The Faraday and electrochemical calculations
Francine Taylor-Campbell, Contributor

Students attending the Gleaner's Editors' Forum for head boys and head girls in high schools at the Gleaner company, on April 5.- Rudolph Brown Photo

POINTS TO NOTE

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

Eg. 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.

Eg. 2Cl_(aq)= Cl2(g) + 2e 4OH- = 2H20 + 02 + 4e

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

Eg. Cu = Cu2+ + 2e-

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

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

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

Eg. 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

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 1hr?

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 0.1 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)

Now that we have looked at the basics, let us now attempt a few examination 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.

ANSWERS

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 five minutes.

(c) What would you expect to observe if the electroplated coin were placed in dilute 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 on to 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 dilute acids.

* Francine Taylor-Campbell is an independent contributor.