Faraday & electro-chemical calculations
Francine Taylor-Campbell, Contributor

YOU SHOULD BE ABLE TO:

• Define the Faraday constant
• Perform calculations involving the mole

POINTS TO NOTE

• The Faraday (F) is the quantity of electricity needed to remove one mole of electrons from the cathode during electrolysis or to deposit one mole of electrons on the anode during electrolysis.
  

• The Faraday Constant is the amount of electric charge carried by one mole of electrons, that is 96,500C.

During electrolysis electrons are taken from the cathode by positive ions called cations.

Example

2H⁺(aq)+ 2e = H_2(g)

Cu²⁺_(aq)+ 2e = Cu_(s)

The ions are said to be discharged.

Electrons are deposited on the anode (+ve) by negative ions called anions.

Example

2Cl^-_(aq)= Cl_2(q) + 2e

4OH^- = 2H₂0 + 0₂ + 4e

Electrons may also leave from the anode if the anode dissolves.

Example

Cu = Cu²⁺ + 2e^-

One Faraday = 96,500 Coulombs, ie,
1F = 96,500 C

The coulomb is the unit of electrical charge and is 1 ampere flowing for 1 second(s).

i.e. coulombs = amps x sec, quantity of electricity = current x time (Q = I x t)

For example: When 2 amps flow for 1 minute, the quantity of electricity flowing (Q)

Q = 2*60 = 120C

NOTE

The Faraday may also be regarded as the charge on 1mole of electrons. Thus F = Le, L = Avogadro's number e = the charge on one electron

Further Examples

What mass of copper would be deposited during electrolysis by 0.5F?

Cu²⁺ + 2e = Cu

2F 64g

thus 0.5F = 16g Cu

Example

What mass of lead, would be produced by a current of 5A, passed through molten lead bromide for 1hr?

C = A x s C = 5 x 60 x 60 = 18,000C

Now Pb²⁺(l) + 2e = Pb₍₁₎

2F 207g

193,000C = 207g Pb

18,000C = (207/ 193,000) x 18,000 = 19.2g Pb

Example

What volumes of (a)H2 (b)O2 would be liberated at R.T.P when 0.1 F is passed through dilute sulphuric acid?

4H⁺ + 4e- = 2H₂

4OH- = 2H₂O + O₂ + 4e-

Calculate the volume of H2, for example, 2H+ + 2e = H₂

2F 1mole = 24 dm³ at R.T.P

Thus 0.1F = (24/2) x 0.1 dm³ = 1.2 dm³ H2 at R.T.P and volume of O₂ = 0.6dm³ at R.T.P (from above equations)

Francine Taylor-Campbell teaches at Jamaica College. Send questions and comments to kerry-ann.hepburn@gleanerjm.com