Name/CG: _________________________________
JC2 H2 Chemistry
Electrochemistry Revision
Duration 25 min
1(a) The following data refer to the ions of the element vanadium in acidic solution.
Ions
VO2+
VO2+
V3+
V2+
Colour of
aqueous ions
Yellow
Blue
Green
Violet
For each of the following situations, use Eocell data to predict what might happen
when the two reagents are mixed together. Write balanced equations for any
reactions that occur.
(i)
excess Zn(s) and VO2+ (aq)
Starting reagent is VO2+ and Zn
To check if Zn can reduce VO2+.
VO2+ + 2H+ + eV3+ + H2O
2+
Zn + 2e .
Zn
E cell
E = +0.34V
E = -0.76V
θ
θ
= E red - E oxid
= +0.34 – (-0.76)
= +1.10V
[1/2]
Since E cell > 0, reaction is feasible.
θ
θ
θ
θ
Zn(s) + 2VO2+ (aq) + 4H+(aq) → Zn2+(aq) + 2V3+ (aq) + 2H2O (l) [1/2]
Colour change of solution from blue to green. [1/2]
To check if Zn can reduce V3+.
V3+ + e- .
Zn2+ + 2e- .
E cell
V2+
Zn
E = -0.26V
E = -0.76V
θ
θ
= E red - E oxid
= -0.26 – (-0.76)
= +0.50V
[1/2]
Since E cell > 0, reaction is feasible.
θ
θ
θ
θ
Zn(s)
+
2V3+ (aq) → Zn2+(aq) + 2V2+ (aq) [1/2]
1
Colour change of solution from green to violet. [1/2]
In conclusion,
Zn(s) + VO2+ (aq) + 2H+(aq) → Zn2+(aq) + V2+ (aq) + H2O (l)
Zn can reduce VO2+ to V2+. Solution changes from blue to green and finally violet
in colour.
(ii)
S4O62- (aq) and V2+(aq)
S4O62-+ 2eV3+ + e- .
E cell
θ
2S2O32V2+
[4]
+0.09 (Reduction)
-0.26 (Oxidation)
= E red - E oxid
= 0.09 – (-0.26)
= +0.35V [1/2]
θ
θ
Since E cell > 0, reaction is feasible.
θ
Solution changes from violet to green. [1/2]
S4O62- (aq) + 2V2+ (aq) →
2S2O32- (aq) +
2V3+ (aq) [1/2]
2
(b)
The magnitudes (but not the sign) of the standard electrode potentials of two
metals B and C are given below:
Magnitude of Eo
Equation
B2+ + 2e
B
0.30
C2+ + 2e
C
0.36
When B2+/B is connected to the standard hydrogen electrode, electrons flow from
the B electrode to the platinum electrode of the standard hydrogen electrode.
When the half-cells of B2+/B and C2+/C are connected, electrons flow from the B
electrode to the C electrode.
Deduce the signs of the Eo values of the B2+/B and C2+/C half-cells. Explain
your answer.
(i)
Since electrons flow from the B electrode (i.e anode/oxidation) [1/2] to the
platinum electrode of SHE and E cell = E red - E oxid> 0
θ
θ
θ
OR
⇒ 0.00 – (– 0.30) > 0 [1/2]
⇒ sign of the E value of the B2+/B is negative [1/2]
θ
Since electrons flow from the B electrode (i.e anode/oxidation) to the C
electrode and E cell = E red - E oxid> 0
⇒ +0.36 – (– 0.30) > 0 [1/2]
⇒ sign of the E value of the C2+/C is positive [1/2]
or other logical reasoning / working
θ
θ
θ
θ
Calculate the e.m.f. of the cell made up of B2+/B and C2+/C half-cells.
(ii)
emf of cell = +0.36 – (– 0.30) = + 0.66 V [1]
no units / + sign – minus ½
(iii)
What would be the effect on the e.m.f. of the cell in (b)(ii) when the
concentration of C2+ is decreased.
C2+ + 2e
•
•
•
C
When [C2+] decreased,
the equilibrium position shifts to the left [1/2] to increase [C2+] by Le
Chatelier’s Principle.
no mark if eqm is not written
Oxidation is more likely to occur.
Electrode potential E
decreases [1/2]
Hence E cell = E
-E
decreases [1].
θ
θ
θ
θ
3
(iv)
Suggest a replacement half-cell for B2+/B half-cell which would reverse the
direction of electron flow in the C2+/C half-cell.
