A2.2 Electrolysis Calculations
These always involve using the equation
Q=I×t\textsf Q= \textsf{I}\times \textsf{t}
Note: I is the current in amps, t is the time in seconds, Q is the number of coulombs of electrical charge that have passed through the electrolytic cell. This charge can be related to the charge on one mole of electrons.
1 mole of electrons has a charge of 96,500 coulombs = 1 Faraday
so
F=Q/96,500\textsf{F}={\textsf{Q}}/{\textsf{96,500}}
You will need to take into account how many F of charge is needed to deposit 1 mole of the substance in question. So you will need the following electrode equation:
e.g.
Ca2++ 2eCa\textsf{Ca}^\textsf{2+}\textsf{+ 2e} \rightarrow \textsf{Ca}
or
Al3+ + 3eAl\textsf{Al} ^\textsf{3+}\textsf{ + 3e} \rightarrow \textsf{Al}
so 1 mole of Ca needs 2 F but 1 mole of Al needs 3F

Worked example
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Suppose a current of
2.5 ×102\textsf{2.5 }\times \textsf{10}^{2}
amps is passed through molten magnesium chloride for 1 hour. What mass of magnesium will be formed at the cathode?
Follow this scheme to answer the question:
1. Calculate Q Q = I x t
=2.5 ×10 2×60 ×60=\textsf{2.5 }\times \textsf{10 }^{2}\times\textsf{60 }\times\textsf{60}
=9 ×10 5 coulombs=\textsf{9 }\times \textsf{10 }^\textsf{5} \ \textsf{coulombs}
2. Calculate the number of moles of
electrons passed (Faradays) F = Q / 96500
=9 ×10 5/96,500=\textsf{9 }\times \textsf{10 }^\textsf{5}{\textsf/\textsf{96,500}}
=93.3 F(to 3 s.f.)=\textsf{93.3}\ \textsf F (\textsf {to 3 s.f.)}
3. Consider the equation
Mg2+ + 2eMg\textsf{Mg}^\textsf{2+} \textsf{ + 2e} \rightarrow \textsf{Mg}
4. So no. of moles of substance formed
= moles of substance / no of moles of e in the equation.
=93.3/2=\textsf{93.3}/\textsf{2}
=46.6=\textsf{46.6}
5. Mass of product = no of moles
×\times
Molar mass
=46.6×24=1120g (3 s.f.)=\textsf{46.6} \times \textsf {24} = \textsf{1120g} \ \textsf{(3 s.f.)}

Author's tip
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Hint: keep all the numbers in the calculator and only round off to three significant figures (3 s.f.) at step 5.
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