Chem1612 2014-N-12 Worksheet With Answers - The University Of Sydney - 2014 Page 10
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28CHEM1612
2010-N-10
November 2010
• How many minutes would be required to obtain 10.0 g of liquid mercury by passing a
2
constant current of 0.17 A through a solution containing Hg
(NO
)
(aq)?
2
3
2
The number of moles of Hg in 10.0 g is:
-1
amount of mercury = mass / molar mass = (10.0 g) / (200.59 g mol
)
= 0.0499 mol
The redox reaction is the 2 electron process below:
2+
-
Hg
(aq) + 2e
2Hg
2
Hence, for each mole of Hg, 1 mol of electrons is required so 0.0499 mol of
electrons are required.
The number of moles of electrons passed by a current I in a time t is given by:
number of moles of electrons = It / F
The time required to pass 0.0499 mol of electrons using I = 0.17 A is therefore:
-1
t = (0.0499 mol) × (96485 C mol
) / (0.17 A) = 28000 s = 470 mins
Answer: 470 mins
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