NCEA Level 3 Chemistry (90696) 2012 — page 1 of 5
Assessment Schedule – 2012
Chemistry: Describe oxidation-reduction processes (90696)
Evidence Statement
Q
Evidence
ONE
(a)(i)
Cu2+(aq) + 2e  Cu(s)
(ii)
(iii)
Achievement
Achievement with Merit
Achievement with Excellence
THREE of:
• Two equations correct.
THREE of:
• Correct cell equations.
THREE of:
• Correct cell equations.
EITHER:
• Correct cell diagram
OR
• E° value calculated with error
in sign only.
•
Correct cell diagram
AND
• E° value correctly calculated
including working and units.
•
OR TWO of
• Identifies suitable substance.
• Idea of salt bridge ions
moving.
• Idea of inert / nonreactive
ions.
• Correct statement relating to
equalising charge.
• Cations move to the right.
• Anions move to the left.
• Identifies suitable substance
AND
• Correct direction of
movement of ions from the
salt bridge to the half cells
with reference to stopping
the build up of charge /
keeping neutral.
•
H2(g)  2H+(aq) + 2e
•
H2(g) + Cu2+(aq)  2H+(aq) + Cu(s)
(b)(i)
Pt | H2 | H+ || Cu2+| Cu
(ii)
E°Cell = E°Red – E°Ox
= +0.34 – (0.00)
= +0.34 V
Correct cell diagram
AND
• E° value correctly calculated
including working and units.
•
(c)
KNO3 in solution or other appropriate ionic substance. (Group 1
soluble salt or acids) (not hydroxides)
• Ionic substance in solution which contains mobile ions that do not
react with the electrodes or species in the solutions.
• NO3¯ ions move left into Hydrogen / oxidation half cell / anode
where positive charge is generated by the formation of H+ in the
reaction.
• K+ ions move right into the copper / reduction half cell / cathode
where the positive charge is lost due to Cu2+ being used in the
reaction.
• This prevents the build up of charge within the half cells, as the
oxidation-reduction reactions precede.
Identifies a suitable
substance and fully discusses
particle movement in the salt
bridge.
NCEA Level 3 Chemistry (90696) 2012 — page 2 of 5
(d)
•
•
Electrons move through the external circuit / wire from the standard
hydrogen electrode (SHE) to the copper half cell / left to right /
negative to positive / anode to cathode. Allow labelled arrow on
wire OR arrow and electron.
In the LHS half cell, H2 goes to H+ producing electrons which move
through the wire to the RHS half cell, where electrons are gained
turning Cu2+ to Cu.
Cu2+ / Cu half cell:
• Gain in the mass of the electrode / amount of Cu / pink deposit
forms
• Concentration of Cu2+ decreases in solution / weakening of the blue
colour.
2H+ / H2 half cell:
• H+ forms at the Pt electrode / Acidity increases / pH drops
• No colour change occurs / no observation
• Bubbles of H2 dissolve.
OR
Either
• Correct electron direction and
one change / observation
OR
• Two correct changes /
observations
Either
• Correct electron direction
and partially describes
changes
OR
• Correct electron direction
and reason for electron flow.
OR
•
Full justification of the
electron direction and fully
describes changes.
NCEA Level 3 Chemistry (90696) 2012 — page 3 of 5
Q
TWO
Evidence
Achievement
Note: reverse diagram is acceptable as long as everything is
reversed.
•
(a)
Fe2+  Fe3+ + e–
MnO4¯ + 8H+ + 5e–  Mn2+ + 4H2O
(b)
(c)
RHS / cathode: purple MnO4– goes to colourless Mn2+(pale
pink). Reduction since e– gained / ON +7 to +2
• LHS / anode: pale green Fe2+ goes to orange Fe3+. Oxidation
since e– lost / ON +2 to +3
• The MnO4– / Mn2+ has the higher E° value.
The reduction / permanganate half cell has the higher value
because the higher the E° number the easier it is to be reduced /
the stronger the oxidant, and the permanganate was reduced.
OR
•
Achievement with Excellence
TWO of:
TWO of:
TWO of:
Either three of:
• Correctly labels both
solutions
• Appropriate electrodes
• External circuit wire drawn
• Correct direction of electron
flow
Either three of:
• Correctly labels both solutions
• Appropriate electrodes
• External circuit wire drawn
• Correct direction of electron
flow
Either three of:
• Correctly labels both solutions
• Appropriate electrodes
• External circuit wire drawn
• Correct direction of electron
flow
OR
• Correct half-equations.
OR
• Correct half-equations.
OR
• Correct half-equations.
AND
Two of
• Correct oxidation and
reduction processes
explained using electron
transfer or oxidation number
• All species linked to colours
• Identifies MnO4– / Mn2+ has
the higher E° value with
explanation linked to
observed reactions.
AND
• Correct oxidation and
reduction processes
explained using electron
transfer or oxidation number
AND
• All species linked to colours
AND
• Identifies MnO4– / Mn2+ has
the higher E° value with
explanation linked to
observed reactions.
•
•
•
Achievement with Merit
ECell = ERed – EOx therefore ERed must be greater than EOx to give
a positive value / spontaneous reaction.
EITHER two of:
• Identifies that MnO4– is
reduced and Fe2+ is oxidised
• Links colour to the species
for two ions OR colours of
four ions poorly linked
• Identifies MnO4– / Mn2+ has
the higher E° value
OR
• Justifies oxidation or
reduction and links colours to
species for one half cell.
NCEA Level 3 Chemistry (90696) 2012 — page 4 of 5
Q
THREE
(a)
Evidence
SO42– = +6
SO2
= +4
I2
= 0
2HOCl + 2H+ + 2e  Cl2 + 2H2O
6H2O + I2  12H+ + 2IO3¯ + 10e–
10HOCl + I2  2H+ + 5Cl2 + 2IO3¯ + 4H2O
(b)(i)
Achievement
TWO of:
•
Achievement with Merit
•
Achievement with Excellence
TWO of:
•
TWO of the three oxidation
•
numbers correct.
•
•
•
TWO of the equations
correct.
•
Assigns correct E° values to
the three corresponding
couples
•
All equations correct
•
All equations correct
OR
2OCl¯ + 4H+ + 2e  Cl2 + 2H2O
6H2O + I2  12H+ + 2IO3¯ + 10e–
10OCl¯ + I2 + 8H+ 5Cl2 + 2IO3¯ + 4H2O
(ii)
E°(HOCl / Cl2) = 1.63 V
E°(SO42– / SO2) = 0.16 V
E°(IO3– / I2)
= 1.21 V
•
•
•
•
I2 reduces HOCl, so E°(HOCl / Cl2) > E°(IO3– / I2)
SO2 reduces HOCl, so E°(HOCl / Cl2) > E°(SO42– / SO2)
The reaction I2 + SO42–  IO3¯ + SO2 does not occur, so SO42–
cannot oxidise I2  IO3¯ so E°(IO3– / I2) > E°(SO42– / SO2).
The reduction half cell has the higher value because the higher
the E° number the easier it is to be reduced / the stronger the
oxidant
OR
•
ECell = ERed – EOx therefore ERed must be greater than EOx to give
a positive value / spontaneous reaction.
OR
• Gives a valid reason for one
correct E° value.
AND
• Assigns correct E° values to
corresponding couples with
partial / brief justifying
statements.
AND
Assigns correct E° values to
corresponding couples with fully
explained valid reasons.
NCEA Level 3 Chemistry (90696) 2012 — page 5 of 5
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
2A
2M+1A
OR
1E+2A
OR
1E+1M
2E