Use the chemical equation to complete the activity.
H2(g) + I2(g) → 2HI(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 VIIA, so it has an oxidation number of –1 in both the reactant I2 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.
pls help me with this question i dont know how to work these out thank you!!
The mistake the student made is assuming that hydrogen always has an oxidation number of +1 and iodine always has an oxidation number of -1. The oxidation number of an element can vary depending on its chemical environment. In this case, hydrogen has an oxidation number of 0 in the reactant H2 and +1 in the product HI, while iodine has an oxidation number of 0 in I2 and -1 in HI. The reaction is indeed a redox reaction because the oxidation states of both hydrogen and iodine change during the reaction.
The mistake the student made is assuming that hydrogen always has an oxidation number of +1 and iodine always has an oxidation number of -1. In reality, the oxidation number of an element can vary depending on its chemical surroundings. In the given chemical equation, hydrogen is in the diatomic form (H2), which means its oxidation state is 0. In the product HI, hydrogen has an oxidation state of +1. Iodine, on the other hand, has an oxidation state of 0 in the reactant I2 and -1 in the product HI. Therefore, both hydrogen and iodine change their oxidation states in this reaction, which makes it a redox reaction.
Thank you for not just giving the answer to this question as it comes directly from a chemistry unit test.
It would help if you let us know what you don't understand. I suspect you just don't know how to determine oxidation states. Here is a link that will get you to a very good site that explains how you do that.
://www.chemteam.info/ChemTeamIndex.html
You want to click on the oxidation-reduction square. You will see that the oxidation states of both H and I change.