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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 |
Answer» Solution :(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 respectivesymbols  step 2 divide the skeleton Eq (a) in to two halfreaction 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 atomsL.H.S of eq (f) and FOUR on theR.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 andeq (f) by5 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 |
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