Electrolysis & 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
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- Q = charge (coulombs, C)
- I = current (amperes, A)
- t = time, (seconds, s)
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- 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 96,485 C mol-1
- Thus, the relationship between the Faraday constant and the Avogadro constant (L) is:
F = L x e
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- F = Faraday’s constant (96,485 C mol-1)
- L = Avogadro’s constant (6.022 x 1023 mol-1)
- e = charge on an electron
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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 (96 485 C) carried by 1 mole of electrons
Answer 1:
- One mole of electrons is used
- Therefore, one Faraday of electricity (96,485 C mol–1) is needed to deposit one mole of sodium.
Answer 2:
- Now, two moles of electrons are used
- Therefore, two Faradays of electricity (2 x 96,485 C mol–1) are required to deposit one mole of magnesium.
Answer 3:
- Two moles of electrons are released
- Therefore, it requires two Faradays of electricity (2 x 96,485 C mol–1) to form one mole of fluorine gas.
Answer 4:
- Four moles of electrons are released
- Therefore it requires four Faradays of electricity (4 x 96,485 C mol–1) to form one mole of oxygen gas.
Worked example
Calculate the amount of magnesium deposited when a current of 2.20 A flows through the molten bromide for 15 minutes.
Answer:
- The magnesium (Mg2+) ion is a positively charged cation that will move towards the cathode.
- Step 1: Write the half-equation at the cathode
Mg2+(aq) + 2e- → Mg(s)
1 mol 2 mol 1 mol
- Step 2: Calculate the charge transferred during the electrolysis
- Q = I x t
- Q = 2.20 x (60 x 15) = 1980 C
- Step 3: Determine the number of coulombs required to deposit one mole of magnesium at the cathode
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- For every one mole of electrons, the number of coulombs needed is 96,485 C mol-1
- In this case, there are two moles of electrons required
- So, the number of coulombs needed is = 2 x 96,485 = 192,970 C mol-1
- Step 4: Calculate the mass of magnesium deposited by simple proportion using the relative atomic mass of Mg
Charge (C) | Amount of Mg deposited (mol) | Amount of Mg deposited (g) |
192,970 | 1 | 24.30 |
1980 | = 0.0103 | 0.0103 x 24.30 = 0.25 |
- Therefore, 0.25 g of magnesium is deposited at the cathode
Worked example
A current of 2.18 A is delivered to an electrolytic cell for 65 min.
How many grams of the following will be obtained
1) Al from AlCl3
2) Ag from AgNO3
3) Cu from CuCl2
Answers:
To work out mass:
- Write half equation
- find mol of solid by = (where n = mol e– per mole of substance in balanced equation)
- find mass of solid = mol x M = x g
Answer 1:
- Au3+ + 3e– → Al
- mol Au = = 0.02937 mol
- g Au = 0.02937 x 196.97 g mol–1 = 5.79 g
Answer 2:
- Ag+ + 1e– → Ag
- mol Ag = = 0.08812 mol
- g Ag = 0.08812 x 107.87 g mol–1 = 9.51 g
Answer 3:
- Cu2+ + 2e– → Cu
- mol Cu = = 0.04406 mol
- g Cu = 0.04406 x 63.55 g mol–1 = 2.80 g