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Revision of Qualitative Analysis, Redox and Electrochemistry #2 – Answers
Question 1: Identify chemicals A to J:
A is CaI2(aq)
G is NH3(g)
B is I2(aq)
H is
NO3−(aq)
C is H2(g)
D is O2(g)
E is PbI2(s)
F is Ca(NO3)2(aq)
J is Ca(OH)2(s)
Question 2:
Write the ionic equation for the reaction between solution A and aqueous bromine:
Br2(aq) + 2I−(aq)  2Br−(aq) + I2(aq)
Question 3:
Write the ionic equation for the reaction between solution A and aqueous lead(II) nitrate:
Pb2+(aq) + 2I−(aq)  PbI(s)
Question 4:
When aqueous sodium hydroxide is added to solution A, a white precipitate is formed. This observation infers
that which cations might be present in solution A?
Al3+, Ca2+, Pb2+, Zn2+
Question 5:
What is observed when yellow solid E is heated with distilled water?
The yellow solid dissolves in the distilled water to form a colourless solution.
Question 6:
Write ionic half-equations to describe Electrolysis 1 of solution A:
Anode:
2I−(aq)  I2(aq) + 2e−
Cathode:
2H+(aq) + 2e−  H2(g)
1
)
Question 7:
a)
Write the ionic half-equation for the reaction at the anode during Electrolysis 2 of solution A:
Anode: 4OH−(aq)  O2(g) + 2H2O(l) + 4e−
b)
Explain why element B is produced at the anode during Electrolysis 1, but colourless gas D is
produced at the anode during Electrolysis 2:
The production of iodine during Electrolysis 1 infers that the solution is concentrated, therefore iodide
ions are preferentially discharged (oxidised) at the anode. The addition of distilled water prior to
Electrolysis 2 results in the formation of a dilute aqueous electrolyte. Consequently, hydroxide ions
are preferentially discharged (oxidised) at the anode.
Question 8:
If an acidified solution of potassium manganate(VII) is added to solution A, the following reactions take place:
ionic half-equation 1: MnO4−(aq) + 8H+(aq) + 5e−  Mn2+(aq) + 4H2O(l)
ionic half-equation 2: 2I−(aq)  I2(aq) + 2e−
a)
Combine ionic half-equation 1 and ionic half-equation 2 together and hence write the overall ionic
equation for the reaction:
Multiply the ionic half-equations by whole numbers to ensure that the number of electrons lost during oxidation
equals the number of electrons gained during reduction:
MnO4−(aq) + 8H+(aq) + 5e−  Mn2+(aq) + 4H2O(l)  2
2I−(aq)  I2(aq) + 2e−  5
Becomes:
2MnO4−(aq) + 16H+(aq) + 10e−  2Mn2+(aq) + 8H2O(l)
10I−(aq)  5I2(aq) + 10e−
Combine:
2MnO4−(aq) + 16H+(aq) + 10e− + 10I−(aq)  2Mn2+(aq) + 8H2O(l) + 5I2(aq) + 10e−
Cancel electrons on the left-hand-side and right-hand-side to give the overall ionic equation:
2MnO4−(aq) + 16H+(aq) + 10I−(aq)  2Mn2+(aq) + 8H2O(l) + 5I2(aq)
b)
In the reaction between potassium manganite(VII) and solution A, clearly explain:
i)
What has been oxidised: Iodide ions (−1) have been oxidised to form molecular iodine (0).
ii)
What has been reduced: Manganese ions (oxidation state +7) within the manganate(VII) ions
have been reduced to form manganese(II) ions (oxidation state +2).
iii) What is the oxidising agent: MnO4−(aq) has oxidised the iodine from –1 to 0.
iv) What is the reducing agent: I−(aq) has reduced the manganese from +7 to +2.
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