Describe you would obtain a sample of lead via electrolysis

Explanation:

ELECTROCHEMISTRY revision notes on electrolysis, cells, experimental methods, apparatus, batteries, fuel cells and industrial applications of electrolysis

5. The electrolysis of molten lead(II) bromide

A simple method of investigating the electrolysis of molten lead(II) bromide is described. The formation of the products of electrolysing molten lead bromide is fully explained with the appropriate electrode equations. What are the products of the electrolysis of molten lead bromide?

Reminders: Electrolysis (of lead bromide) is a way of splitting up (decomposition) of the compound (lead bromide) using electrical energy. The electrical energy comes from a d.c. (direct current) battery or power pack supply. A conducting liquid, containing ions, called the electrolyte (molten lead bromide), must contain the compound (lead bromide) that is being broken down. The electricity must flow through electrodes dipped into the electrolyte to complete the electrical circuit with the battery. Electrolysis can only happen when the circuit is complete, and an electrical current (electricity) is flowing, then the products of electrolysing molten lead(II) bromide are released on the electrode surfaces where they can be collected. Electrolysis always involves a flow of electrons in the external wires and electrodes and a flow of ions in the electrolyte and there is always a reduction at the negative cathode electrode (which attracts positive ions, cations) and an oxidation at the positive anode electrode (which attracts negative ions, anions) and it is the ions which are discharged to give the products. These revision notes on the electrolysis of molten lead bromide and other molten salts should prove useful for the new AQA chemistry, Edexcel chemistry & OCR chemistry GCSE (9–1, 9-5 & 5-1) science courses.

This is a good teacher demonstration in the school laboratory – brown vapour and silvery lump provide good evidence of what's happened. At the end of the experiment its best to pour the molten salt onto cold ceramic surface. Let the residue cool and break it up to find the silvery lump of lead - to prove you do get lead from the electrolysis of lead bromide, the orange-brown vapour of bromine is pretty obvious and pretty obnoxious! PLEASE do in a fume cupboard!

The electrode reactions and products of the electrolysis of the molten ionic compound lead bromide (the electrolyte) are illustrated by the theory diagram above.

This is quite a simpler electrolysis situation where the ionic compound lead bromide on melting provides a highly concentrated mixture of positive lead ions and negative bromide ions.

The half-equations for the electrolysis of lead(II) bromide.

(a) The negative cathode electrode reaction for the electrolysis of molten lead(II) bromide

The positive lead(II) ions are attracted to the negative electrode and are discharged to form molten lead

Pb2+(l) + 2e– ==> Pb(l)

positive ion reduction by electron gain

This is a reduction reaction because the lead ions gain electrons.

(b) The positive anode electrode reaction for the electrolysis of molten lead bromide

The negative bromide ions are attracted to the positive anode electrode and discharged to form bromine vapour.

2Br–(l) – 2e– ==> Br2(g)

or 2Br–(l) ==> Br2(g) + 2e–

negative ion oxidation by electron loss

This is an oxidation reaction because the bromide ions lose electrons.

Extra comments on the electrolysis of lead bromide and other molten ionic compounds

1. Overall equation for the electrolysis of molten lead bromide: PbBr2(l) ==> Pb(l) + Br2(g)

2. Electrolysis of molten bromide salts(l) or their concentrated aqueous solution(aq) or conc. hydrobromic acid(aq) to make bromine

SUMMARY OF PRODUCTS FROM THE ELECTROLYSIS OF LEAD(II) BROMIDE

with inert carbon electrodes

Electrolytenegative cathode productnegative electrode

cathode half-equation

positive anode productpositive electrode

anode half-equation

molten lead(II) bromide

PbBr2(l)

molten leadPb2+(l) + 2e– ==> Pb(l)bromine vapour

2Br–(l) – 2e– ==> Br2(g)

or 2Br–(l) ==> Br2(g) + 2e–

Electrolysis of other molten ionic compounds

Each provides a positive ion and a negative ion, and this molten mixture of ions constitutes the electrolyte.

answer7,570,82356 cubic centimeters;