[8]
θ
E cell = E red - 0.36 > 0
Choose half-cell with Eθ> 0.36V
θ
Ag+/ Ag [1] or other correct half-cell
(c)
Explain the following observations as fully as you can.
The electrolysis of dilute potassium bromide chloride produces oxygen gas at
the anode, whereas the electrolysis of concentrated potassium bromide chloride
results in the formation of a orange-red a yellowish-green colouration at the
anode region.
[4]
Cl2 + 2e
2Cl4H+ (l) + O2 (g) + 4e
2 H2O (l)
E = +1.36 V
E = +1.23 V
For dilute potassium bromide:
Both Cl- and H2O migrate to anode
H2O is preferentially discharged as it has a less positive E [1/2] value compared
to Clθ
Oxidation: 2 H2O (l) → 4H+ (l) + O2 (g) + 4e [1/2]
Hence effervescence of O2 gas is observed
For concentrated potassium bromide:
Both Cl- and H2O migrate to anode
Cl- is preferentially discharged even though it is has a more positive E value
compared to H2O because the concentration of Cl- is very high. [1/2]
θ
Oxidation:
2Cl- (aq)
→ Cl2 (aq) + 2e [1/2]
Hence effervescence of pale yellow due to Cl2 is observed.
4
RJC Prelim 05/P2/Q2 (18 min)
2
Aluminium/air batteries are used as back-up power supplies in many telephone
exchanges. The structure of one such battery is shown below:
porous electrode Z
terminal X
terminal Y
aluminium alloy
electrolyte
The electrolyte used is aqueous potassium hydroxide. In the reaction, oxygen in
air is reduced while aluminium is oxidised as shown by the following equation:
Al (s) + 4OH (aq) ⎯→ Al(OH)4 (aq) + 3e
−
(a)
−
−
Is terminal Y negative or positive? Give a reason for your answer.
[2]
Terminal Y is negative.
This is because electrons are produced by the oxidation of aluminium and leave
the battery via terminal Y.
(b)
Write a balanced equation with state symbols for the reaction that takes place at
the porous electrode Z.
[1]
O2 (g) + 2H2O (l) + 4e ⎯→ 4OH (aq)
−
(c)
−
Why must electrode Z be porous?
[1]
Z must be porous so that oxygen may diffuse through the electrode and come
into contact with the electrolyte where it is reduced.
(d)
Sometimes, electrode Z becomes clogged up so that it is no longer porous. This
causes the aluminium/air battery to be potentially explosive. With the help of a
suitable equation, explain why this is so.
[2]
Instead of oxygen being reduced, water may be reduced at electrode Z forming
hydrogen gas: 2H2O (l) + 2e ⎯→ H2 (g) + 2OH (aq)
H2 gas is produced in a sealed container, so potentially explosive.
−
(e)
−
The aluminium used is specially alloyed so that it is not easily oxidised by air.
Explain why ordinary aluminium cannot be used.
[1]
5
Ordinary aluminium is protected by a dense, impervious layer of aluminium
oxide. This layer prevents aluminium from being oxidised so that it cannot act
as an anode.
(f)
Besides aluminium, zinc may also be used. Give two advantages of using
aluminium rather than zinc.
[2]



(g)
Zn2+ + 3e–
Zn EO = – 0.76V
Al3+ + 3e–
Al EO = – 1.66V
Aluminium has a more negative standard reduction potential/ bigger standard
oxidation potential than zinc so that the operating voltage of aluminium/air
battery is higher than zinc/ air battery.
Aluminium produces more electron per mole of metal so that more electrons
can be produced from aluminium/air battery than zinc/ air battery for the same
mass of metal used.
Aluminium is lighter than zinc
(Accept other sensible answers, but not aluminium is cheaper than Zn, less
corrosive than Zn)
When aqueous potassium hydroxide is replaced by aqueous sodium chloride,
it is noticed that a white solid is formed around the aluminium alloy electrode.
Name the white solid and explain why it is formed.
[2]
The white solid is aluminium hydroxide.
There are insufficient hydroxide ions to dissolve the insoluble amphoteric
aluminium hydroxide. Al(OH)3 is formed due to the precipitation reaction
between Al3+ (produced by oxidation of Al) and OH (produced by reduction of
oxygen.)
−
(h)
The aluminium/air battery with aqueous potassium hydroxide as an electrolyte
was used as an electrical source to electroplate an article with copper. If the
mass of aluminium electrode decreased by 5.4 g, calculate the mass of copper
that had plated out of the solution.
