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Permanagante (VIII) ion in basic solution oxidses iodide ion I^(-) to produce moleculariodine I_(2) and managanes (IV) oxide (MnO_(23)) write balaced ionic equation to repersent this redox reaction |
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Answer» Solution :Step 1 write the skeletal equatoin for the given reaction `MnO_(4)^(-)(aq)+I^(-)(aq)rarrMnO_(2)(s)+I_(2)(s)` step 2 write the O.N of all the elements above their respective symbol  step 3 find oiut the species which hve been oxidised and reduced and split the given skeletal equation in to two rection step 4 To balance oxidation half eq(ii) (a) balance all atoms other than O and H atoms since there are two I atoms on the R.H.S of Eq (ii) but only one on ethe L.H.S therefore multiply `I^(-)` ion by 2 we have (b) balance O.N by adding electrons the O.N of I in `I^(-)` ion is -1 while that in `I_(2)` is 0 thus each `I^(-)` ions loses one electron since there are two `I^(-)` ions on the L.H.S therefore add `2e^(-)` to the R.H.S of Eq (iv) we have `2I^(-)(aq)rarrI_(2)(s)+2e^(-)` (c ) balance charge not needed since charge on EITHER siede of equ(v) is balaned thus eq (v) represent the balanced oxidation half equation step 5 balance teh reduction half equation (iii) (a) balance all atoms other tha O and H not needed because Mn is already balanced ltvbrgt (b) balanced O.N by adding electrons the O.N of Mn in `Mn_(4)^(-)` on the L.H.S of equ (iii) is +7 while the O.N of Mn in `MnO_(2)` is +4 on the R.H.S therefore and `3e^(-)` to R.H.S of eq (iii) we have `MnO_(4)^(-)(aq)+3e^(-)rarrMnO_(2)(s)` (c ) balance charge by adding `OH^(-)` ions sicne the reaction occurs in the basic medium Therefore add add 4 `OH^(-)` to the R.H.S of Eq (vi) we have (d) balance O atoms The R.H.S of eq (vii) has six o atom s while the L.H.S has only four therefore add `2H_(2)O` to the L.H.S of E(vii) we `MnO_(4)^(-)(aq)+2H_(2)O(aq)+2H_(2)O(l)+3e^(-)rarrMnO_(2)(s)+4OH^(-)(aq)` by doing so H atoms are automatically balanced therefore eq (VIII) represents the balance reduction half equatoin step 6 to balance the electrons lost in Qe (v) and gained eq (viii) multiply eq (v) by 3 and eq (viii) by 2 and add we have `6I^(-)(aq)rarr3I_(2)(s)+6e^(-)` `2MnO_(4)^(-)(aq)+4H_(2)O(l)+6e^(-)rarr2MnO_(2)(s)+8OH^(-)(aq)` `2MnO_(4)^(-)(aq)+6I^(-)(aq)+4H_(2)Orarr2MnO_(2)(s)+3I_(2)(s)+8OH^(-)(aq)` this represent the final balanced redox equation step 7 VERIFICATION total charge on L.H.S of eq (ix) =2(-1) +6(-1)=-8 total charge of R.H.S of Eq (ix) =8 since the magnitude of charge on either side of Eq (ix) isequal therefore eq (ix) represent the correct balance redox equation |
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