IB Chemistry – HL
FA Topic 9.2 and 19.1 and 19.2
Name:____________________
Multiple Choice: Choose the best answer. 1 mark each.
__D__1.
Which statement is correct about how current is conducted in an electrolytic
cell?
A.
Electrons move through the electrolyte and the external circuit.
B.
Ions move through the electrolyte and through the external cicuit.
C.
Ions move through the salt bridge and electrons move through the external circuit.
D.
Ions move through the electrolyte and electrons move through the external circuit.
__C__2.
A.
B.
C.
D.
Which statement about the electrolysis of molten sodium chloride is correct?
A.
A yellow-green gas would be produced at the negative electrode.
B.
A silvery metal is produced at the positive electrode.
C.
Chloride ions are attracted to the positive electrode and undergo oxidation.
D.
Sodium ions are attracted to the negative electrode and undergo oxidation.
__B__3.
Which labels of the electrolytic cell are correct?
1
2
3
4
site of reduction site of oxidation
conductor
electrolyte
site of oxidation site of reduction
electrolyte
conductor
site of reduction site of oxidation
electrolyte
conductor
site of oxidation site of reduction
conductor
electrolyte
__D__4. What species are produced at the positive and negative electrodes during the
electrolysis of molten sodium chloride?
A.
B.
C.
D.
Positive electrode
Na+(l)
Cl−(l)
Na(l)
Cl2(g)
Negative electrode
Cl2(g)
Na+(l)
Cl2(g)
Na(l)
__B__5. What happens at the positive electrode in a voltaic cell and in an electrolytic cell?
A.
B.
C.
D.
Voltaic cell
Oxidation
Reduction
Oxidation
Reduction
Electrolytic cell
Reduction
Oxidation
Oxidation
Reduction
IB Chemistry – HL
FA Topic 9.2 and 19.1 and 19.2
Name:____________________
_D___6.
Consider the overall reaction taking place in a voltaic cell.
Ag2O(s) + Mg(s) + H2O(l) → 2Ag(s) + Mg(OH)2(s)
What is the role of magnesium in the cell?
A.
The positive electrode and the oxidizing agent.
B.
The positive electrode and the reducing agent.
C.
The negative electrode and the oxidizing agent.
D.
The negative electrode and the reducing agent.
_A__7. Consider the following standard electrode potentials.
Zn2+(aq) + 2e–
Zn(s)
Cl2(g) + 2e–
2Cl–(aq)
EO = –0.76 V
EO = +1.36 V
Mg2+(aq) + 2e–
Mg(s)
EO = –2.37 V
What will happen when zinc powder is added to an aqueous solution of magnesium chloride?
A.
No reaction will take place.
B.
Chlorine gas will be produced.
C.
Magnesium metal will form.
D.
Zinc chloride will form.
__A__8.
Which statement is correct about the value of Eο?
A.
The more positive the value of Eο, the greater the driving force for reduction.
B.
The more negative the value of Eο, the greater the driving force for reduction.
C.
The more positive the value of Eο, the greater the rate of reaction.
D.
The more negative the value of Eο, the greater the rate of reaction.
__B__9.
Consider these standard electrode potentials.
Mg2+(aq) + 2e–
Mg(s)
EO = –2.36 V
Zn2+(aq) + 2e–
Zn(s)
EO = –0.76 V
What is the cell potential for the voltaic cell produced when the two half-cells are
connected?
A.
–1.60 V
C.
–3.12 V
B.
+1.60 V
D.
+3.12 V
__C_10.
What is the cell potential, in V, for the reaction that occurs when the
following two half-cells are connected?
Fe2+(aq) + 2e–
Fe(s)
EO = –0.44 V
Cr2O72–(aq) + 14H+(aq) + 6e–
A.
B.
2Cr3+(aq) + 7H2O(l)
C.
+ 1.77
D.
+ 2.65
+ 0.01
+ 0.89
EO = +1.33 V
__ C ___11. Which equation represents the reduction process occurring in the standard
hydrogen electrode?
A.
H2(g) → 2H+(aq) + 2e–
C.
2H+(aq) + 2e– → H2(g)
B.
H+(aq) + OH–(aq) → H2O(l)
D.
