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Balance the following redox reactions by ion electron method : (a) MnO_(4(aq))^(-)+I_((aq))^(-)toMnO_(2(s))+underset(("in basic medium"))(I_(2(s))) (b) MnO_(4(aq))^(-)+SO_(2(g))toMn_((aq))^(2+)+underset(("in acidic solution"))(HSO_(4(aq))^(-)) ( c) H_(2)O_(2(aq))+Fe_((aq))^(2+)toFe_((aq))^(3+)+underset(("in acidic solution"))(H_(2)O_((l))) (d) Cr_(2)O_(7)^(2-)+SO_(2(g))toCr_((aq))^(3+)+underset(("in acidic solution"))(SO_(4(aq))^(2-)) |
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Answer» Solution :(a) `MnO_(4(aq))^(-)+I_((aq))^(-)toMnO_(2(s))+underset("(BASIC medium)")(I_(2(s)))` Step-1 : Write the half reaction. OHR : `overset(-1)(I_((aq))^(-))tooverset(0)(I_(2(s)))` RHR : `overset(-7)(MnO_(4(aq))^(-))tooverset(+4)(MnO_(2(aq)))` Step-2 : In OHR BALANCE the I and then balance the charge while adding electrons. OHR : `2I_((aq))^(-)toI_(2(s))+2e^(-)` Step-3 : In RHR, oxidation number of Mn is reducing from +7 to +4. Therefore add `3e^(-)` to left side and balance the electric charge. `MnO_(4(aq))^(-)+3e^(-)toMnO_(2(aq))+4OH^(-)` Step-4 : Adding `2H_(2)O`, for the balancing of oxygen. `MnO_(4(aq))^(-)+2H_(2)O+3e^(-)toMnO_(2(aq))+4OH^(-)` Step-5 : In both reaction, balancing the electron O.H.R. is multiply by 3 and R.H.R. is multiply by 2 and adding the reaction.  (b) `MnO_(4(aq))^(-)+SO_(2(g))toMn_((aq))^(2+)+underset(("Acidic medium"))(HSO_(4(aq))^(-))` Step-1 : Half reaction. OHR : `overset(+4)(SO_(2(g)))tooverset(+6)(HSO_(4(aq))^(-))` RHR : `overset(-7)(MnO_(4(aq))^(-))toMn_((aq))^(2+)` Step-2 : In O.H.R. and R.H.R., balancing the oxygen atom, `H_(2)O` is added and as it is acidic medium `H^(+)` is added. `SO_(2(g))+2H_(2)O_((l))toHSO_(4(aq))^(-)+3H_((aq))^(+)` `MnO_(4(aq))^(-)+8H_((aq))^(+)toMn_((aq))^(+2)+4H_(2)O` Step-3 : To balance the electric charge `e^(-)` is added. `SO_(2(g))+2H_(2)O_((l))toHSO_(4(aq))^(-)+3H_((aq))^(+)+2e^(-)` `MnO_(4(aq))^(-)+8H_((aq))^(+)+5e^(-)toMn_((aq))^(+2)+4H_(2)O_((l))` Step-4 : In both the half reaction, balancing the `e^(-)` O.H.R. is multiplyed by 5 and R.H.R. multiplied and added by 2 and adding the reaction.  ( c) `H_(2)O_(2(aq))+Fe_((aq))^(2+)toFe_((aq))^(3+)+underset(("Acidic medium"))(H_(2)O_((l)))` Step-1 : Half reaction. O.H.R. : `Fe_((aq))^(2+)toFe_((aq))^(3+)` R.H.R. : `overset(-1)(H_(2)O_(2(aq)))tooverset(-2)(H_(2)O_((l)))` Step-2 : In R.H.R., oxygen atom is balanced by `H_(2)O` and as it is acidic medium `H^(+)` is added. `H_(2)O_(2(aq))^(+)+2H_((aq))^(+)to2H_(2)O_((l))` Step-3 : Add `e^(-)` to balance the electric charge. O.H.R. : `2Fe_((aq))^(2+)toFe_((aq))^(3+)+e^(-)` R.H.R. : `H_(2)O_(2(aq))+2H_((aq))^(+)+2e^(-)to2H_(2)O_((l))` Step-4 : For balancing half reaction both of side have to multiply by 2 in O.H.R.  (d) `Cr_(2)O_(7)^(2-)+SO_(2(g))toCr_((aq))^(3+)+underset(("Acidic solution"))(SO_(4(aq))^(2-))` Step-1 : Write half reactions. O.H.R. : `overset(+4)(SO_(2(g)))tooverset(+6)(SO_(4(aq))^(2-))` R.H.R. : `underset(+6)(Cr_(2)O_(7)^(2-))tounderset(+3)(Cr_((aq))^(3+))` Step-2 : Multiply by 2 for balance Cr in R.H.R. `Cr_(2)O_(7)^(2-)to2Cr^(+3)` Step-3 : For balancing H and O we have to add `H_(2)OandH^(+)` in acidic medium in both of the side. O.H.R. : `SO_(2(g))+2H_(2)OtoSO_(4(aq))^(2-)+4H_((aq))^(+)` R.H.R. : `Cr_(2)O_(7(aq))^(-2)+4H^(+)to2Cr_((aq))^(+3)+7H_(2)O` Step-4 : Balance charges of half reaction both of the side by electron. O.H.R. : `SO_(2(g))+2H_(2)OtoSO_(4(aq))^(2-)+4H_((aq))^(+)+2e^(-)` R.H.R. : `Cr_(2)O_(7(aq))^(-2)+14H_((aq))^(+)+6e^(-)to2Cr_((aq))^(+3)+7H_(2)O_((l))` Step-5 : For balancing electrons, both of the side multiply 3 on O.H.R. 
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