Answer:
Explanations:
Given the unbalanced chemical reaction as shown below:
[tex]Pb\left(OH\right)_4^{2-}+ClO^-→PbO_2+Cl^-[/tex]
Separate into half reactions
[tex]\begin{gathered} Pb(OH)_4^{2-}→PbO_2(Oxidation) \\ ClO^-\rightarrow Cl^-(Reduction) \end{gathered}[/tex]
Balance the atoms in each half reactions except hydrogen and oxygen
[tex]\begin{gathered} Pb(OH)_4^{2-}\rightarrow PbO_2 \\ ClO^-\rightarrow Cl^- \end{gathered}[/tex]
Balance the oxygen atoms by adding water (H2O)
[tex]\begin{gathered} Pb(OH)_4^{2-}\rightarrow PbO_2+2H_2O \\ ClO^-\rightarrow Cl^-+H_2O \end{gathered}[/tex][tex]\begin{gathered} Pb(OH)_4^{2-}\rightarrow PbO_2+2H_2O \\ ClO^-+2H^+\rightarrow Cl^-+H_2O \end{gathered}[/tex]
The hydrogen atoms has been balanced in the last equation
Balance the charges
[tex]\begin{gathered} Pb(OH)_4^{2-}\rightarrow PbO_2+2H_2O+2e^- \\ ClO^-+2H^++2e^-\rightarrow Cl^-+H_2O \end{gathered}[/tex]
Cancel the charge and add the half reactions to have:
[tex]Pb(OH)_4^{2-}+ClO^-+2H^+\rightarrow PbO_2+Cl^-+3H_2O[/tex]
