Answer :
To find the cell potential of the electrochemical cell given the half-reactions:
1. Identify the half-reactions and their standard reduction potentials:
- For the silver half-reaction: [tex]\( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)[/tex], the standard reduction potential is [tex]\( +0.80 \text{ V} \)[/tex].
- For the iron half-reaction: [tex]\( \text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \)[/tex], the reduction potential for the reverse reaction ( [tex]\( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \)[/tex] ) is [tex]\( -0.77 \text{ V} \)[/tex]. Therefore, the oxidation potential for the iron reaction is the same [tex]\( -0.77 \text{ V} \)[/tex].
2. Determine which reaction is the anode and which is the cathode:
- The half-reaction with the higher reduction potential occurs at the cathode, where reduction takes place. In this case, the silver reaction ([tex]\( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)[/tex]) has a reduction potential of [tex]\( +0.80 \text{ V} \)[/tex].
- The iron reaction ([tex]\( \text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \)[/tex]) occurs at the anode, where oxidation takes place, with an effective reduction potential of [tex]\( -0.77 \text{ V} \)[/tex].
3. Calculate the cell potential:
- Use the formula for the cell potential:
[tex]\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \][/tex]
- Plugging in the given values:
[tex]\[ E^\circ_{\text{cell}} = 0.80 \text{ V} - (-0.77 \text{ V}) = 0.80 \text{ V} + 0.77 \text{ V} = 1.57 \text{ V} \][/tex]
Therefore, the cell potential is [tex]\( \boxed{1.57 \text{ V}} \)[/tex].
Given the options [tex]\(A. -0.44 V\)[/tex], [tex]\(B. -1.24 V\)[/tex], [tex]\(C. 0.44 V\)[/tex], [tex]\(D. 1.24 V\)[/tex], none of them match the calculated cell potential of [tex]\(1.57 \text{ V}\)[/tex], making it clear there is likely a problem with the given multiple-choice options or the way the values were combined.
However, if required to choose one of the options provided and there was a typo in them, the calculation definitively satisfies that [tex]\( \boxed{1.57 \text{ V}} \)[/tex] would be the correct approach. In scenarios where options need fitting, verification or problem context correct suiting might provide [tex]\(C\)[/tex] as potential match being a different understanding. Such mismatch scene typically checked for data entry or context clearly to align.
1. Identify the half-reactions and their standard reduction potentials:
- For the silver half-reaction: [tex]\( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)[/tex], the standard reduction potential is [tex]\( +0.80 \text{ V} \)[/tex].
- For the iron half-reaction: [tex]\( \text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \)[/tex], the reduction potential for the reverse reaction ( [tex]\( \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \)[/tex] ) is [tex]\( -0.77 \text{ V} \)[/tex]. Therefore, the oxidation potential for the iron reaction is the same [tex]\( -0.77 \text{ V} \)[/tex].
2. Determine which reaction is the anode and which is the cathode:
- The half-reaction with the higher reduction potential occurs at the cathode, where reduction takes place. In this case, the silver reaction ([tex]\( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)[/tex]) has a reduction potential of [tex]\( +0.80 \text{ V} \)[/tex].
- The iron reaction ([tex]\( \text{Fe} \rightarrow \text{Fe}^{3+} + 3e^- \)[/tex]) occurs at the anode, where oxidation takes place, with an effective reduction potential of [tex]\( -0.77 \text{ V} \)[/tex].
3. Calculate the cell potential:
- Use the formula for the cell potential:
[tex]\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \][/tex]
- Plugging in the given values:
[tex]\[ E^\circ_{\text{cell}} = 0.80 \text{ V} - (-0.77 \text{ V}) = 0.80 \text{ V} + 0.77 \text{ V} = 1.57 \text{ V} \][/tex]
Therefore, the cell potential is [tex]\( \boxed{1.57 \text{ V}} \)[/tex].
Given the options [tex]\(A. -0.44 V\)[/tex], [tex]\(B. -1.24 V\)[/tex], [tex]\(C. 0.44 V\)[/tex], [tex]\(D. 1.24 V\)[/tex], none of them match the calculated cell potential of [tex]\(1.57 \text{ V}\)[/tex], making it clear there is likely a problem with the given multiple-choice options or the way the values were combined.
However, if required to choose one of the options provided and there was a typo in them, the calculation definitively satisfies that [tex]\( \boxed{1.57 \text{ V}} \)[/tex] would be the correct approach. In scenarios where options need fitting, verification or problem context correct suiting might provide [tex]\(C\)[/tex] as potential match being a different understanding. Such mismatch scene typically checked for data entry or context clearly to align.
