# A2.2 Electrolysis Calculations

These always involve using the equation
$\textsf Q= \textsf{I}\times \textsf{t}$
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
$\textsf{F}={\textsf{Q}}/{\textsf{96,500}}$
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.
$\textsf{Ca}^\textsf{2+}\textsf{+ 2e} \rightarrow \textsf{Ca}$
or
$\textsf{Al} ^\textsf{3+}\textsf{ + 3e} \rightarrow \textsf{Al}$
so 1 mole of Ca needs 2 F but 1 mole of Al needs 3F

### Worked example 🖊

Suppose a current of
$\textsf{2.5 }\times \textsf{10}^{2}$
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 ​$=\textsf{2.5 }\times \textsf{10 }^{2}\times\textsf{60 }\times\textsf{60}$ $=\textsf{9 }\times \textsf{10 }^\textsf{5} \ \textsf{coulombs}$​ 2. Calculate the number of moles of electrons passed (Faradays) F = Q / 96500 ​$=\textsf{9 }\times \textsf{10 }^\textsf{5}{\textsf/\textsf{96,500}}$ $=\textsf{93.3}\ \textsf F (\textsf {to 3 s.f.)}$ 3. Consider the equation ​$\textsf{Mg}^\textsf{2+} \textsf{ + 2e} \rightarrow \textsf{Mg}$ 4. So no. of moles of substance formed = moles of substance / no of moles of e in the equation. ​$=\textsf{93.3}/\textsf{2}$ $=\textsf{46.6}$ 5. Mass of product = no of moles $\times$ Molar mass ​$=\textsf{46.6} \times \textsf {24} = \textsf{1120g} \ \textsf{(3 s.f.)}$

### Author's tip 📌

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