O2(g) + 4H+(aq) + 4e– → 2H2O(l)
IB Chemistry – HL
FA Topic 9.2 and 19.1 and 19.2
Name:____________________
“Any time you have two different metals that are physically or electrically connected and immersed in seawater, they become a
battery. Some amount of current flows between the two metals. The electrons that make up that current are supplied by one of the
metals giving up bits of itself--in the form of metal ions--to the seawater. This is called galvanic corrosion and, left unchecked, it
quickly destroys underwater metals.
The most common casualty of galvanic corrosion is a bronze or aluminum propeller on a stainless steel shaft, but metal struts, rudders,
rudder fittings, outboards, and stern drives are also at risk. The way we counteract galvanic corrosion is to add a third metal into the
circuit, one that is quicker than the other two to give up its electrons. This piece of metal is called a sacrificial anode, and most often it
is zinc. In fact, most boaters refer to sacrificial anodes simply as zincs.
It would be hard to overstate the importance of maintaining the anodes on your boat. When an anode is missing or largely wasted
away, the metal component it was installed to protect begins to dissolve--guaranteed.”
Excerpt from: http://www.boatus.com/boattech/casey/sacrificialzincs.asp?WT.src=share_EM
This block is a block of zinc on
the iron hull of the boat.
1.
Explain the chemistry of the zinc block and how it prevents the iron from oxidizing and what happens to zinc block
over time.
[3]
.............................................................................................................................................................................................
Zinc is more reactive than iorn (and other metals on boat)/Zinc more willing to oxidize/give up electrons;
Prevents the less reactive metals from converting from solid state to ion state;
Zinc block will lose mass/disintegrate over time
2.
Sodium metal can be obtained by the electrolysis of molten sodium chloride.
(i)
Explain why it is very difficult to obtain sodium from sodium chloride by any other method.
(2)
sodium is a very powerful reducing agent/high in electrochemical series;
any chemical reducing agent would need to be even higher in ECS to reduce Na+ / OWTTE;
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IB Chemistry – HL
FA Topic 9.2 and 19.1 and 19.2
(ii)
Name:____________________
Explain why an aqueous solution of sodium chloride cannot be used to obtain sodium metal by
electrolysis.
(2)
H+ ions gain electrons more readily than Na+ / hydrogen is evolved instead;
hydrogen is below Na in ECS;
if sodium were to be formed it would react with the water in the solution / OWTTE;
3. During the electrolysis of copper sulfate solution using platinum electrodes 0.636 g of copper is formed.
a. Determine the half-reactions occurring at each of the electrodes.
Cathode: Cu2+ + 2e-  Cu (s)
(2)
Anode: 4OH-  O2 (g) + 2H2O (l) + 4e(I told you guys to memorize those reactions for electrolysis of water!)
b. Calculate the volume of oxygen (measured at STP) produced at the anode.
0.636 𝑔 𝐶𝑢 𝑥
1 𝑚𝑜𝑙 𝐶𝑢
2 𝑒𝑙𝑒𝑐𝑡𝑜𝑛𝑠
1 𝑚𝑜𝑙 𝑂2
𝑥
𝑥
63.55 𝑔 𝐶𝑢
1𝑚𝑜𝑙 𝐶𝑢
4 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
𝑥
22.7 𝑑𝑚3 𝑂2
=
1 𝑚𝑜𝑙 𝑂2
(2)
.11359 dm3 O2 = 0.114 dm3 O2
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4.
(a)
(i) Draw an annotated diagram of a voltaic cell composed of a nickel electrode in 1.0 mol dm–3 nickel nitrate
solution and a magnesium electrode in 1.0 mol dm–3 magnesium nitrate solution. State the direction of
electron flow on your diagram.
(4)
correctly labelled electrodes and solutions;
labelled salt bridge;
voltmeter; direction of electron flow;
Allow bulb or ammeter
(ii)
Deduce half-equations for the oxidation and reduction reactions.
(2)
Oxidation:Mg(s) → Mg2+(aq) + 2e–;
Reduction:Ni2+(aq) + e– → Ni(s);
Ignore state symbols.
Award [1 max] if equations not labelled reduction or oxidation or
labelled the wrong way round.
Allow e instead of e–.
Penalize equilibrium sign or reversible arrows once only.
IB Chemistry – HL
FA Topic 9.2 and 19.1 and 19.2
(iii)
Name:____________________
Using Table 24 of the Data Booklet, calculate the cell potential for this cell.
