Question
Explain electrolysis of lead bromide.
✓
Answer
A steel rod is connected to the negative terminal and a graphite rod to positive terminal of a battery. A silica crucible is filled to about two thirds with solid lead bromide and the rods are dipped into into it and is then melted by heating it over a Bunsen burner.
Lead bromide contains :
$PbBr _2 \rightleftharpoons Pb ^{+2}+2 Br ^{-}$
Reaction at cathode : Lead ions are positively charged they get discharged at cathode. It receives two electrons from cathode and changes to atom of lead and get deposited at the cathode.
$Pb^{+2} + 2e^-→ Pb$
Reaction at the anode : Bromide ions, being negatively charged discharge at anode. It loses its only elctron and becomes an atom of bromine. Thus the atoms of bromine combine in pairs to form molecules of bromine which escape as red vapours, at the anode.
$Br^-+ e^-→ Br$
$Br + Br → Br_2$
Lead bromide contains :
$PbBr _2 \rightleftharpoons Pb ^{+2}+2 Br ^{-}$
Reaction at cathode : Lead ions are positively charged they get discharged at cathode. It receives two electrons from cathode and changes to atom of lead and get deposited at the cathode.
$Pb^{+2} + 2e^-→ Pb$
Reaction at the anode : Bromide ions, being negatively charged discharge at anode. It loses its only elctron and becomes an atom of bromine. Thus the atoms of bromine combine in pairs to form molecules of bromine which escape as red vapours, at the anode.
$Br^-+ e^-→ Br$
$Br + Br → Br_2$
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