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To determine if the given reaction:
[tex]\[ \text{Zn} + 2\text{Fe}^{3+} \longrightarrow \text{Zn}^{2+} + 2\text{Fe}^{2+} \][/tex]
will occur spontaneously, we need to calculate the standard cell potential ([tex]\( E_{\text{cell}} \)[/tex]) for the reaction. The standard reduction potentials for the species involved are given as:
[tex]\[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \][/tex]
[tex]\[ E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.80 \text{ V} \][/tex]
To find the cell potential, we must first identify which species gets oxidized and which species gets reduced. In the reaction:
[tex]\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^- \quad \text{(oxidation)} \][/tex]
[tex]\[ 2\text{Fe}^{3+} + 2\text{e}^- \rightarrow 2\text{Fe}^{2+} \quad \text{(reduction)} \][/tex]
Zinc ([tex]\(\text{Zn}\)[/tex]) is being oxidized, and iron ([tex]\(\text{Fe}^{3+}\)[/tex]) is being reduced.
To calculate the standard cell potential ([tex]\(E_{\text{cell}}\)[/tex]), we use the following formula:
[tex]\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \][/tex]
Here:
- The cathode is where reduction occurs. For this reaction, iron is being reduced ([tex]\(\text{Fe}^{3+} \rightarrow \text{Fe}^{2+}\)[/tex]), so:
[tex]\[ E_{\text{cathode}} = +0.80 \text{ V} \][/tex]
- The anode is where oxidation occurs. For this reaction, zinc is being oxidized ([tex]\(\text{Zn} \rightarrow \text{Zn}^{2+}\)[/tex]), so:
[tex]\[ E_{\text{anode}} = -0.76 \text{ V} \][/tex]
Now, substituting these values into the formula, we get:
[tex]\[ E_{\text{cell}} = +0.80 \text{ V} - (-0.76 \text{ V}) \][/tex]
[tex]\[ E_{\text{cell}} = +0.80 \text{ V} + 0.76 \text{ V} \][/tex]
[tex]\[ E_{\text{cell}} = 1.56 \text{ V} \][/tex]
Since the standard cell potential ([tex]\(E_{\text{cell}}\)[/tex]) is positive (1.56 V), the reaction is spontaneous. Therefore, the reaction [tex]\( \text{Zn} + 2\text{Fe}^{3+} \longrightarrow \text{Zn}^{2+} + 2\text{Fe}^{2+} \)[/tex] will occur.
[tex]\[ \text{Zn} + 2\text{Fe}^{3+} \longrightarrow \text{Zn}^{2+} + 2\text{Fe}^{2+} \][/tex]
will occur spontaneously, we need to calculate the standard cell potential ([tex]\( E_{\text{cell}} \)[/tex]) for the reaction. The standard reduction potentials for the species involved are given as:
[tex]\[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \][/tex]
[tex]\[ E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.80 \text{ V} \][/tex]
To find the cell potential, we must first identify which species gets oxidized and which species gets reduced. In the reaction:
[tex]\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^- \quad \text{(oxidation)} \][/tex]
[tex]\[ 2\text{Fe}^{3+} + 2\text{e}^- \rightarrow 2\text{Fe}^{2+} \quad \text{(reduction)} \][/tex]
Zinc ([tex]\(\text{Zn}\)[/tex]) is being oxidized, and iron ([tex]\(\text{Fe}^{3+}\)[/tex]) is being reduced.
To calculate the standard cell potential ([tex]\(E_{\text{cell}}\)[/tex]), we use the following formula:
[tex]\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \][/tex]
Here:
- The cathode is where reduction occurs. For this reaction, iron is being reduced ([tex]\(\text{Fe}^{3+} \rightarrow \text{Fe}^{2+}\)[/tex]), so:
[tex]\[ E_{\text{cathode}} = +0.80 \text{ V} \][/tex]
- The anode is where oxidation occurs. For this reaction, zinc is being oxidized ([tex]\(\text{Zn} \rightarrow \text{Zn}^{2+}\)[/tex]), so:
[tex]\[ E_{\text{anode}} = -0.76 \text{ V} \][/tex]
Now, substituting these values into the formula, we get:
[tex]\[ E_{\text{cell}} = +0.80 \text{ V} - (-0.76 \text{ V}) \][/tex]
[tex]\[ E_{\text{cell}} = +0.80 \text{ V} + 0.76 \text{ V} \][/tex]
[tex]\[ E_{\text{cell}} = 1.56 \text{ V} \][/tex]
Since the standard cell potential ([tex]\(E_{\text{cell}}\)[/tex]) is positive (1.56 V), the reaction is spontaneous. Therefore, the reaction [tex]\( \text{Zn} + 2\text{Fe}^{3+} \longrightarrow \text{Zn}^{2+} + 2\text{Fe}^{2+} \)[/tex] will occur.
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