[2]
Al ≡ 3e
Cu2+ (aq) + 2e ⎯→ Cu (s)
Hence 2 Al ≡ 3 Cu
−
−
No.of mol of Al used = 5.4/27 = 0.2
No of mol of Cu formed = 3/2 x 0.2 = 0.300
Mass of Cu = 0.300 x 63.5 g = 19.1 g
6
NJC Prelim 05/P2/Q2 (15 min)
3(a) Iron is the most widely used d-block metal in the construction industry, mainly
because it is very abundant in the earth’s crust. However, we are also familiar
with the red-brown solid known as rust, which weakens the structural strength of
iron. The control of corrosion of iron is hence of great economic importance.
The scheme below outlines the stages in the rusting process.
Stage I
Fe(s)
(i)
moist air
Fe(OH)2(s)
prolonged
exposure to air
Fe2O3.xH2O(s)
red-brown rust
Why do you think that the structural strength of iron is weakened by the
formation of rust?
Formation of rust, an ionic compound, will weaken the structural strength
of iron, as rust is brittle.
(ii)
Stage I of rusting is an electrochemical process.
air
C
water
A
iron
Taking A to be the anode, and C to be the cathode of the electrochemical
cell reaction of rusting in the above diagram,
1
write ion-electron equations for the process at
A : Fe (s) → Fe
2+
(aq) + 2e─
─
C : O2 + 2H2O + 4e → 4OH
─
2
By means of an arrow, show the direction of electron flow in the
above cell.
3
Show, by a cross (X) on the diagram, one spot where you would
expect most rust to be accumulated.
7
(iii)
Use information from your Data Booklet, estimate a value for the potential
of the cell in (a)(ii), giving reasons.
Ecello = 0.44 – (- 0.40) = +0.84V
The value of the potential of the cell in (a)(ii) will be less than +0.84V
as the conditions of the cell is not of standard conditions.
[6]
(b)
Unlike iron, aluminium has resistance to atmospheric corrosion because of the
oxide layer on its surface.
(i)
State one common use of aluminium which makes use of this property.
Cans for soft drinks, car bodies.
(ii)
This oxide layer can be thickened by an electrolytic process.
On the space below, draw a fully-labelled diagram to show how this
process can take place.
[4]
Graphite
Cathode
Aluminium
Anode
dil H2SO4 (aq)
8
Homework
Duration: 9 min
1
Predict the products formed at the anode, and at the cathode, when the following
liquids are electrolyzed using inert electrodes.
(a)
NaBr(l)
For electrolysis:
At anode(+):
Br- is oxidized to Br2. Hence product is Br2.
At cathode(-):
Na+ is reduced to Na. Hence product is Na.
(b)
NaBr(aq)
•
•
•
•
•
•
(c)
Species present in dilute sodium chloride: Na+, Br-, H2O
At the anode (+)
Both H2O and Br- migrate to anode
Preferential oxidation of Br- as EӨ of Br2 / Br- (+1.07 V) is less positive than EӨ of
O2 / H2O (+1.23V)
Hence product is Br2
At the cathode (-)
Both H2O and Na+ migrate to cathode
Preferential reduction of H2O as EӨ of H2O / H2 (-0.83 V) is more positive than EӨ
of Na+ / Na (-2.71V)
Hence product is H2
CuF2(aq)
•
•
•
•
•
•
Species present in dilute sodium chloride: Cu2+, F-, H2O
At the anode (+)
Both H2O and F- migrate to anode
Preferential oxidation of H2O as EӨ of O2 / H2O (+1.23V) is less positive than EӨ
of F2 / F-- (+2.87 V)
Hence product is O2
At the cathode (-)
Both H2O and Cu2+ migrate to cathode
Preferential reduction of Cu2+ as EӨ of Cu2+ / Cu (+0.34 V) is more positive than
EӨ of H2O / H2 (-0.83V)
Hence product is Cu
[6]
9
Duration: 6 min
2(a) By means of a fully labelled diagram, describe how the standard electrode
potential of Cr3+/Cr2+ system can be measured by using standard hydrogen
electrode.
