Answer :
Sure, let's break down the given reaction and answer the questions step-by-step.
The given reaction is:
[tex]\[ Cu_{(s)} + Ag^{+} \rightarrow Cu^{2+} + Ag_{(s)} \][/tex]
### ai. Identifying the Species that is Reduced
In a redox reaction, reduction refers to the gain of electrons by a species. Here, we need to identify which species undergoes reduction.
Silver Ion ([tex]$Ag^+$[/tex]):
- Initially present as [tex]$Ag^+$[/tex]
- Ends as [tex]$Ag_{(s)}$[/tex]
The reduction equation for silver ion is:
[tex]\[ Ag^+ + e^- \rightarrow Ag_{(s)} \][/tex]
Thus, [tex]$Ag^+$[/tex] (silver ion) is the species that is reduced.
### oi. Writing the Half-Cell Equations
Next, let's write the half-cell equations for the oxidation and reduction processes separately.
Reduction Half-Cell Equation:
- We already identified that the silver ion ([tex]$Ag^+$[/tex]) is reduced to silver solid ([tex]$Ag_{(s)}$[/tex]).
[tex]\[ Ag^+ + e^- \rightarrow Ag_{(s)} \][/tex]
Oxidation Half-Cell Equation:
- Copper Solid ([tex]$Cu_{(s)}$[/tex]):
- Initially present as [tex]$Cu_{(s)}$[/tex]
- Ends as [tex]$Cu^{2+}$[/tex]
The oxidation equation for copper solid is:
[tex]\[ Cu_{(s)} \rightarrow Cu^{2+} + 2e^- \][/tex]
### b. Writing the Overall Reaction
To find the overall reaction, we combine both half-cell equations. Ensure the electrons lost in oxidation match the electrons gained in reduction.
From the half-cell equations:
1. Oxidation (Copper):
[tex]\[ Cu_{(s)} \rightarrow Cu^{2+} + 2e^- \][/tex]
2. Reduction (Silver):
[tex]\[ Ag^+ + e^- \rightarrow Ag_{(s)} \][/tex]
Since the number of electrons must be equal on both sides, we need to multiply the reduction equation by 2 to balance the electrons:
[tex]\[ 2(Ag^+ + e^- \rightarrow Ag_{(s)}) \][/tex]
This results in:
[tex]\[ 2Ag^+ + 2e^- \rightarrow 2Ag_{(s)} \][/tex]
Now combining both balanced half-cell equations:
[tex]\[ Cu_{(s)} \rightarrow Cu^{2+} + 2e^- \][/tex]
[tex]\[ 2Ag^+ + 2e^- \rightarrow 2Ag_{(s)} \][/tex]
The overall balanced reaction is:
[tex]\[ Cu_{(s)} + 2Ag^+ \rightarrow Cu^{2+} + 2Ag_{(s)} \][/tex]
### Summary:
1. Species that is reduced: [tex]$Ag^+$[/tex] (Silver Ion)
2. Half-Cell Equations:
- _Oxidation_: [tex]$Cu_{(s)} \rightarrow Cu^{2+} + 2e^-$[/tex]
- _Reduction_: [tex]$Ag^+ + e^- \rightarrow Ag_{(s)}$[/tex]
3. Overall Reaction:
[tex]\[ Cu_{(s)} + 2Ag^+ \rightarrow Cu^{2+} + 2Ag_{(s)} \][/tex]
I hope this answers your question comprehensively!
The given reaction is:
[tex]\[ Cu_{(s)} + Ag^{+} \rightarrow Cu^{2+} + Ag_{(s)} \][/tex]
### ai. Identifying the Species that is Reduced
In a redox reaction, reduction refers to the gain of electrons by a species. Here, we need to identify which species undergoes reduction.
Silver Ion ([tex]$Ag^+$[/tex]):
- Initially present as [tex]$Ag^+$[/tex]
- Ends as [tex]$Ag_{(s)}$[/tex]
The reduction equation for silver ion is:
[tex]\[ Ag^+ + e^- \rightarrow Ag_{(s)} \][/tex]
Thus, [tex]$Ag^+$[/tex] (silver ion) is the species that is reduced.
### oi. Writing the Half-Cell Equations
Next, let's write the half-cell equations for the oxidation and reduction processes separately.
Reduction Half-Cell Equation:
- We already identified that the silver ion ([tex]$Ag^+$[/tex]) is reduced to silver solid ([tex]$Ag_{(s)}$[/tex]).
[tex]\[ Ag^+ + e^- \rightarrow Ag_{(s)} \][/tex]
Oxidation Half-Cell Equation:
- Copper Solid ([tex]$Cu_{(s)}$[/tex]):
- Initially present as [tex]$Cu_{(s)}$[/tex]
- Ends as [tex]$Cu^{2+}$[/tex]
The oxidation equation for copper solid is:
[tex]\[ Cu_{(s)} \rightarrow Cu^{2+} + 2e^- \][/tex]
### b. Writing the Overall Reaction
To find the overall reaction, we combine both half-cell equations. Ensure the electrons lost in oxidation match the electrons gained in reduction.
From the half-cell equations:
1. Oxidation (Copper):
[tex]\[ Cu_{(s)} \rightarrow Cu^{2+} + 2e^- \][/tex]
2. Reduction (Silver):
[tex]\[ Ag^+ + e^- \rightarrow Ag_{(s)} \][/tex]
Since the number of electrons must be equal on both sides, we need to multiply the reduction equation by 2 to balance the electrons:
[tex]\[ 2(Ag^+ + e^- \rightarrow Ag_{(s)}) \][/tex]
This results in:
[tex]\[ 2Ag^+ + 2e^- \rightarrow 2Ag_{(s)} \][/tex]
Now combining both balanced half-cell equations:
[tex]\[ Cu_{(s)} \rightarrow Cu^{2+} + 2e^- \][/tex]
[tex]\[ 2Ag^+ + 2e^- \rightarrow 2Ag_{(s)} \][/tex]
The overall balanced reaction is:
[tex]\[ Cu_{(s)} + 2Ag^+ \rightarrow Cu^{2+} + 2Ag_{(s)} \][/tex]
### Summary:
1. Species that is reduced: [tex]$Ag^+$[/tex] (Silver Ion)
2. Half-Cell Equations:
- _Oxidation_: [tex]$Cu_{(s)} \rightarrow Cu^{2+} + 2e^-$[/tex]
- _Reduction_: [tex]$Ag^+ + e^- \rightarrow Ag_{(s)}$[/tex]
3. Overall Reaction:
[tex]\[ Cu_{(s)} + 2Ag^+ \rightarrow Cu^{2+} + 2Ag_{(s)} \][/tex]
I hope this answers your question comprehensively!
