Balance the following equations in basic solution: PbO 2 +KCl yields KClO+KPb(OH

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PbO2 + 2e- => Pb(OH)3- on left Pb has 4+ ox state. On right, Pb has 2+ ox state, so add2e-. Balance number of hydrogen with water and oxygen withhydroxide PbO2 + 2e- + 2H2O => Pb(OH)3- +OH- Next half reaction Cl- = > ClO- chlorine goes from -1 ox state to +1 ox state, so add 2 electronson right Cl- => ClO- + 2e- balance number of H with water and O with hydroxide Cl- + 2OH- => ClO- + 2e- + H2O Add both bolded equations PbO2 + 2e- + 2H2O + Cl- + 2OH- => Pb(OH)3- + OH- +ClO- + 2e- + H2O Eliminate species that appear on both sides (H2O, OH-,e-) PbO2 + H2O + Cl- + OH- = >Pb(OH)3- +ClO-

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