How hydrochloric acid reduce #PbCl_4# to lead metal, and oxidize chloride to chlorine gas?

1 Answer
Apr 10, 2017

Answer:

#PbO_2(s) + 4HCl(aq) rarr PbCl_2(s) +2H_2O(l) + Cl_2(g)#

Explanation:

#P(+IV)# is reduced to #Pb(+II)#:

#Pb^(4+) + 2e^(-) rarr Pb^(2+)# #(i)#

Chloride ion is oxidized to chlorine gas:

#Cl^(-) rarr 1/2Cl_2(g) + e^(-)# #(ii)#

And thus #(i) + 2xx(ii):#

#Pb^(4+) + 2Cl^(-) + 2e^(-) rarr Pb^(2+) +Cl_2(g) +2e^-#

Which is balanced with respect to mass and charge.

We can write a more conventional stoichiometric equation, by substituting the ions for the parent compounds.......

#PbO_2 + 4HCl rarr PbCl_2(s)darr +Cl_2(g)uarr +2H_2O#

Is this balanced with respect to mass and charge?