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Balance the following equation in basic medium. Cr(OH)_(3)+IO_(3)^(-)toCrO_(4)^(2-)+I^(-) |
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Answer» Solution :Step 1. The GIVEN equation is `Cr(OH)_(3)+IO_(3)^(-)toCrO_(4)^(2-)+I^(-)` Step 2. WRITING the oxidation number of ATOMS, we have `overset(+3-1+2)(Cr(OH)_(3))+overset(+5-2)(IO_(3)^(-))tooverset(+6-2)(CrO_(4)^(2-))+overset(-1)(I^(-))` Obviously, Cr(OH)3 is undergoing oxidation (the O.N. of Cr is increasing from +3 to +6 ) and `IO_(3)^(-)` is undergoing reduction (the O.N. of I is decreasing from +5 to -1). Step 3. The given reaction can be split up in the following two half reactions. `Cr(OH)_(3)toCrO_(4)^(2-)` (oxidation half reaction) `IO_(3)^(-)toI^(-)` (reduction half reaction) Step 4. (a) Balancing of oxidation half reaction : (i) The atoms other than H and O, i.e., Cr are already balanced. (ii) The given reaction proceeds in basic medium. Therefore, O atoms should be balanced by adding `OH^(-)`. They can be balanced as shown below. `Cr(OH)_(3)+OH^(-)toCrO_(4)^(2-)` The right hand side is deficient in 4H atoms. H atoms can be balanced by adding four `H_(2)O` molecules on the right and four more `OH^(-)` on the left as shown below. `Cr(OH)_(3)+OH^(-)+4HO^(-)toCrO_(4)^(2-)+4H_(2)O` or `Cr(HO)_(3)+5OH^(-)toCrO_(4)^(2-)+4H_(2)O` (iii) The right hand side is deficient in three negative CHARGES. Therefore, charge can be balanced by adding three electrons on the right. `Cr(OH)_(3)+5OH^(-)toCrO_(4)^(2-)+4H_(2)O+3e^(-)` This is the balanced oxidation half reaction. (b) Balancing of reduction half reaction (i) The iodine atoms are already balanced. (ii) Oxygen atoms can be balanced by adding three `OH^(-)` ions on the right. The left hand side is deficient in three H atoms. Therefore, adding three `H_(2)O` molecules on the left and three `OH^(-)` on the right, we have `IO_(3)^(-)+3H_(2)OtoI^(-)+3OH^(-)+3OH^(-)` `IO_(3)^(-)+3H_(2)OtoI^(-)+6OH^(-)` (iii) The charge can be balanced by adding 6 electrons on the left. `IO_(3)^(-)+3H_(2)OtoI^(-)+6OH^(-)` This is the balanced reduction half reaction. Step 5. The oxidation half reaction contains 3 electrons on the right while the reduction half reaction contains 6 electrons on the left. Electrons can be CANCELLED by multiplying the oxidation half reaction by 2 and adding it to the reduction half reaction as shown below. `{:([Cr(OH)_(3)+5OH^(-)toCrO_(4)^(2-)+4H_(2)O+3e^(-)]xx2),(IO_(3)^(-)+3H_(2)O+6e^(-)toI^(-)+6OH^(-)),(bar(2Cr(OH)_(3)+IO_(3)^(-)+100H^(-)+3H_(2)Oto2CrO_(4)^(2-)+I^(-)+6OH^(-)+8H_(2)O)),(or2Cr(OH)_(3)+IO_(3)^(-)+4OH^(-)to2CrO_(4)^(2-)+I^(-)+5H_(2)O):}` This is the final balanced equation in basic medium. |
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