Chem1612 2014-N-12 Worksheet With Answers - The University Of Sydney - 2014 Page 4
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28CHEM1612
2013-N-11
November 2013
Marks
o
• What is the electrochemical potential of the following cell at 25
C?
3
Fe│FeSO
(0.010 M)║(FeSO
(0.100 M)│Fe
4
4
o
As this is a concentration cell, E
= 0 V. The cell notation corresponds to the
0.100 M solution being the cathode, where reduction occurs, and the 0.010 M
solution being the anode, where oxidation occurs. The two half cells are:
2+
-
Anode:
Fe(s) à Fe
(aq, 0.010 M) + 2e
2+
-
Cathode:
Fe
(aq, 0.100 M) + 2e
à Fe(s)
2+
2+
Overall:
Fe
(aq, 0.100 M) à Fe
(aq, 0.010 M)
The potential is given by the Nernst equation for this two electron reaction:
����
E = E° -
lnQ
����
!��
!��
(��.������ �� ��
������
)(������ ��)
(��.������)
= (0 V) –
ln
= +0.0296 V
!��
�� × ���������� ������
(��.������)
Answer: +0.0296 V
• Calculate the mass of aluminium which can be produced with the same quantity of
2
electricity that is used to produce 1.00 kg of copper metal.
-1
As the molar mass of Cu is 63.55 g mol
, 1.00 kg corresponds to:
number of moles = mass / molar mass
3
-1
= 1.00 × 10
g / 63.55 g mol
= 15.7 mol.
2+
Reduction of Cu
requires 2 mol of electrons. Hence, the number of electrons
requires to produce 15.7 mol is:
number of moles of electrons = 2 × 15.7 mol = 31.5 mol
3+
Reduction of a mole of Al
requires 3 mol of electrons. Hence, the number of
moles of aluminium produced by 31.5 mol of electrons is:
number of moles of aluminium = 31.5 / 3 mol = 10.5 mol
-1
As the molar mass of aluminium is 26.98 g mol
, this corresponds to:
-1
mass = number of moles × molar mass = 10.5 mol × 26.98 g mol
= 283 g.
Answer: 283 g
ANSWER CONTINUES ON THE NEXT PAGE
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