Chem1612 2014-N-12 Worksheet With Answers - The University Of Sydney - 2014

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CHEM1612
2014-N-12
November 2014
2
• In the electrolytic production of Al, what mass of Al can be deposited in 2.00 hours
by a current of 1.8 A?
The number of moles of electrons passed in 2.00 hours by a current of 1.8 A is:
number of moles of electrons = It / F
-1
= (1.8 A)(2.00 × 60.0 × 60.0 s) / 96485 C mol
= 0.13 mol
3+
Aluminium is produced from Al
O
which contains Al
. 3 electrons are needed
2
3
to produce each Al so 3 mol of electrons are needed to produce 1 mol of Al. This
quantity of electrons will therefore deposit:
number of moles of Al = 0.13 / 3 mol = 0.045 mol
-1
As Al has a molar mass of 26.98 g mol
, this quantity corresponds to:
mass of Al = number of moles × molar mass
-1
= 0.045 mol × 26.98 g mol
= 1.2 g
Answer: 1.2 g
2
• What products would you expect at the anode and the cathode on electrolysis of a
1 M aqueous solution of NiI
? Explain your answers.
2
At the cathode, there are two possible reduction reactions:
2+
-
o
Ni
(aq) + 2e
à Ni(s)
E
= -0.24 V
+
-
-
o
2H
O(l) + 4H
(aq) + 4e
à H
(g) + OH
(aq)
E
= -0.41 V
2
2
2+
Reduction of Ni
(aq) is easier, even without considering an overpotential for
water.
At the anode, there are two possible oxidation reactions:
-
-
o
2I
(aq) à I
(g) + 2e
E
= -0.62 V
2
+
-
o
2H
O(l) à O
(g) + 4H
(aq) + 4e
E
= -0.82 V
2
2
Both reactions will have an overpotential but oxidation of iodine is easier and
this will probably occur.
Overall, Ni(s) will be produced at the cathode and I
(g) will be produced at the
2
anode.

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