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Based on standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for E\degree value).
(i) $\mathrm{Cu}+\mathrm{Zn}^{2+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Zn}$
(ii) $\mathrm{Mg}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
(iii) $\mathrm{Br}^{2+} 2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}^2+2 \mathrm{B}^{\mathrm{r}-}$
(iv) $\mathrm{Fe}+\mathrm{Cd}^{2+} \rightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$
(i) $\mathrm{Cu}+\mathrm{Zn}^{2+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Zn}$
(ii) $\mathrm{Mg}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
(iii) $\mathrm{Br}^{2+} 2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}^2+2 \mathrm{B}^{\mathrm{r}-}$
(iv) $\mathrm{Fe}+\mathrm{Cd}^{2+} \rightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$
Solution:
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Verified Answer
(i) $\mathrm{Cu}+\mathrm{Zn}^{2+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Zn}$
Here $\mathrm{Cu}$ undergoes oxidation so it acts as anode and $\mathrm{Zn}$ acts as the cathode. So from the table. For example $\mathrm{E}^{\circ}$ anode $=-0.76 \mathrm{~V}$
For anode $\mathrm{E}^{\circ}$ anode $=0.52 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=-0.24 \mathrm{~V}$
As the EMF of the cell is negative the given reaction will not occur spontaneously if they were to form a cell placed as electrodes.
(ii) $\mathrm{Mg}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
Similarly, we can say that $\mathrm{Mg}$ undergoes oxidation and Fe undergoes reduction,
$\mathrm{E}^{\circ}$ cathode $=-0.44 \mathrm{~V}$
$\mathrm{E}^{\circ}$ anode $=-2.36 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=+1.92 \mathrm{~V}$
Positive EMF implies that the reaction will give out energy and attain stability, thus it will occur spontaneously. So the given redox reaction will occur.
(iii) $\mathrm{Br}_2+2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_2+2 \mathrm{Br}^{-}$
Here $\mathrm{Br}$ undergoes reduction thus acting as cathode and $\mathrm{Cl}$ acting as the anode.
For cathode $\mathrm{E}^{\circ}$ cathode $=1.09 \mathrm{~V}$
For anode $\mathrm{E}^{\circ}$ cathode $=1.36 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=-0.25$
The negative potential prevents easy reaction, so the redox reaction will not occur.
(iv) $\mathrm{Fe}+\mathrm{Cd}^{2+} \rightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$
$\mathrm{Fe}$ is the cathode and $\mathrm{Cd}$ is the anode
For cathode $\mathrm{E}^{\circ}$ cathode $=-0.44 \mathrm{~V}$
For anode $\mathrm{E}^{\circ}$ anode $=-0.40 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=-0.04 \mathrm{~V}$
The negative potential prevents easy reaction, so the redox reaction will not occur.
Here $\mathrm{Cu}$ undergoes oxidation so it acts as anode and $\mathrm{Zn}$ acts as the cathode. So from the table. For example $\mathrm{E}^{\circ}$ anode $=-0.76 \mathrm{~V}$
For anode $\mathrm{E}^{\circ}$ anode $=0.52 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=-0.24 \mathrm{~V}$
As the EMF of the cell is negative the given reaction will not occur spontaneously if they were to form a cell placed as electrodes.
(ii) $\mathrm{Mg}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
Similarly, we can say that $\mathrm{Mg}$ undergoes oxidation and Fe undergoes reduction,
$\mathrm{E}^{\circ}$ cathode $=-0.44 \mathrm{~V}$
$\mathrm{E}^{\circ}$ anode $=-2.36 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=+1.92 \mathrm{~V}$
Positive EMF implies that the reaction will give out energy and attain stability, thus it will occur spontaneously. So the given redox reaction will occur.
(iii) $\mathrm{Br}_2+2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_2+2 \mathrm{Br}^{-}$
Here $\mathrm{Br}$ undergoes reduction thus acting as cathode and $\mathrm{Cl}$ acting as the anode.
For cathode $\mathrm{E}^{\circ}$ cathode $=1.09 \mathrm{~V}$
For anode $\mathrm{E}^{\circ}$ cathode $=1.36 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=-0.25$
The negative potential prevents easy reaction, so the redox reaction will not occur.
(iv) $\mathrm{Fe}+\mathrm{Cd}^{2+} \rightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$
$\mathrm{Fe}$ is the cathode and $\mathrm{Cd}$ is the anode
For cathode $\mathrm{E}^{\circ}$ cathode $=-0.44 \mathrm{~V}$
For anode $\mathrm{E}^{\circ}$ anode $=-0.40 \mathrm{~V}$
$\mathrm{E}^{\circ}$ cell $=-0.04 \mathrm{~V}$
The negative potential prevents easy reaction, so the redox reaction will not occur.
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