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Part D of this question asks us to "combine the half-reactions to produce a balanced redox equation". From the oxidation reaction, I got 2 electrons on the product side and from the reduction half-reaction, I got 6 electrons on the reactant side. The correct answer for part d multiplies the oxidation half-reaction by 3 and leaves the original reduction half-reaction. How do I identify how many electrons I gained and lost, and why do I multiply the oxidation half-reaction by 3?
For the first question, I believe you identify how many electrons are gained and lost by the least common multiple of electrons in the two half reactions (in this problem 6 electrons). Since the least common multiple of electrons is 6, you would have to multiply the oxidation half reaction by 3 to match and cancel out the 6 electrons needed from the reduction half reaction.
In a balanced redox equation that combines the oxidation and reduction half reactions, you want to make sure that there are no leftover electrons, and the electrons in the products of the reduction half reaction and in the reactants of the oxidation half reaction cancel out. Thus, you would need to multiply the half reactions by a factor that gives you an equal number of electrons in both. In this case, you can multiply the oxidation half reaction by 3, which would give it an equal number of electrons to the reduction half reaction. Combining the two half reactions now will give you the overall balanced redox equation with no leftover electrons on either the product or reactants sides.
After looking at the overall net charge of all the species on both sides of the reaction, you determine how many electrons you need to add in order to have the charges equal on both sides of the reaction. From here, you need to multiply the oxidation half reaction by 3 to turn the 2 electrons into 6 electrons. As a result, when you add the oxidation and reduction half reaction, the electrons will cancel out.
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