Using the standard electrode potentials given in table \(8.1\) (NCERT), predict if the reaction between the following is feasible:
(a) \(\mathbf{F e}^{3+}(\mathrm{aq})\) and \(\mathrm{I}^{-}(\mathrm{aq})\)
(b) \(\mathbf{A g}^{+}(a q)\) and \(\mathrm{Cu}(\mathrm{s})\)
(c) \(\mathrm{Fe}^{3+}(\mathrm{aq})\) and \(\mathrm{Cu}(\mathrm{s})\)
(d) \(\mathrm{Ag}(\mathrm{s})\) and \(\mathrm{Fe}^{3+}(\mathrm{aq})\)
(e) \(\mathrm{Br}_2(\mathrm{aq})\) and \(\mathrm{Fe}^{2+}(\mathrm{aq})\)

Options:

None

Solution:
A reaction is feasible if E.M.F. of the cell is \(+\) ve.
(a) \(\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^0=+0.77 \mathrm{~V}\) \(\mathrm{E}_{\mathrm{I}_2 / 2 \mathrm{I}^{-}}^{\mathrm{o}}=+0.54 \mathrm{~V}\). Since we have to determine a reaction between \(\mathrm{Fe}^{3+}\) and \(\mathrm{I}^{-}\), the reaction should be \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}\) at cathode
\(2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 \mathrm{e}^{-}\)at anode We know that \(\mathrm{EMF}=\mathrm{E}_{\text {cathode }}^0-\mathrm{E}_{\text {anode }}^0=0.77-0.54\) \(=0.23 \mathrm{~V}\) as it is positive the reaction is feasible and the following reaction would occur.
\(2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2\)
(b) \(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=+0.80 \mathrm{~V}\);
\(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V}\). Since we have to determine a reaction between \(\mathrm{Ag}^{+}\)and \(\mathrm{Cu}\), the reaction should be \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}\) at cathode
\(\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}\)at anode.
We know that \(\mathrm{EMF}=\mathrm{E}_{\text {cathode }}^0-\mathrm{E}_{\text {anode }}^0=0.80-0.34\) \(=0.46 \mathrm{~V}\) as it is positive the reaction is feasible and the following reaction would occur.
\(2 \mathrm{Ag}^{+}+\mathrm{Cu} \longrightarrow 2 \mathrm{Ag}+\mathrm{Cu}^{2+}\)
(c) \(\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\mathrm{O}}=+0.77 \mathrm{~V}\) \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0=+0.34 \mathrm{~V}\). Since, we have to determine \(\mathrm{a}\) reaction between \(\mathrm{Fe}^{3+}\) and \(\mathrm{Cu}\), the reaction should be \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}\) at cathode
\(\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}\)at anode . We know that \(\mathrm{EMF}=\mathrm{E}_{\text {cathode }}^0-\mathrm{E}_{\text {anode }}^{\circ}=0.77-0.34\) \(=0.43 \mathrm{~V}\) as it is positive the reaction is feasible and the following reaction would occur.
\(2 \mathrm{Fe}^{3+}+\mathrm{Cu} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Cu}^{2+}\)
(d) \(\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\mathrm{O}}=+0.77 \mathrm{~V}\)
\(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=+0.80 \mathrm{~V}\). Since we have to determine a reaction between \(\mathrm{Fe}^{3+}\) and \(\mathrm{Ag}\), the reaction should be \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}\) at cathode
\(\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-}\)at anode
We know that EMF \(=\mathrm{E}_{\text {cathode }}^0-\mathrm{E}_{\text {anode }}\) a \(=0.77-0.80=-0.03 \mathrm{~V}\) as it is negative the reaction is not feasible.
(e) \(\quad \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^0=+0.77 \mathrm{~V}\)
\(\mathrm{E}_{\mathrm{Br}_2 / 2 \mathrm{Br}^{-}}^{\mathrm{o}}=+1.09 \mathrm{~V}\). Since, we have to determine a reaction between \(\mathrm{Fe}^{2+}\) and \(\mathrm{Br}_2\), the reaction should be \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}\)at anode
\(\mathrm{Br}_2+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Br}^{-}\)at cathode .
We know that EMF \(=\mathrm{E}_{\text {cathode }}^0-\mathrm{E}_{\text {anode }}^0\) \(=1.09-0.77=0.32 \mathrm{~V}\) as it is positive the reaction is feasible and the following reaction would occur.
\(2 \mathrm{Fe}^{2+}+\mathrm{Br}_2 \longrightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{Br}^{-}\)

Topic: Chemistry, Redox Reactions