2)
• Oxidant and reductant
identified (e3)
• Oxidant and reductant
identified (e3)
• Oxidation AND reduction
processes explained (m4).
• One oxidation AND both
reduction processes explained
(e4).
Allow follow on
Acidified
• colourless I– to yellow (solution) / brown (solution) / black
precipitate of I2
• purple MnO4– to colourless / pale pink Mn2+
–
–
• I is reductant, MnO4 is oxidant
• I– oxidation number (ON) changes from -1 to 0
OR electrons lost
• MnO4– oxidation number (ON) changes from +7 to +2
OR electrons gained
Without acid
• purple MnO4– to brown / black solid MnO2
• MnO4– oxidation number changes from +7 to +4
OR electrons gained.
• ONE correct observed change
OR
• observations linked to 2 of
Mn2+, MnO2, I2
(a2)
• Oxidant and reductant
identified (e3)
• One oxidation or reduction
process explained (a4).
NCEA Level 3 Chemistry (90696) 2009 — page 2 of 4
Q
TWO
(a)(i)
Evidence
Achievement
• Note: reverse diagram is acceptable as long as everything is reversed
Achievement with Merit
Achievement with
Excellence
THREE of
• One half cell correct.
OR
Solutions correct + salt
bridge.
OR
Solutions correct +
correct electron flow
(a1)
FOUR of:
FOUR of:
• Diagram correct
including solutions but
lacking one label
(m1)
• Diagram labelled
correctly
(e1)
• Correct cell diagram
(e2)
• Correct cell diagram
(e2)
• Correct cell diagram
(e2)
• E° correct
(m3)
• Eo correct
(m3)
• Eo correct (unit required)
(e3)
• Correct half cell chosen
(m4)
• Correct half cell chosen
(m4)
• Correct half cell chosen
and justified
(e4)
• Both observations
OR
One observation
supported by one
statement
OR
Two supporting
statements for 1 cell.
(a5)
• Both observations,
each supported by one
statement
OR
One observation
supported by two
statements.
(m5)
• Both observations, each
supported by two
statements.
(e5)
(OR C electrodes)
(ii)
(iii)
(b)(i)
(ii)
Pt | Br–, Br2 || Cr2O72–, Cr3+ | Pt
(or Pt | Cr3+, Cr2O72– || Br2, Br–|Pt )
o
• E cell = E°RHE – E°LHE = 1.33 – (+1.07) = +0.26 V
(or consistent with 2nd cell)
= 1.07 – 1.33
•
= -0.26 V)
• Cu2+ / Cu used because largest E/potential difference
• OR Cu2+ / Cu used with E°cell for three possible cells calculated
• Items positively linked to cell (a)
Observation: orange/red produced
Statements:
E°(Cr2O72–/Cr3+)>E°(Br2/Br–)
2–
(Cr2O7 is a stronger oxidising agent than Br2)
Cr2O72– oxidises Br1- to Br2
electron flow:
from Br/Br- electrode/anode
OR to Cr2O72– electrode/cathode
Items positively linked to cell (b)
Observation: orange/red colour fades
Statements:
E°(Cu2+/Cu)<E°(Br2/Br–)
(Cu is a stronger reducing agent than Br–)
Cu reduces Br2 to Br1electron flow:
from Cu electrode/anode
OR to Br/Br- electrode/cathode
NCEA Level 3 Chemistry (90696) 2009 — page 3 of 4
Q
THREE
(a)
Evidence
• Fe2+
Achievement
Achievement with
Merit
Achievement with
Excellence
ONE of
TWO of
m1, e2 with e3, m4
3 e and 1 m
• Fe2+ recognised with one
comparison partially
explained
(m1)
• Fe2+ recognised with
one comparison
partially explained
(m1)
• Fe2+ recognised with
both comparisons
explained.
(e1)
EO for (Cu2+/Cu) compared against both others in an appropriate way
Eg
- As E°(Cu2+/Cu)> E°(Fe2+/Fe)
therefore Cu2+ is a stronger oxidising agent than Fe2+
therefore Cu2+ can oxidise Fe to Fe2+
- As E°(Cu2+/Cu)< E°(Fe3+/Fe2+)
therefore Cu2+ is a weaker oxidising agent than Fe3+
therefore Cu2+ cannot oxidise Fe2+ to Fe3+
Eg
- For Fe reacting with Cu2+
i.e.
Fe + Cu2+  Fe2+ + Cu
cell diag is: Fe|Fe2+||Cu2+|Cu
E°cell = E°RHE – E°LHE / 0.34V - -0.44V > 0
Hence spontaneous reaction oxidising Fe to Fe2+
- For Fe2+ reacting with Cu2+
i.e.
Fe2+ + Cu2+  Fe3+ + Cu (not balanced)
cell diag is: (Pt | Fe2+, Fe3+||Cu2+ | Cu)
E°cell = E°RHE – E°LHE / 0.34V – 0.77V < 0
Hence Fe2+ not oxidised further to Fe3+
Looking for:
• indication of what reaction being considered
• comparison or calculation using EO values
• indication of whether reaction occurs
NCEA Level 3 Chemistry (90696) 2009 — page 4 of 4
(b)
• Eo(HOCl / Cl2) > Eo (BrO3– / Br2) > Eo(SO42– / SO2)
• Strongest reductant: SO2
• Justification
- SO2 reduces HOCl, so SO2 is a stronger reductant than Cl2
so Eo(HOCl/Cl2) > Eo(SO42–/SO2)
- The reaction Br2 + SO42-  BrO3- + SO2 does not occur
Br2 does not reduce SO42-, so Br2 is a weaker reductant than SO2
so Eo(BrO3–/Br2)>Eo(SO42–/SO2)
-Br2 reduces HOCl, so Br2 is a stronger reductant than Cl2
so Eo(HOCl/Cl2)>Eo(BrO3– / Br2)
• Eo order correct (e2)
• Eo order correct (e2)
• Eo order correct (e2)
• SO2 correct (e3)
• SO2 correct (e3)
• SO2 correct (e3)
• Eo order for one reaction
explained
OR
Relative strengths of
oxidants / reductants
explained for one
reaction
(m4)
• Eo order for one
reaction explained
OR
Relative strengths of
oxidants / reductants
explained for one
reaction
(m4)
• Eo order explained
OR
SO2 explained
(e4)
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
2A
2M
2E+1A
NOTE:
Lower case a, m, e may be used throughout the paper to indicate contributing evidence for overall grades for questions.
Only upper case A, M and E grades shown at the end of each full question are used to make the final judgement.