A2.2 Electrolysis Calculations

These always involve using the equation Q=I×t\textsf Q= \textsf{I}\times \textsf{t}Q=I×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}}F=Q/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++ 2e→Ca\textsf{Ca}^\textsf{2+}\textsf{+ 2e} \rightarrow \textsf{Ca}Ca2++ 2e→Ca
        or        Al3+ + 3e→Al\textsf{Al} ^\textsf{3+}\textsf{ + 3e} \rightarrow \textsf{Al}Al3+ + 3e→Al
   
so 1 mole of Ca needs 2 F                 but 1 mole of Al needs 3F
Worked example 🖊
 
Suppose a current of 2.5 ×102\textsf{2.5 }\times \textsf{10}^{2}2.5 ×102
  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}=2.5 ×10 2×60 ×60

=9 ×10 5 coulombs=\textsf{9 }\times \textsf{10 }^\textsf{5} \ \textsf{coulombs}=9 ×10 5 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}}=9 ×10 5/96,500

=93.3 F(to 3 s.f.)=\textsf{93.3}\ \textsf F (\textsf {to 3 s.f.)} =93.3 F(to 3 s.f.)
 
3. Consider the equation
​Mg2+ + 2e→Mg\textsf{Mg}^\textsf{2+} \textsf{ + 2e} \rightarrow \textsf{Mg}Mg2+ + 2e→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}=93.3/2
 
 =46.6=\textsf{46.6}=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.)}=46.6×24=1120g (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.

A2.1 Metal Reduction and Reactivity

A2.3 Alloys

Last modified 2yr ago

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	$=\textsf{46.6} \times \textsf {24} = \textsf{1120g} \ \textsf{(3 s.f.)}$