Faraday's Law & Avogadro (Cambridge (CIE) A Level Chemistry)
Revision Note
Faraday's Law
The amount of substance that is formed at an electrode during electrolysis is proportional to:
The amount of time where a constant current to passes
The amount of charge, in coulombs, that passes through the electrolyte (strength of electric current)
The relationship between the current and time is:
Q = I x t
Q = charge (coulombs, C)
I = current (amperes, A)
t = time, (seconds, s)
The amount or the quantity of electricity can also be expressed by the faraday (F) unit
One faraday is the amount of electric charge carried by 1 mole of electrons or 1 mole of singly charged ions
1 faraday is 96 500 C mol-1
Thus, the relationship between the Faraday constant and the Avogadro constant (L) is:
F = L x e
F = Faraday’s constant (96 500 C mol-1)
L = Avogadro’s constant (6.02 x 1023 mol-1)
e = charge on an electron
Worked Example
Determine the amount of electricity required for the following reactions:
1) To deposit 1 mol of sodium
Na+ (aq) + e– → Na (s)
2) To deposit 1 mol of magnesium
Mg2+ (aq) + 2e– → Mg (s)
3) To form 1 mol of fluorine gas
2F– (aq) → F2 (g) + 2e–
4) To form 1 mole of oxygen
4OH– (aq) → O2 (g) + 2H2O (l) + 4e–
Answers:
One Faraday is the amount of charge (96 500 C) carried by 1 mole of electrons
Answer 1:
As there is one mole of electrons, one faraday of electricity (96 500 C) is needed to deposit one mole of sodium.
Answer 2:
Now, there are two moles of electrons, therefore, two faradays of electricity (2 x 96 500 C) are required to deposit one mole of magnesium.
Answer 3:
Two moles of electrons are released, so it requires two faradays of electricity (2 x 96 500 C) to form one mole of fluorine gas.
Answer 4:
Four moles of electrons are released, therefore it requires four faradays of electricity (4 x 96 500 C) to form one mole of oxygen gas.
Determining Avogadro's Constant by Electrolysis
The Avogadro’s constant (L) is the number of entities in one mole
L = 6.02 x 1023 mol-1
For example, four moles of water contains 2.41 x 1024 (6.02 x 1023 x 4) molecules of H2O
The value of L (6.02 x 1023 mol-1) can be experimentally determined by electrolysis using the following equation:
L =
Finding L experimentally
The charge on one mole of electrons is found by using a simple electrolysis experiment using copper electrodes
Apparatus set-up for finding the value of L experimentally
Method
The pure copper anode and pure copper cathode are weighed
A variable resistor is kept at a constant current of about 0.17 A
An electric current is then passed through for a certain time interval (e.g. 40 minutes)
The anode and cathode are then removed, washed with distilled water, dried with propanone, and then reweighed
Results
The cathode has increased in mass as copper is deposited
The anode has decreased in mass as the copper goes into solution as copper ions
Often, it is the decreased mass of the anode which is used in the calculation, as the solid copper formed at the cathode does not always stick to the cathode properly
Let’s say the amount of copper deposited in this experiment was 0.13 g
Calculation:
The amount of charge passed can be calculated as follows:
Q = I x t
= 0.17 x (60 x 40)
= 408 C
To deposit 0.13 g of copper (2.0 x 10-3 mol), 408 C of electricity was needed
The amount of electricity needed to deposit 1 mole of copper can therefore be calculated using simple proportion using the relative atomic mass of Cu
Calculating the amount of charge required to deposit one mole of copper table
Charge (C) | Amount of Cu deposited (mol) | Amount of Cu deposited (g) |
---|---|---|
408 | = 488.5 | 0.13 |
1 | 63.5 |
Therefore, 199 292 C of electricity is needed to deposit 1 mole of Cu
The half-equation shows that 2 mol of electrons are needed to deposit one mol of copper:
Cu2+ (aq) + 2e- → Cu (s)
So, the charge on 1 mol of electrons is:
Q =
Given that the charge on one electron is 1.60 x 10-19 C, then L equals:
L =
L = = 6.23 x 1023 mol-1
The experimentally determined value for L of 6.23 x 1023 mol-1 is very close to the theoretical value of 6.02 x 1023 mol-1
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