Chem1612 2014-N-12 Worksheet With Answers - The University Of Sydney - 2014 Page 3
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28CHEM1612
2014-N-13
November 2014
2+
What is [Cd
] when E
reaches 0.015 V?
cell
!��
!��
(��.������ �� ��
������
)(������ ��)
��+
[����
���� ]
����
E
= (0.04 V) –
= +0.015 V
cell
��+
!��
(�� × ���������� �� ������
)
[����
���� ]
so:
��!
[����
���� ]
= 7.0
��!
[����
���� ]
2+
If the change from the initial concentrations is x, [Cd
(aq)] = (0.200 – x) M and
2+
[Fe
(aq)] = (0.800 + x) M:
��.������ – ��
= 7.0
x = -0.075 M
��.������!��
2+
So that [Cd
(aq)] = (0.200 - 0.075) M = 0.125 M.
Answer: 0.125 M
What are the equilibrium concentrations of both ions?
����
Using E° =
lnK = 0.04 V gives K = 22.5
����
so:
��!
[����
���� ]
= 22.5
��!
[����
���� ]
2+
If the change from the initial concentrations is x, [Cd
(aq)] = (0.200 – x) M and
2+
[Fe
(aq)] = (0.800 + x) M:
��.������ – ��
= 22.5
x = 0.158 M
��.������!��
2+
2+
So that [Cd
(aq)] = (0.200 – 0.158) M = 0.042 M and [Fe
(aq)] = (0.800 + 0.158)
M = 0.958 M
2+
2+
[Cd
] = 0.042 M
[Fe
] = 0.958 M
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