oxidation charge of 2IO3- Solution .IO3 + I -+ ___? I2 + ___ (i think the other

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.IO3 + I -+ ___? I2 + ___ (i think the other I has a charge of -1....you just forgot to put it maybe...) Sometimes, the problem will tell you that this can happen in either an acidic medium or a basic medium. Since the problem didn't tell you where the reaction occurred, we'll assume that this reaction happened in acidic medium. 1. Separate the equation into two half-reactions. Oxidation: I- --> I2 Reduction: IO3 --> I2 In oxidation, there is an increase in oxidation number. In reduction, there is a decrease in oxidation number. It's obvious that I- underwent an increase in oxidation number (-1 to 0), so it was oxidized. If you know which got oxidized, for sure, the other half must get reduced. 2. Balance the oxidation half-reaction. I- --> I2 Ok. We have one I atom on the left and two on the right. To compensate, we need to put 2 before I- on the left side. 2I- --> I2 The equation is now balanced in terms of atoms. The charges are not yet balanced because on the left, the charge is -1 and the charge is 0. To compensate for the lack, we put two electrons on the right side. 2I- --> I2 + 2e- The atoms and charges are now balanced. 3. Balance the reduction half-reaction. IO3 --> I2 The atoms are not yet balanced. There's one I atom and 3 O atoms on the left while on the right, there's only 2 I atoms and no O atoms. We'll balance the I atoms first. We put 2 before IO3 on the left side to balance the I atoms. 2IO3 --> I2 Now, let's balance the O atoms. In acidic medium, we put H2O molecules if we want to balance the O atoms. To balance the H atoms, we put H+ ions. Since there are 6 O atoms on the left and none on the right, we'll place 6 H2O molecules on the right. 2IO3 --> I2 + 6H2O Let's now balance the H atoms. There are 12 H atoms on the right and none on the left. To compensate, let's put 12 H+ ions on the left. 2IO3 + 12H+ --> I2 + 6H2O The atoms are now balanced. The charges aren't. Let's now put 12 electrons on the left to balance the charge. 2IO3 + 12H+ + 12e- --> I2 + 6H2O 4. We now combine the two half-reactions to produce a single equation. But before that, we need to multiply both half-reactions by a certain number so that the number of electrons lost will be equal to number of electrons gained. In short, both equations SHOULD have the SAME number of electrons. 6(2I- --> I2 + 2e- ) 1(2IO3 + 12H+ + 12e- --> I2 + 6H2O) which gives 12I- --> 6I2 + 12e- 2IO3 + 12H+ + 12e- --> I2 + 6H2O Combining the two half-reactions gives 12I- + 2IO3 + 12H+ + 12e- --> 6I2 + 12e- + I2 + 6H2O Cancelling and/or combining like terms on both sides gives 12I- + 2IO3 + 12H+ -->7I2 + 6H2O (note: in the products side, we have 6I2 + I2....combine them and you get 7I2.)

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