The Faraday and electrochemical calculations
Francine Taylor-Campbell, Contributor

Students of Denbigh High School, Clarendon, are busy with their studies, on Thursday, January 11. - Nathaniel Stewart/Freelance Photographer

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.

E.g. 2H+(aq) + 2e = H2(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.

E.g. 2Cl- (aq) = Cl 2(g) + 2e

4OH- = 2H20 + 0 2 + 4e

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

E.g. Cu = Cu2+ + 2e-

One Faraday = 96,500 Coulombs i.e. 1F = 96,500 C

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

That is coulombs = amps x sec, quantity of electricity = current x time (Q = I x t)

E.g. 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 1 mole of electrons.

Thus F = Le L = Avogadro's number e = the charge on one electron.

Further examples:

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

Cu²+ + 2e = Cu
2F 64g
Thus 0.5F = 16g Cu

2. What mass of lead would be produced by a current of 5A, passed through molten lead bromide for one hour?

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

Now Pb2+(l) + 2e = Pb(l)

2F 207g

193,000C = 207g Pb

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

3. What volumes of (a) H2 (b) O2 would be liberated at R.T.P. when 0.1F 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 H 2 at R.T.P

And volume of O 2 = 0.6dm3 at R.T.P ( from above equations)

Past paper questions

Examples of exam questions;

Q1 . A current of 3 amperes was passed for one minute and 36 seconds through an electrolyte containing aluminium ions. Calculate the number of moles of aluminium deposited at the cathode.

Answer

1. First of all, the amount of electricity flowing through the electrolyte needs to be determined. Q = current x time(in sec) Q = 3 x (60 + 36) = 3 x 96 = 288C

Aluminium ions are discharged according to the equation Al3+ + 3e == Al.

This means that to deposit 1 mole of aluminium, that is 27g, requires 3F or 289500C.

1 mol Al ===== 289500C

X mol Al ===== 288C, X = 288/289500 == 0.00099mol = 0.001 mol aluminium.

Q2. A solution of copper (II) nitrate was used to electroplate a silver coin with copper. (a) Write an equation to represent the reaction taking place at the anode.

(b) Originally you were given a silver coin that weighed 1.00g. Calculate the total mass of the coin after electrolysis, when a current of 5A was passed through the copper (II) nitrate solution for 5 minutes.

(c) What would you expect to observe if the electroplated coin were placed into diluted sulphuric acid?

ANSWERS

2. To electroplate the silver coin, the anode is made of copper, which dissolves in a solution of copper ions to be later deposited onto the silver coin.

At the anode: Cu (s) === Cu2+(aq) + 2e

(b) Quantity of electricity == 5 x (5x60) = 1500C

1 mol Cu or 63.5g is deposited by 2F or 193000C.

X mol Cu is deposited by 1500C.

X == 0.00777 mol Cu

Mass of Cu deposited = 0.00777 x 63.5 = 0.494 g

The total mass of the silver coin would be 1.00 + 0.494 = 1.494g.

(c) The electroplated coin would not react with sulphuric acid since the copper coat does not react with diluted acids.

Francine Taylor-Campbell is an independent contributor.