IB Chemistry 2 –SL – C block
Topic 9: Redox Test 2015
Paper 2: Short Answer Name:____________________
1.
(i) Define oxidation in terms of oxidation numbers.
.....................
increase in the oxidation number;
(ii) Describe using a labeled diagram, the essential components of an electrolytic cell.
Annotated diagram of cell showing: power supply/battery; electrolyte; cathode/negative electrode and anode/positive electrode;
[1]
[3]
(3)
(iii) Explain why solid magnesium chloride does not conduct electricity but molten magnesium chloride does. [2]
(solid) ions in a lattice / ions cannot move;
(molten) ions mobile / ions free to move;
(iv) Molten magnesium chloride undergoes electrolysis in an electrolytic cell. For each electrode deduce the half-equation and state whether oxidation or reduction takes place. Deduce the equation of the overall cell reaction including state symbols. [5] reduction occurs at the cathode/negative electrode and oxidation occurs at the anode/positive electrode;
Cathode/negative electrode: Na
+
+ e
–
Anode/positive electrode: 2Cl
–
→ Cl
2
→ Na;
+ 2e
–
/ Cl
–
→ Cl
2
2
+ e
–
;
Award [1 max] if the two electrodes are not labelled/labelled incorrectly for the two half-equations.
Overall cell reaction: Na
+
(l) + Cl
–
(l) → Na(l) + Cl
2
(g)
Award [1] for correct equation and [1] for correct state symbols.
Allow NaCl(l) instead of Na
+
(l) and Cl
–
(l).

IB Chemistry 2 –SL – C block
Topic 9: Redox Test 2015
2. Consider the following three reactions of unknown halogens:
I. X2(aq) + 2Y− (aq) → 2X− (aq) + Y2(aq)
II. X2(aq) + 2Z− (aq) → 2X− (aq) + Z2(aq)
III. Y2(aq) + 2Z−aq) → 2Y− (aq) + Z2(aq)
(a) Identify the species that is the strongest oxidizing agent.
....................X.........................................................
(b) Identify the species that is the strongest reducing agent.
.......................Y-....................................................................................................
(c) Explain how species Y is behaving in reactions I and III.
[1]
[1]
[2]
I. Y is oxidized and is therefore the reducing agent
III. Y is reduced and is therefor the oxidizing agent
(d) When substance Z has an oxidation state of zero, it is a brownish-yellow color. When substance X
2
is formed it is a reddish gas. Describe and explain what will be observed when a yellow-brown aqueous solution of Z is added to an aqueous solution of sodium and X ions.
[2]
No reaction so solution will stay the same (perhaps a bit diluted);
Z is not a strong enough oxidizing agent to remove the electron from X ions/X is a better reducing agent therefore it will keep it’s additional electron and remain an ion
3.
Chlorine can be made by reacting concentrated hydrochloric acid with potassium permanganate.
2KMnO
4
(aq) + 16HCl(aq) → 2MnCl
2
(aq) + 2KCl(aq) + 5Cl
2
(aq) + 8H
2
O(aq)
(a) State the oxidation number of manganese in KMnO
4
and in MnCl
2
.
KMnO
4
VII (+7) .............. MnCl
2
...II.(+2) .........................
(b) Deduce which species has been oxidized in this reaction and state the change in oxidation number that it has undergone.
....
Chlorine -1 to 0 ............................................................................................
[2]
.....................................................................................................................................................[2]
4. An iron ore sample weighing 0.8565 g is dissolved in HCL and the iron is obtained as Fe +2 (aq).
This solution is then titrated with 27.43 cm 3 of 0.04212 moldm -3 K
2
Cr
2
O
7
(aq) (M=294.20). a) The half reactions are as follows:
Fe 2+  Fe 3+ and Cr
2
O
7
2 Cr 3+
Determine the overall balanced equation for this reaction, clearly identify the oxidation and

IB Chemistry 2 –SL – C block
Topic 9: Redox Test 2015 reduction reactions. [3]
.........
oxidation: Fe 2+  Fe 3+ + 1 e-................................................................................
-
+
Cr
2
O
7
2 2Cr 3+ +7H
2
O
........Overall: 6 Fe 2+ + 14H + + Cr
2
O
7
2 6Fe 3+ + 2Cr 3+ +7H
2
O ..............
...........................................................................................................................................................
........................................................................................................................................................... b) Calculate the moles of Fe in the sample. [3]
( 27.43 cm 3 x 1/1000 * 0.04212 moldm 3 * 6 mol Fe/1 mol = mol Fe
................................................. .006932 mol Fe ...........................................................
............................................................................................................................................................
............................................................................................................................................................ c) Calculate the % Fe, by mass, in the ore sample. [2]
??? mol Fe * 55.65 g Fe/1mol Fe)/ 0.8565 g *100 = 45.04% Fe
.................... .38577 g Fe/0.8565g
. ............................................................................................
............................................................................................................................................................
............................................................................................................................................................
............................................................................................................................................................
............................................................................................................................................................
5. An excess of manganese(II) sulfate in alkaline solution was added to 100.0 cm 3 of a sample of water taken from a local lake. The oxygen in the water reacts with manganese(II) sulfate according to the equation:
2Mn 2+ (aq) + 4OH − aq) + O
2
(g) → 2MnO
2
(s) + 2H
2
O(l)
The solution was then acidified and excess potassium iodide was then added which released iodine according to the equation:
MnO
2
(s) + 2I − (aq) + 4H +
→ Mn 2+ (aq) + I
2
(aq) + 2H
2
O(l)
The iodine formed was then titrated with 3.20 x 10 -3 starch as an indicator: mol dm -3 sodium thiosulfate solution using

IB Chemistry 2 –SL – C block
Topic 9: Redox Test 2015
I
2
(aq) + 2S
2
O
3
2(aq) → S
4
O
6
2(aq) + 2I (aq)
The volume of thiosulfate solution required to react completely with the iodine was 25.80 cm 3 . a) Calculate the concentration of dissolved oxygen in ppm.
.................................
6.61ppm
........................................................................................
[2]
............................................................................................................................................................
............................................................................................................................................................ b) After 5 days’ incubation at 20°C another sample of water from the same source had a dissolved oxygen concentration of 2.50 ppm. Calculate the BOD. [1]
...............................
6.61-2.50 =4.11
.......................................................
............................................................................................................................................................ c) Comment on the health of the local lake. [2]
Not so good, because in 5 days consumed about 2/3 of the oxygen available; indicates that there is a lot of demand for oxygen from microbes in the water, thus poor health.
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