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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+

What Is The Following Reactions Using The Half Reaction Method1 PbOH42 ClO PbO2 Cl2Tl2O3 NH2OH TlOH N23Cr2O72 CH3OH HCO2H Cr3 class=

Sagot :

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]

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