Balance the equation for redox reactions. Show as a net ionic equation. Identify

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(I)- (aq) + (MnO4)- (aq) = (IO4)- (aq) + Mn+2 (aq) We separate the entire equation into two half-equations. One will be for the reduction half and the other, oxidation. Reduction: MnO4- --> Mn2+ Oxidation: I- --> IO4- Recall that reduction involves the gain of electrons and decrease in oxidation number. Oxidation, on the other hand, involves the loss of electrons and increase in oxidation number. To know which species had an increase/decrease in oxidation number, assign oxidation numbers to each species in the reaction. You will know it from there. We now balance the two half-equations. Let's start with the oxidation half. I- --> IO4- As you can see here, the charges are balanced, but not the number of O atoms. To compensate for this, we put 4 H2O molecules on the left side. I- + 4H2O --> IO4- Now, the O atoms are balanced. We now balance the H atoms by putting 8 H+ ions on the right side. I- + 4H2O --> IO4- + 8H+ The charges on both sides are not yet balanced. To balance the charge, place 8 electrons on the right side. I- + 4H2O --> IO4- + 8H+ + 8e- Next, we balance the reduction half. MnO4- --> Mn2+ Both the total charge and the number of atoms are not balanced. Let us balance the number of atoms first. It's obvious that the O atoms are not balanced. There are 4 O on the left and none on the right. To balance, we place 4 H2O molecules on the right side. MnO4- --> Mn2+ + 4H2O We place 8 H+ ions to balance the H. MnO4- + 8H+ --> Mn2+ + 4H2O Then, place 6 electrons on the left to balance the charges. MnO4- + 8H+ + 6e- --> Mn2+ + 4H2O Both equations are now balanced. However, the number of electrons lost is not equal to the number of electrons gained. To equalize this, multiply both equations by a number that would equalize the number of both. 6[I- + 4H2O --> IO4- + 8H+ + 8e-] 8[MnO4- + 8H+ + 6e- --> Mn2+ + 4H2O] Then, add both equations to come up with a full equation. 6I- + 24H2O --> 6IO4- + 48H+ + 48e- 8MnO4- + 64H+ + 48e- --> 8Mn2+ + 32H2O 6I- + 8MnO4- + 16H+ --> 6IO4- + 8Mn2+ + 8H2O The equation can still be reduced to lowest terms (because all the numbers here have a common factor), so.... 3I- (aq) + 4MnO4-(aq) + 8H+(aq) --> 3IO4-(aq) + 2Mn2+(aq) + 4H2O(l)

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