(2)
-0.26 – (–2.37) = 2.11 V
correct data;
answer with unit;
Award [1] for –2.11 V or correct working of wrong values.
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5. State TWO of the conditions and TWO of the materials needed to make a standard hydrogen electrode.
(2)
any two: 101 kPa/1 atm pressure, solutions 1.0 moldm3, 25C/298K, pure substances
any two: H2 gas, platinum electrode covered in platinum black, HCl
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6.
Use the following equations to answer the questions below.
Cr2O72- (aq) + 14H+ (aq) + 6e-  2 Cr3+ (aq) + 7H2O
E = +1.33V
F2(g) + 2e-  2F-(aq)
E = +2.87V
Br2(g) + 2e-  2Br-(aq)
E = +1.09V
a. Deduce which halide will be oxidized by potassium dichromate (IV) in an acidified solution.
(2)
Br- will lose e- to Cr2O72- ; because Br- is not as strong as Cr2O72- in retaining the electron; Cr2O72- is not a strong
enough oxidizing agent to remove electrons from Fb. Deduce the cell potential for the situation you determined in part (a).
(1)
+1.33 – (+1.09)= +0.24V
............................................................................................................................................................................................
c. Determine the overall balance equation for the reaction you determined in part (a)
Oxidation: Br- Br2 + 2eReduction: Cr2O72- +14 H+ + 6e-  2 Cr3+ + 7H2O
Overall: Cr2O72- +14 H+ + 6Br-  2 Cr3+ + 7H2O + 3Br2
(all aqueous)
(3)
IB Chemistry – HL
FA Topic 9.2 and 19.1 and 19.2
Name:____________________
d. Using values in Table 1 and Table 2 of the data booklet, calculate the Gibb’s free energy change (G in kJmol-1)
for the reaction you determined in part (a).
(2)
O
∆G = -nF E
= -6(9.65x104) x .24 = 138960 C*V/mol=138960 J * 1kJ/1000J = 138.960 kJ = 140 kJ/mol (2 sig fig)
O
0.24 = E
n = moles of electrons = 6 (from complete balanced equation)
F= 9.65 X104 (from data booklet)
(FYI 1 volt-coulomb is equal to 1 joule)
7.
The Winkler method was used to determine the BOD of the pond across the street from SP. Initially 100.00 cm 3 of water was
collected and found to have 7.81 ppm of dissolved oxygen. The sample was sealed and placed in a 20°C incubator for 5 days. After
5 days, it was found that it took 21.75 cm3 of sodium thiosulfate with a concentration of 3.80x10 -3 moldm-3 in the final titration.
The Winkler method involves the following reactions:
I.
2Mn2+ (s) + O2 (aq)  2MnO2+ (s)
II.
MnO2+ (s) + 4H+ (aq)  Mn4+ (aq) + 3 H2O (l)
III.
Mn4+(aq) + 2I- (aq)  Mn2+ (aq) + I2 (aq)
IV.
I2 (aq) + 2S2O32- (aq)  S4O62-(aq) + 2I-(aq)
Using this information, determine the following.
a. The amount of dissolved oxygen that remained in the water sample after five days.
21.75cm3 S2O32- x
𝑔 𝑂𝑥𝑦𝑔𝑒𝑛
𝑥
𝑡𝑜𝑡𝑎𝑙 𝑔𝑟𝑎𝑚𝑠
1 𝑑𝑚3
3.80∗10−3 𝑚𝑜𝑙𝑠 S2O32−
1 𝑚𝑜𝑙 𝑂2
𝑥
𝑥
3
1000 𝑐𝑚
1𝑑𝑚3
4 𝑚𝑜𝑙𝑠 S2O32−
106 = 𝑝𝑝𝑚
6.612𝑥10−4
𝑥106
100 𝑐𝑚3
𝑥
32.00𝑔 𝑂2
=
1 𝑚𝑜𝑙 𝑂2
[3]
6.612x10-4 g O2
= 6.612g= 6.61 ppm (3 sig fig)
b. Determine the BOD for the pond across the way.
[1]
............7.81 – 6.61 = 1.20 (ppm) = BOD..........................................................
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c. Deduce the health of the pond and explain your answer using the data.
[2]
Healthy because not a significant amount of oxygen was consumed in the 5 days/similar oxygen values/small difference;
Indicates that there is not great demand (from microorganisms) to consume the oxygen, therefore the pond is healthy.