[4]
Voltmeter
Pt (s)
Cr2+ ions (1 mol dm-3) &
Cr3+ ions (1 mol dm-3)
[diagram [2], anything missing or wrong -1/2 till zero]
(b)
Use the Data Booklet to predict the outcome of mixing acidified aqueous
hydrogen peroxide with
(i)
aqueous bromine,
Br2 is reduced to BrH2O2 is oxidised to O2
E cell = 1.07 – 0.68 = +0.39 V [1/2] > 0 and reaction is feasible
Br2(aq) + H2O2 → 2Br-(aq) + O2(g) + 2H+(aq) [1] no state symbol – minus ½
Reddish-brown aqueous bromine is decolourised [1/2] and effervescence of
O2 gas is seen [1/2].
θ
(ii)
aqueous iron(II) sulfate.
Include Eocell, balanced equation and observation in your answer.
Fe2+ is oxidised to Fe3+
H2O2 is reduced to H2O
E cell = 1.77 – 0.77 = +1.00 V [1/2] > 0 and reaction is feasible
2Fe2+ (aq) + H2O2 + 2H+(aq) → 2Fe3+(aq) + 2H2O(l) [1] no s.s. – minus ½
Green Fe2+ (aq) solution turns to yellow Fe3+(aq) [1/2].
θ
[4]
10
Duration: 15 min
3(a) An electrochemical cell has an overall e.m.f. of +1.33 V at 25°C. The overall cell
reaction is shown below.
2CO (g) + O2 (g) → 2CO2 (g)
The cell diagram is given as:
Ni (s) | CO (g), CO2 (g) | CO32- (aq) || CO32- (aq) | O2 (g), CO2 (g) | Ag (s)
The reaction occurring at the silver electrode is:
2CO2 (g) + O2 (g) + 4e- → 2CO32- (aq)
(i)
Deduce the reaction occurring at the nickel electrode and calculate the
standard electrode potential for the reaction involved.
Anode:
2CO (g) +
Eθcell = Eθred – Eθoxid
+1.33 = +0.69 – Eθoxid
Eθoxid = 0.69 – 1.33
= –0.64 V [1]
(ii)
Eθ = +0.69 V
2CO32- (aq) →
4CO2 (g)
4e-
+
[1]
Draw a labelled diagram to show the laboratory experimental set-up of the
electrochemical cell under standard conditions. Indicate the direction of
the electron flow in your sketch.
Electron flow
V
CO2 (g) and CO (g)
at 298K and 1atm
Ni
electrode
2-
[CO3 (aq)] = 1moldm
-3
salt bridge
CO2 (g) and O2 (g)
at 298K and 1atm
Ag
electrode
2-
[CO3 (aq)] = 1moldm
-3
[3m, anything missing – minus ½]
11
(iii)
Suggest a replacement half cell, involving a gas, for CO32- (aq) | O2 (g),
CO2 (g) | Ag (s), which would reverse the direction of the electron flow in
the CO32- (aq) | CO2 (g), CO (g) | Ni(s) half cell.
Your answer needs to state both the electrode and the reagents of your
new half-cell.
[6]
To reverse electron flow, the selected half cell should now undergo
oxidation & the other half cell with the Ni electrode will undergo reduction
(Eθred = –0.64 V)
Eθcell = Eθred – Eθoxid
Eθcell = –0.64 – Eθoxid > 0
⇒ Eθoxid < –0.64 V
Pt (s) | H2O (l) | OH- (aq) | H2 (g)
Or
(b)
[2]
Pt(s) [1]
KOH(aq), H2(g) [1]
A modified version of the Leclanche cell is now widely used as a dry cell battery
to power portable gadgets.
Two possible half-cell reactions in the dry cell are:
[Zn(NH3)4]2+ (aq) + 2e-
Zn (s) + 4NH3 (aq)
NH4+ (aq) + MnO2 (s) + H2O (l) + e(i)
Eθ = -1.03 V
Mn(OH)3 (s) + NH3 (aq) Eθ = +1.00 V
Calculate the standard electrode potential, Eθ of the dry cell.
Eθcell = Eθred – Eθoxid
= 1.00 – (-1.03)
= +2.03 V [1]
12
(ii)
The actual e.m.f. of the cell is +1.05 V. Give a reason as to why this value
is different from the one calculated in (b)(i).
Due to non-standard conditions / non-standard concentrations. [1]
(iii)
Determine the amount of time in seconds required for a dry cell containing
6.50 g of MnO2 to produce a constant current of 2.0 × 10 3 A.
[4]
No. of mol of MnO2 = 6.50 / 86.9 = 0.0748 mol [1]
−
1 mol of MnO2 ≡ 1 mol of e0.0748 =
t = 3.61 × 106 s [1]
13