Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To identify if each reaction is an oxidation or a reduction, it is important to understand the fundamental concepts:
1. Oxidation is the loss of electrons.
2. Reduction is the gain of electrons.
Given this understanding, let's evaluate each reaction:
1. [tex]$Be^{2+}(aq) + 2e^{-} \rightarrow Be(s)$[/tex]
- Here, the beryllium ion [tex]\(Be^{2+}\)[/tex] is gaining 2 electrons to become neutral beryllium metal [tex]\(Be\)[/tex]. Since it is gaining electrons, this reaction is a reduction.
2. [tex]$Cd(s) \rightarrow Cd^{2+}(aq) + 2e^{-}$[/tex]
- In this reaction, cadmium metal [tex]\(Cd\)[/tex] is losing 2 electrons to form a cadmium ion [tex]\(Cd^{2+}\)[/tex]. Since it is losing electrons, this reaction is an oxidation.
3. [tex]$Co^{3+}(aq) + e^{-} \rightarrow Co^{2+}(aq)$[/tex]
- Here, the cobalt ion [tex]\(Co^{3+}\)[/tex] is gaining 1 electron to become a cobalt ion [tex]\(Co^{2+}\)[/tex]. Since it is gaining electrons, this reaction is a reduction.
4. [tex]$Al(s) \rightarrow Al^{3+}(aq) + 3e^{-}$[/tex]
- In this reaction, aluminum metal [tex]\(Al\)[/tex] is losing 3 electrons to form an aluminum ion [tex]\(Al^{3+}\)[/tex]. Since it is losing electrons, this reaction is an oxidation.
Now, let's categorize these reactions:
### Reduction:
- [tex]$Be^{2+}(aq) + 2e^{-} \rightarrow Be(s)$[/tex]
- [tex]$Co^{3+}(aq) + e^{-} \rightarrow Co^{2+}(aq)$[/tex]
### Oxidation:
- [tex]$Cd(s) \rightarrow Cd^{2+}(aq) + 2e^{-}$[/tex]
- [tex]$Al(s) \rightarrow Al^{3+}(aq) + 3e^{-}$[/tex]
In conclusion, the reactions have been identified as follows:
Reduction Reactions:
- [tex]$Be^{2+}(aq) + 2e^{-} \rightarrow Be(s)$[/tex]
- [tex]$Co^{3+}(aq) + e^{-} \rightarrow Co^{2+}(aq)$[/tex]
Oxidation Reactions:
- [tex]$Cd(s) \rightarrow Cd^{2+}(aq) + 2e^{-}$[/tex]
- [tex]$Al(s) \rightarrow Al^{3+}(aq) + 3e^{-}$[/tex]
1. Oxidation is the loss of electrons.
2. Reduction is the gain of electrons.
Given this understanding, let's evaluate each reaction:
1. [tex]$Be^{2+}(aq) + 2e^{-} \rightarrow Be(s)$[/tex]
- Here, the beryllium ion [tex]\(Be^{2+}\)[/tex] is gaining 2 electrons to become neutral beryllium metal [tex]\(Be\)[/tex]. Since it is gaining electrons, this reaction is a reduction.
2. [tex]$Cd(s) \rightarrow Cd^{2+}(aq) + 2e^{-}$[/tex]
- In this reaction, cadmium metal [tex]\(Cd\)[/tex] is losing 2 electrons to form a cadmium ion [tex]\(Cd^{2+}\)[/tex]. Since it is losing electrons, this reaction is an oxidation.
3. [tex]$Co^{3+}(aq) + e^{-} \rightarrow Co^{2+}(aq)$[/tex]
- Here, the cobalt ion [tex]\(Co^{3+}\)[/tex] is gaining 1 electron to become a cobalt ion [tex]\(Co^{2+}\)[/tex]. Since it is gaining electrons, this reaction is a reduction.
4. [tex]$Al(s) \rightarrow Al^{3+}(aq) + 3e^{-}$[/tex]
- In this reaction, aluminum metal [tex]\(Al\)[/tex] is losing 3 electrons to form an aluminum ion [tex]\(Al^{3+}\)[/tex]. Since it is losing electrons, this reaction is an oxidation.
Now, let's categorize these reactions:
### Reduction:
- [tex]$Be^{2+}(aq) + 2e^{-} \rightarrow Be(s)$[/tex]
- [tex]$Co^{3+}(aq) + e^{-} \rightarrow Co^{2+}(aq)$[/tex]
### Oxidation:
- [tex]$Cd(s) \rightarrow Cd^{2+}(aq) + 2e^{-}$[/tex]
- [tex]$Al(s) \rightarrow Al^{3+}(aq) + 3e^{-}$[/tex]
In conclusion, the reactions have been identified as follows:
Reduction Reactions:
- [tex]$Be^{2+}(aq) + 2e^{-} \rightarrow Be(s)$[/tex]
- [tex]$Co^{3+}(aq) + e^{-} \rightarrow Co^{2+}(aq)$[/tex]
Oxidation Reactions:
- [tex]$Cd(s) \rightarrow Cd^{2+}(aq) + 2e^{-}$[/tex]
- [tex]$Al(s) \rightarrow Al^{3+}(aq) + 3e^{-}$[/tex]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.