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Balance the following reactions by ion electron method :Cl_(2) + OH^(-) rarr ClO_(3)+Cl + H_(2)O |
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Answer» Solution :Step I `:` Oxidation`: Cl_(2) rarr ClO_(3)^(-)` Reduction`: Cl_(2) rarr CL^(-)` Step II `:` Oxidation `Cl_(2) rarr 2ClO_(3)^(-)` Reduction `:` `Cl_(2) rarr2Cl^(-)` Step III `:` Oxidation `: Cl_(2) +12OH^(-) rarr 2ClO_(3)^(-) + 6 H_(2) O` Reduction `:` `Cl_(2) rarr 2Cl^(-)` Sep IV `:` Oxidation `:` `Cl_(2) +12OH^(-) rarr 2ClO_(3) + 6 H_(2) O + 10E` Reduction `:Cl_(2) + 2e rarr 2Cl^(-)` Step V `:` Oxidation`:``Cl_(2) = 12OH^(-) rarr 2ClO_(3)^(-) + 6H_(2)O+10e ` Reduction `: Cl_(2) + 2e rarr 2Cl^(-) ] xx5` `6Cl_(2) = 12OH^(-) rarr 2ClO_(3)^(-) + 10 Cl ^(-) + 6H_(2) O` or `3Cl_(2) + 6 OH^(-) rarr ClO_(3) + 5Cl^(-) + 3 H_(2)O`is the balanced reaction. Step III onwards may be replaced as `:` Step III `:` Oxidation`: Cl_(2) +6H_(2)O rarr 2ClO_(3) + 12H^(+)` Reduction `: Cl_(2) rarr 2Cl^(-)` Step IV `:` Oxidation `: Cl_(2) +6 H_(2) O rarr 2ClO_(3) +12H^(+) + 10e` Reduction `: Cl_(2) + 2e rarr 2Cl^(-)` Step V `:` Oxidation`: Cl_(2) + 6 H_(2) rarr 2ClO_(3) ^(-) + 12H ^(+) + 10e` Reduction `: Cl_(2) + 2e rarr 2eCl^(-) ] xx5` `6 Cl_(2) + 12OH^(-) rarr 2eCl^(-) + 10Cl^(-) + 6 H_(2)O` or , `3Cl_(2) + 3H_(2) O rarr ClO_(3)^(-) + 5Cl^(-) + 6 H^(+)` To REMOVE `H^(+)` ion, and equal NUMBER of `OH^(-)` IONS in both sides. `3Cl_(2) + 3H_(2) O + 6 OH^(-) rarr ClO_(3) + 5Cl^(-) +6OH^(-)` or, `3Cl_(2) + 3H_(2) O + 6 OH^(-) rarr ClO_(3) + 5Cl+ 6 H_(2) O ` or, `3Cl_(3) + 6OH^(-) rarr ClO_(3)^(-) + 5Cl^(-) +3H_(2)O` is the balanced reaction. |
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