Question

What is the following reactions using the half reaction method?1.) Pb(OH)42- + ClO- PbO2 + Cl-2.)Tl2O3 + NH2OH TlOH + N23.)Cr2O72- + CH3OH HCO2H + Cr3+

Answer 1

Explanations:

Given the unbalanced chemical reaction as shown below:

`Pb\left(OH\right)_4^(2-)+ClO^-→PbO_2+Cl^-`

Separate into half reactions

`\begin{gathered} Pb(OH)_4^(2-)→PbO_2(Oxidation) \\ ClO^-\rightarrow Cl^-(Reduction) \end{gathered}`

Balance the atoms in each half reactions except hydrogen and oxygen

`\begin{gathered} Pb(OH)_4^(2-)\rightarrow PbO_2 \\ ClO^-\rightarrow Cl^- \end{gathered}`

Balance the oxygen atoms by adding water (H2O)

`\begin{gathered} Pb(OH)_4^(2-)\rightarrow PbO_2+2H_2O \\ ClO^-\rightarrow Cl^-+H_2O \end{gathered}`
`\begin{gathered} Pb(OH)_4^(2-)\rightarrow PbO_2+2H_2O \\ ClO^-+2H^+\rightarrow Cl^-+H_2O \end{gathered}`

The hydrogen atoms has been balanced in the last equation

Balance the charges

`\begin{gathered} Pb(OH)_4^(2-)\rightarrow PbO_2+2H_2O+2e^- \\ ClO^-+2H^++2e^-\rightarrow Cl^-+H_2O \end{gathered}`

Cancel the charge and add the half reactions to have:

`Pb(OH)_4^(2-)+ClO^-+2H^+\rightarrow PbO_2+Cl^-+3H_2O`