Balance the following ionic equations
(i) `Cr_(2)O_(7)^(2-)+H^(+)+I^(-)rarrCr^(3)+I_(2)+H_(2)O`
(ii) `Cr_(2)O_(7)^(2-)+H^(+)rarrCr^(3+)+Fe^(3+)+H_(2)O`
(iii) `MnO_(4)^(-)+SO_(3)^(2-)+H^(+)rarrMn^(2+)+SO_(4)^(2-)+H_(2)O`
(iv) `MnO_(4)^(-)+H^(+)+Br^(-)rarrMn^(2+)+Br_(2)+H_(2)O`

Explanation

(i) step 1 wrtie the O.N of all atoms above their respecitve symbols

step2 divide the gven skeleton e (a) in to two half reaction equatoi
reduction half equation : `Cr_(2)O_(7)rarrCr^(3+)`
oxdation half equation `E^(-)rarrI_(2)`
step 3 to balance reduction half equation (b) following the method discussed under sample problem the balanced reduction half equation is `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7 H_(2)O`
step 4 To balance the oxidation half eqution
step 5 To balance the elcrons gained in eq
step 1 write the O.N of all atoms above their respective symbols

step 2 divide the skeleton Eq (a) in to two half reaction equation
reduction half equatio `MnO_(4)^(-) rarr Mn^(2+)`
oxidatoin half equation `SO_(3)^(2-)rarr SO_(4)^(2-)`
step 3 to balance reduction half eq (b) following the method disucessed under
step 4 To balane oxidation half Eq
balance o atoms by adding `H_(2)O` molecule since there are three o atoms L.H.S of eq (f) and four on the R.H.S add 1 `H_(2)O` to he L.H.S of Eq
step 5 To balance electrons multiply Eq
`2MnO_(4)^(-)+16 H^(+)+10e^(-)rarrr2Mn^(2+)+8H_(2)O`
`5MnO_(4)^(-)+5H_(2)Orarr5SO_(4)^(2+)+10H^(+)10 E^(-)`
`2 MnO_(4)^(-)+5SO_(3)^(2-)+6H^(+)rarr2Mn^(2+)+5SO_(4)^(2-)+3H_(2)O`
this represent the correct balanced redox equation
(iv) step 1 write the O.N of all the atoms above their respective symbols

step 2 divide skeleton eq (a) into two half reaction equation
reduction half equation `MnO_(4)^(-)rarrMn^(+)`
oxidation half equation `Br^(-)rarrBr_(2)`
step 3 To balance reduction half equation (b) following the method discussed under
`MnO_(4)^(-) +8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
step 4 To balance oxidatoin half equatio (c ) balance atoms other O and H since there are two Br atoms on R.H.S and only one on L.H.S of eq (c ) therefore multiple `Br^(-)` on L.H.S of Eq (c ) by 2 we have
`2Br^(-)rarrBr_(2)+2e^(-)`
This represent balnced oxidation half equatoin
step 5 To balance the number of eletrons gaind in Eq and lost in Eq (f) multiply Eq (d) by 2 and eq (f) by 5 and add together we have
`2MnO_(4)^(-)+16H^(+)+10 E^(-)rarr2 Mn^(2)+8 H_(2)O`
`10 br^(-)rarr 5 Br_(2)+10e^(-)`
`2MnO_(4)^(-)+10 Br^(-)+16 H^(+)rarr2 Mn^(2+)+5Br_(2)+8H_(2)`
This represents the correcty balanced ionic equation