Question

(a) What should be the physical state of Lead bromide if it is to conduct current?

(b) What particles are present in pure Lead bromide?

(c) Write the equations for the reactions which take place at electrodes during the electrolysis of Lead bromide.

Answer 1

a) State of Lead bromide$\left({\mathrm{PbBr}}_2\right)$ that conducts electricity:

• The conductivity of the metal is due to the movement of free electrons.
• The electrons are free either in a molten state or in solution form.

Thus, Lead bromide should be in a molten state or liquid state to conduct electricity.

b) Lead ions $\left({\mathrm{Pb}}^{2+}\right)$ and Bromide ions $\left({\mathrm{Br}}^{-}\right)$ are present in pure Lead bromide.

c) Reactions that take place at the electrodes during the electrolysis of lead bromide are as follows;

At the anode:

$\underset{\mathrm{Bromide}\mathrm{ion}}{2{\mathrm{Br}}^{-}(\mathrm{l}})\to \underset{\mathrm{Bromine}}{{\mathrm{Br}}_2\left(\mathrm{g}\right)}+2{\mathrm{e}}^{-}$

At the cathode:

$\underset{\mathrm{Lead}\mathrm{ion}}{{\mathrm{Pb}}^{2+}(\mathrm{l}})+2{\mathrm{e}}^{-}\to \underset{\mathrm{Lead}}{\mathrm{Pb}\left(\mathrm{l}\right)}$

Overall reaction:

$\underset{\mathrm{Lead}\mathrm{Bromide}}{{\mathrm{PbBr}}_2\left(\mathrm{l}\right)}\to \underset{\mathrm{Lead}}{\mathrm{Pb}\left(\mathrm{l}\right)}+\underset{\mathrm{Bromine}}{{\mathrm{Br}}_2\left(\mathrm{g}\right)}$