Let us balacne the chemical equations by ion electron method
(a) `Cr_(2)O_(7)^(2-)+H^(+)+I^(-)rarrCr^(3+)+I_(2)+H_(2)O`
step I Separtion of the equation in to the half reactions
(i)Write O.N of all the aotms involved in the skeletion equation
`(Cr_(2)O_(7))^(2-)+overset(+1)(H)^(+)+overset(-1)(I)rarroverset(+3)Cr+overset(0)I_(2)+overset(+1)H_(2)overset(-2)O`
(ii) Identify the atoms which undergo lnchange in O.N
`(overset(+6)Cr_(2)overset(-2)O_(7))^(2-)+H^(+)+overset(-1)(I)rarr(overset(+3)Cr)^(3+)+I_(2_(0))+H_(2)O`
(iii) Find out teh species involved in oxidation and reduction half rections
StepII Balancing of oxidaiton half rection.
`2I^(-)rarrI_(2)+2e^(-)`
stepIII Balancing of redcution half reaction
The reduction half reaction is : `(overset(+6)Cr_(2)O_(7))^(2-)rarr(overset(+3)Cr)^(3+)`
(i) The decrease in O.N per Cr atom is 3 and the total decrease in O.N for two Cr atoms is 6 Therefore, add `6e^(-)` on the reactant side
(ii) Balance Cr atoms on both sides of the equation
`(Cr_(2))O_(7)^(2-)+6e^(-)rarr2Cr^(3+)`
(iii) In order to balance O atoms add seven `H_(2)O` molecules on the product side and then to balance H atoms add `14^(+)` on the reactant side.
`(Cr_(2)O_(7))^(2-)+6e^(-)+14H^(+)rarr2Cr^(3+)+7H_(2)O`
Step(iv) Adding the two half reactions
In order to equate the electrons multiply equation (i) by 3 and then add to equation (ii) in order to get the final equation
`{:(2I^(-)rarrI_(2)+2e^(-)xx3),(Cr_(2)O_(7)^(2-)+6e^(-)+14H^(+)rarr2Cr^(3+)+7H_(2)O),(Cr_(2)O_(7)^(-)+6I^(-)+14H^(+)rarr3I_(2)+2Cr^(3+)+7H_(2)O):}`
`(b) (Cr_(2)O_(7))^(2-)+Fe^(2+)+H^+rarrCr^(3+)+Fe^(3+)+H_(2)O`
For balancing consult example18.20 (Text part)
`(c )MnO_(4)^(-)+SO_(3)^(2-)+H^(+)rarrMn^(2+)+SO_(4)^(2-)+H_(2)O` ltbr step I Separarion of the equation in to two half rections
(i) Write the O.N off all the atoms involved in the skeleton in the skeleton equation
`overset(+7)(Mnoverset(-2)O_(4))^(-)+overset(+4)Soverset(-2)O_(3)]^(2-)+overset(+12)[H]^(+)rarroverset(+2)[Mn]^(2+)+overset(+6)Soverset(-2)O_(4)]^(2-)+overset(+1)H_(2)O`
(ii) Identify the atoms which undergo change in O.N
`(overset(+7)MnO_(4))^(-)+(overset(+4)SO_(3))^(2-)+H^(+)rarr(overset(+2)[Mn])^(2+)+(overset(+6)SO_(4))^(2-)+H_(2)O`
(iii) Find out the species involved in oxidation and reduction half reactions
StepII Balancing the oxidation half reaction
The oxidation half rection is :`(overset(+4)SO_(2))^(2-)rarr(overset(+6)SO_(4))^(2-)`
(i)As the increase in O.N is 2 therefore add two `e^(-)` on the product side to balance lthe change in O.N `SO_(3)^(2)rarr(SO_(4))^(2-)+2e^(-)`
(ii) In order to balance oxygen atoms, add one molecule of `H_(2)O` on the reactant side and two `H^(+)` ions on the product side.
`(SO_(3))^(2)+H_(2)Orarr(SO_(4))^(2-)+2e^(-)+2H^(+)`
step III Balancing the reduction half reaction
The reduction half recation is :
`overset(+7)(MnO_(4))^(-)rarroverset(+23)(Mn)`
(i)The decrease in O.N of Mn is 5 . Therefore, add `5e^(-)` on the reactant side to balance change in O.N
`MnO_(4)^(-)+5e^(-)rarrMn^(2+)`
(i) The decrease inO.N of Mn is 5.Therefore, add `5e^(-)` on the reactant side to balacne change in O.N `MnO_(4)^(-)+5e^(-)rarroverset(+2)(Mn)`
(ii) In order to balance the no of O atoms , add four `H_(2)O` ,molecules on the product side and then too balance H atoms add `8H^(+)` on the reactatn side.
`MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
step IV Adding the twoi half reactions
In oder to equate elctons multiply oxidation half reaction by 5 and reduction half reaction by 2.Then ad up the two equations
`[SO_(3)^(2-)+H_(32)OrarrSO_(4)^(2-)+2H^(+)+2e^(-)]xx5`
`{:([MnO_(4)^(-)+8H^(+)+Se^(-)rarrMn^(2+)+4H_(2)O]xx2),(Add:5SO_(3)^(2-)+5H_(2)O+2MnO_(4)^(-)rarr5sO_(4)^(2-)+10H^(+)):}`
`+16H^(+)+Oe^(-) 2Mn^(2+)+8H_(2)O`
or `5SO_(3)^(2+-)+2MnO_(4)^(-)+6H^(+)rarr5SO_(4)^(2-)+2Mn^(2+)+2H_(2)O`
`(b)MnO_(4)^(-)+H^(+)+Br^(-)rarrMn^(2+)+Br_(2)+H_(2)O`
please balance the equation of your own.
