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Permangante (VII) ion, in basic solution oxidizes iodide ion I- to produce molecular iodine (I_2) and manganes (IV) oxide (MnO_2). Write a balanced ionic equation to represent this redox reaction. |
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Answer» Solution :Step -1 write the SKELETAL equation for the GIVEN reaction `MnO_4^(-)(aq)+1^(-)(aq)toMnO_2(s)+l_2(s)....(i)` Step-2: Write the O.N. of all the elements above their respective symbols.  `Mn_n^(+7)O_4^(2-)toM_n^(+4)O_2` (ON of Mn from +7t +4 i.e., 3 unit charge)....(ii) `I^(-)toI_2^(0)` (ON of iodine from -1 to 0)......(iii) Oxidation half equation `I^(-) (aq) to I_2 (s)`..........(IV) Reduction half equation `MnO_4^(-)(aq)toMnO_2(s)`......(v) Step 4: To balance oxidation half Eq. (a) Balance all atoms `2l^(-) (aq)toI_2(s)` (b) Balance O.N. by adding electrons `2l^(-) (aq)toI_2(s)+2e^(-)` .....(vi) Charge on either side of Eq (v) is balanced. THus eq (vI) represents the balanced oxidation half equation. Step 5: Balance the reduction half equation (v) Balance O.N. by adding electrons. `MnO_4^(-)(aq)+2H_2O(l)+2e^(-)toMnO_2(s)+4OH^(-) (aq)`...(vii) Step 6: To balance the electrons lost in Eq. (vi) and GAINED in Eq. (vii) Equation (vi) `times 3+` Equation (vii) `times 2` we have, `6I^(-) (aq)to3I_2(s)+6e^(-)` `2MnO_4^(-) (aq) +4H_2O(l)+6e^(-) to2MnO_2(s)+8OH^(-) (aq)` `overline(underline(2MnO_4^(-) (aq) +6I^(-) (aq)+4H_2O(I)to2MnO_2(s)+3I_2(s)+8OH^(-) (aq)))` |
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