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What are the methods for balancing the redox reaction |
| Answer» There are two ways of balancing Redox reactions:\tOxidation number method\tHalf equation methodOxidation method: The steps to be followed-\tWrite the skeletal equation of reactants and products.\tIndicate the oxidation number of all the elements involved in the reaction.\tCalculate the increase or decrease in oxidation number per atom. Also, identify the oxidizing and reducing agents.\tMultiply the formula of oxidizing agent and reducing agent by suitable integers, so as to equalize the total increase or decrease in oxidation number as calculated in step c.\tBalance all atoms other than H and O.\tFinally balance H and O atoms by adding water molecules using hit and trial method.\tIn case of Ionic reactions:\tFor acidic medium\tFirst balance O atoms by adding water molecules to the deficient side.\tBalance H+ ions to the side deficient in H atoms.\tFor Basic medium\tFirst balance oxygen atom by adding water molecules to the deficient side.\tThen to balance hydrogen, add water molecules equal to the number of deficiency of H atoms.\tAlso add equal number of OH-\xa0ions to opposite side of the equation.Example: Permagnate ion reacts with bromide ion in basic medium to give manganese dioxide and Bromate ion .Step1: the skeletal ionic equation is :MnO4-\xa0(aq) +Br-\xa0(aq) ---> MnO2\xa0+BrO3-Step 2: assign oxidation numbers for Mn and BrStep3: calculate the increase and decrease in oxidation number and make the change equal :Step: 4 as the reaction occurs in basic medium, and the ionic charges are not equal on both sides, add 2OH-\xa0ions on the right to make it equal.Step5: finally count the hydrogen atoms and add appropriate number of water molecules on the left side to achieve balanced Redox reaction.Half reaction method or Ion electron method\tWrite the skeletal equation and indicate the oxidation number of all the elements which occur in skeletal equation\tFind out the species that are oxidized and reduced.\tSplit the skeletal equation into two half reactions: oxidation half reaction and reduction half reaction\tBalance the two-half equation separately by rules described below:\tIn each half reaction first balance the atoms of element that has undergone a change in oxidation number.\tAdd electrons to whatever side is necessary to make up the difference in oxidation number in each half reaction.\tBalance the charge by adding H+\xa0ions, if the reaction occurs in acidic medium .For basic medium, add OH-\xa0ions if the reaction occurs in basic medium.\tBalance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms\tIn the acidic medium, H atoms are balanced by adding H\xa0+\xa0ions to the side deficient in H atoms.\tHowever, in the basic medium H atoms are balanced by adding water molecules equal to number to H atoms deficient.\tAdd equal number of OH-\xa0ions to opposite side of equation.\tThe two half reactions are then multiplied by suitable integers .so that the total number of electrons gained in half reaction becomes equal to total number of electrons lost in another half reaction.\tThen the two half reactions are added up.\tTo verify the balancing, check whether the total charge on either is equal or not.Example: Let us consider the skeletal equation:Fe2+\xa0+ Cr2O72-\xa0--> Fe3+\xa0+Cr3+Step 1: Separate the equation in to two halves:Oxidation half reaction: Fe2\xa0-->\xa0Fe3+\xa0Reduction half reaction: Cr2O72-\xa0\xa0--> Cr3+Step 2: Balance the atoms other than hydrogen and oxygen in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms .For the reduction half reaction, we multiply the Cr3+\xa0by 2 to balance Cr atoms.Step 3: For reactions occurring in acidic medium, add water molecules to balance oxygen atoms and hydrogen ions are balanced by adding H atoms. Thus, we get:Cr2O72-\xa0+\xa014 H+ + 6e-\xa0--> 2 Cr3+\xa0+ 7H2OStep 4: Add electrons to one side of the half reaction to balance the charges .if needed make the number of electrons equal in two half reactions by multiplying one or both half reaction by suitable coefficient.The oxidation half reaction is thus written again to balance the charge .Now in the reduction half reaction there are 12 positive charges on the left hand side and only 6 positive charge on right hand side .Therefore, we add six electrons to left hand side .Cr2O72-\xa0+\xa014 H+ + 6e-\xa0--> 2 Cr3+\xa0+ 7H2OTo equalize the number of electrons in both reactions, we multiply oxidation half reaction by 6 and write as:6Fe2+ --> 6Fe3+\xa0+6e-Step 5: We add the two half reactions to achieve the overall reaction and cancel the electrons on each side .This give us net ionic equation:6Fe2+ + Cr2O72-\xa0+ 14 H+\xa0--> 2Cr3++6Fe3+\xa0+7H2OStep6: Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number atoms and the charges. | |