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

Question

1930 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-10T10:04:05

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

$P(+IV)$ 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?

How hydrochloric acid reduce PbCl_4PbCl_4 to lead metal, and oxidize chloride to chlorine gas?