What is being oxidized in each of the following reactions?
(a) Sr(s) + F2(g) SrF2(s)
F2
SrF2
Sr
.(b) H2(g) + I2(g) 2 HI(g)
H2
HI
I2
.(c) 2 Li(s) + Pb2+(aq) 2 Li+(aq) + Pb(s)
Pb2+
Li+
Pb
Li
thanks:) can you give an explanation. I know a is Sr but I don't get why. If it doesnt give you the charges then how are you suppose to know?
In each of these reactions, oxidation and reduction occur. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. To determine which species is being oxidized or reduced, you need to consider changes in oxidation numbers.
Let's go through each reaction and identify the oxidized species:
(a) Sr(s) + F2(g) -> SrF2(s)
In this reaction, Sr starts with an oxidation number of 0 since it is an uncombined element. F2(g) starts with an oxidation number of 0 as well. In SrF2(s), the oxidation number of Sr is +2, and the oxidation number of F is -1.
To determine which species is oxidized, we observe that the oxidation number of Sr increases from 0 to +2. This indicates that Sr has lost electrons and has been oxidized. Hence, in this reaction, Sr is being oxidized.
(b) H2(g) + I2(g) -> 2 HI(g)
In this reaction, H2 starts with an oxidation number of 0 since it is an uncombined element. I2(g) also starts with an oxidation number of 0. In HI(g), the oxidation number of H is +1, and the oxidation number of I is -1.
To determine which species is oxidized, we observe that the oxidation number of H increases from 0 to +1. This indicates that H has lost electrons and has been oxidized. Therefore, in this reaction, H2 is being oxidized.
(c) 2 Li(s) + Pb2+(aq) -> 2 Li+(aq) + Pb(s)
In this reaction, Li starts with an oxidation number of 0 since it is an uncombined metal. Pb2+(aq) has an oxidation number of +2. In the product, Li+(aq) has an oxidation number of +1, and Pb(s) has an oxidation number of 0.
To determine which species is oxidized, we observe that the oxidation number of Pb decreases from +2 to 0. This means that Pb has gained electrons and has been reduced. Thus, in this reaction, Pb2+ is being oxidized.
When you have ionic compounds, like SrF2 and Li+(aq), you can determine the oxidation number of elements based on their known charges. For example, Sr's charge in SrF2 is +2, and Li's charge in Li+(aq) is +1. However, when you have uncombined elements, their oxidation number is taken as 0.
In summary:
(a) Sr is being oxidized.
(b) H2 is being oxidized.
(c) Pb2+ is being oxidized.