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Identify the species oxidized and the species reduced in the following electron-transfer reaction.

[tex]\[
\begin{array}{l}
Sn(s) + Br_2(l) \longrightarrow Sn^{2+}(aq) + 2Br^{-}(aq)
\end{array}
\][/tex]

Species oxidized: \_\_\_\_\_\_\_\_

Species reduced: \_\_\_\_\_\_\_\_

As the reaction proceeds, electrons are transferred from \_\_\_\_\_\_\_\_ to \_\_\_\_\_\_\_\_.

Sagot :

GinnyAnswer
Let's analyze the given chemical reaction step-by-step to determine the species that are oxidized and reduced:

[tex]\[ \text{Sn (s)} + \text{Br}_2 (\text{l}) \rightarrow \text{Sn}^{2+} (\text{aq}) + 2 \text{Br}^- (\text{aq}) \][/tex]

1. Determine the initial oxidation states:
- For elemental tin (Sn) in its solid state, the oxidation state is 0.
- For bromine (Br[tex]\(_2\)[/tex]) in its diatomic liquid form, the oxidation state of each bromine atom is also 0.

2. Determine the final oxidation states:
- For [tex]\(\text{Sn}^{2+}\)[/tex], the tin ion has an oxidation state of +2.
- For [tex]\(\text{Br}^-\)[/tex], each bromide ion has an oxidation state of -1.

3. Identify changes in oxidation states:
- Tin (Sn): The oxidation state changes from 0 in [tex]\(\text{Sn (s)}\)[/tex] to +2 in [tex]\(\text{Sn}^{2+}\)[/tex]. This indicates that tin loses 2 electrons.
- Bromine (Br[tex]\(_2\)[/tex]): The oxidation state changes from 0 in [tex]\(\text{Br}_2 (\text{l})\)[/tex] to -1 in [tex]\(\text{Br}^-\)[/tex]. Each bromine atom gains 1 electron, and since there are 2 bromine atoms, a total of 2 electrons are gained.

4. Determine the species oxidized and reduced:
- Oxidation (loss of electrons): Tin (Sn) loses electrons, hence it is oxidized. So, the species oxidized is [tex]\( \text{Sn} \)[/tex].
- Reduction (gain of electrons): Bromine (Br[tex]\(_2\)[/tex]) gains electrons, hence it is reduced. So, the species reduced is [tex]\( \text{Br}_2 \)[/tex].

5. Determine the electron transfer:
- Electrons are transferred from the species that is oxidized to the species that is reduced.
- Therefore, electrons are transferred from [tex]\( \text{Sn} \)[/tex] to [tex]\( \text{Br}_2 \)[/tex].

Summarizing the findings:
- Species oxidized: [tex]\( \text{Sn} \)[/tex]
- Species reduced: [tex]\( \text{Br}_2 \)[/tex]

- As the reaction proceeds, electrons are transferred from [tex]\( \text{Sn} \)[/tex] to [tex]\( \text{Br}_2 \)[/tex].

So, the complete solution to the problem is:
[tex]\[ \text{Species oxidized: } \text{Sn} \\ \text{Species reduced: } \text{Br}_2 \][/tex]
- As the reaction proceeds, electrons are transferred from [tex]\( \text{Sn} \)[/tex] to [tex]\( \text{Br}_2 \)[/tex].
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