H2(g) + 12(g) + 2H1(g)
A student makes the following statements:
- Hydrogen always has the same oxidation number, so it has an oxidation number of +1 in both the reactant H2 and product Hi.
- Iodine is in Group VIlA, so it has an oxidation number of -1 in both the reactant 12 and
product HI.
- Neither hydrogen nor iodine changes oxidation states, so the reaction is not a redox reaction.
In three to five sentences, describe the mistake that the student made, and determine whether or not the reaction is a redox reaction. Explain your answer.
It would have been SOOOOOO much easier if you had written what you do (or don't) know about this.
First the equation is not written properly. It should be
H2(g) + I2(g) = 2HI(g) (Note the I and not 1. Note = sign not +)
H and I don't always have the same oxidation state. On the left H is 0 and so is I. On the right H is +1 and I is -1.
The fact that BOTH H and I changed oxidation state means it was a redox reaction.
Now you can put all of that into sentence form. I'm a chemist and not a writer.
It seems like the student has made a mistake in their analysis. While it is true that hydrogen typically has an oxidation number of +1, in the reactant H2 it is actually in its elemental form and has an oxidation number of 0. In the product Hi, each hydrogen atom still has an oxidation number of +1.
For iodine, it is indeed in Group VIIA and typically has an oxidation number of -1. In the reactant, 12 represents diatomic iodine molecules, where each iodine atom has an oxidation number of 0. In the product HI, each iodine atom has an oxidation number of -1.
Therefore, both hydrogen and iodine change their oxidation states in this reaction. Hydrogen goes from 0 to +1, while iodine goes from 0 to -1. This indicates that the reaction is indeed a redox reaction.