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Write the skeleton equation for each of the following processes and balance them by ion electron method : (i) Permagnate ion oxidizes oxalate ions in acidic medium to carbon dioxide and gets reduced itself to Mn^(2+) ion . (ii) Bromine and hydrogen peroxide react to give bromate ions and water . (iii) Chlorine reacts with base to give chlorate ion, chloride ion and water. |
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Answer» Solution :(i) The skeleton equation for theprocess : `MnO_(4)^(-) + C_(2)O_(4)^(2-) + H^(+) to Mn^(2+)+ CO_(2) + H_(2)O ` Step (1) : Indicating oxidation number : `overset(+7)Mnoverset(-2)(O_(4)^(-)) + overset(+3)C_(2) O_(4)^(2-) to Mn^(2+) + overset(+4)Coverset(-2)O_(2) + overset(+1)H_(2) overset(-2)O` Step (2) : Writing oxidation and reduction half reaction : `overset(+3)C_(2)O_(4)^(2-) to 2overset(+4)C O_(2)`(Oxidation half ) `overset(+7)MnO_(4)^(-) to Mn^(2+) ` (Re duction half ) Step (3) : Adding electrong to make the DIFFERENCE in O.N. `overset(+3)C_(2) O_(4)^(2-) to 2overset(+4)CO_(2) + 2e^(-)` `overset(+7)MnO_(4)^(-) + 5e^(-) to Mn^(2+)` Step (4) : Balancing .O. atom by adding `H_(2)O` molecules `C_(2)O_(4)^(2-) to 2CO_(2) + 2e^(-)` `MnO_(4)^(-) + 5e^(-) to Mn^(2+) + 4H_(2)O ` Step (5) : Balancing H atom by adding `H^(+)` ions `C_(2)O_(4)^(2-) to 2CO_(2) + 2e^(-)` ` MnO_(4)^(-) + 5e^(-) + 8H^(+) to Mn^(2+) + 4H_(2)O` Step (6) Multiply the oxidation half reaction by 2 and reduction half reaction by 5 to equalize the electrons lost and gained and add the two half reactions `[ C_(2)O_(4)^(2-) to 2CO_(2) + 2e^(-) ] xx 5` `([MnO_(4)^(-) + 5e^(-) + 8H^(+) to Mn^(2+) + 4H_(2)O] xx 2)/(2MnO_(4)^(-) + 5C_(2)O_(4)^(2-) + 16H^(+) to 10CO_(2)+ 2Mn^(2+) + 8H_(2)O)` (ii) The skeleton equation for the given process is `Br_(2) + H_(2)O_(2) to BrO_(3)^(-) + H_(2)O`(in acidic medium ) Step (1) : Indicate TEH oxidation number of each atom `overset(0)Br_(2) + overset(+1)H_(2)overset(-2)O_(2) to overset(+5)Broverset(-2)(O_(3)) + overset(+1) H_(2)overset(-2)O` Thus, Br and O changes their oxidation numbers. Step (2) :Write the oxidation and reduction half reactions `overset(0) Br_(2) to 2 (overset(+5)BrO_(3)^(-))`(Oxidation half ) `H_(2) overset(-1)O_(2) to 2H_(2) overset(-2)O `(Reduction half ) Step (3) : Addition of electrons to make up for the difference in O.N. `overset(0)Br_(2) to 2 (overset(+5)BrO_(3)^(-)) + 10e^(-)` `H_(2)O_(2)^(-1) + 2e^(-) to 2H_(2)O^(-2)` Step (4) : Balance .O. ATOMS by addding `H_(2)O` molecules `Br_(2) + 6H_(2)O to 2BrO_(3)^(-) + 10e^(-)` `H_(2)O_(2) + 2e^(-) to 2H_(2)O` Step (6) : Equalize the electrons by multiplying THEREDUCTION half with 5 and add the two half reactions `Br_(2) + 6H_(2)O to 2BrO_(3)^(-) + 10e^(-) + 12H^(+)` `([H_(2)O_(2) + 2e^(-) + 2H^(+) to 2H_(2)O] xx 5)/(Br_(2) + 5H_(2)O_(2) to 2BrO_(3)^(-) + 4H_(2)O + 2H^(+))` (iii) The skeleton equation for the given process : `Cl_(2) + OH^(-) to Cl^(-) + ClO_(3)^(-) + H_(2)O ` Step (1) : Indicate teh oxidation number of each atom `overset(0)Cl_(2)+overset(-2)OH^(-) to Cl^(-) + overset(+5)ClO_(3)^(-) + overset(+1)H_(2) overset(2-)O` Thus, chlorine is the only element which undergoes the change in oxidation number. It decreases its oxidation number from 0 to `-1` and also increases its oxidation number from 0 to 5 Step (2) : Write the oxidation and reduction half reactions `Coverset(0)l_(2) to 2overset(+5)ClO_(3)^(-)`(Oxidation half) `Coverset(0)l_(2) to 2Cl^(-)` (Reduction half) Step (3) : Add electrons to make up for the difference in O.N . `overset(0)Cl_(2) to 2overset(+5)ClO_(3)^(-) + 10 e^(-)` `overset(0)Cl_(2) + 2e^(-) to 2Cl^(-)` Step (4) : Balance O atoms by adding `H_(2)O` molecules `Cl_(2) + 6H_(2)O to 2(ClO_(3))^(-) + 10 e^(-)` `Cl_(2) + 2e to Cl` Step (5) : Since medium is basic , balance H atoms by adding `H_(2)O` molecules to the side falling short of H atoms and equal number of `OH^(-)` ions to the other side . `Cl_(2) + 6H_(2)O + 12OH^(-) to 2ClO_(3)^(-) + 10e^(-) + 12H_(2)O` ` Cl_(2) + 2e^(-) to 2Cl^(-)` Step (6) : Multiply the reduction halfreaction by 5 and add two half reactions `Cl_(2) + 5H_(2) O + 2OH^(-) to 2ClO_(3)^(-) + 10e^(-) + 12H_(2)O` `oversetul([Cl_(2) + 2e^(-) to 2Cl^(-)] xx 5)(Cl_(2) + 5Cl_(2) + 12OH^(-) to)2 ClO_(3)^(-) + 10Cl^(-) + 6H_(2)O` or, `6Cl_(2) + 12OH^(-) to 2ClO_(3)^(-) + 10Cl^(-) + 6H_(2)O` or, `3Cl_(2) + 6OH^(-) to ClO_(3)^(-) + 5Cl^(-) + 3H_(2)O` |
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