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Nguyễn Hiếu
Asked: 2024-07-16T20:13:36+00:00

cân bằng các PT oxi hóa khử sau: 1/ KI+KClO3+H2SO4 -> K2SO4+I2+KCl+H2O 2/ Cl2 +KOH -> KCl + KClO + H2O

cân bằng các PT oxi hóa khử sau:
1/ KI+KClO3+H2SO4 -> K2SO4+I2+KCl+H2O
2/ Cl2 +KOH -> KCl + KClO + H2O

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    2 Answers

    1. Vaocap3

      1. ${\rm{6K}}\mathop {\rm{I}}\limits^{{\rm{ – 1}}} {\rm{ + K}}\mathop {{\rm{Cl}}}\limits^{{\rm{ + 5}}} {{\rm{O}}_{\rm{3}}}{\rm{ + 3}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{ }} \to {\rm{ 3}}{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{ + }}{\mathop {{\rm{3I}}}\limits^{\rm{0}} _{\rm{2}}}{\rm{ + K}}\mathop {{\rm{Cl}}}\limits^{{\rm{ – 1}}} {\rm{ + 3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$

      3 x    $2\mathop {\rm{I}}\limits^{{\rm{ – 1}}}  \to 2\mathop {\rm{I}}\limits^{\rm{0}}  + 2e$

      1 x    $\mathop {{\rm{Cl}}}\limits^{{\rm{ + 5}}}  + 6e \to \mathop {{\rm{Cl}}}\limits^{{\rm{ – 1}}} $

      2. $\mathop {{\rm{Cl}}}\limits^{\rm{0}} {}_{\rm{2}}{\rm{  + 2KOH }} \to {\rm{ K}}\mathop {{\rm{Cl}}}\limits^{{\rm{ – 1}}} {\rm{  +  K}}\mathop {{\rm{Cl}}}\limits^{{\rm{ + 1}}} {\rm{O  +  }}{{\rm{H}}_{\rm{2}}}{\rm{O}}$

      1 x     $\mathop {{\rm{Cl}}}\limits^{\rm{0}}  \to \mathop {{\rm{Cl}}}\limits^{{\rm{ + 1}}} {\rm{ + 1e}}$

      1 x     $\mathop {{\rm{Cl}}}\limits^{\rm{0}}  + 1e \to \mathop {{\rm{Cl}}}\limits^{{\rm{ – 1}}} $

    2. avodarttak

      `1)KI + KClO_3 + H_2SO_4 -> K_2SO_4 + I_2 + KCl + H_2O`

      `2I^{-1} -> I_2^0 + 2e | xx 3`

      `Cl^{+5} + 6e -> Cl^{-1} | xx 1`

      `6KI + KClO_3 + 3H_2SO_4 -> 3K_2SO_4 + 3I_2 + KCl + 3H_2O`

      `2)Cl_2 + KOH -> KCl + KClO + H_2O`

      `Cl^0 + 1e -> Cl^{-1} | xx1`

      `Cl^0 -> Cl^{+1} + 1e | xx1`

      `Cl_2 + 2KOH -> KCl + KClO + H_2O`