Using the standard electrode potentials given in the table 3.1(in NCERT), predict if the reaction between the following is feasible:
(i) Fe³⁺ (aq) and I^– (aq)
(ii) Ag⁺ (aq) and Cu(s)
(iii) Fe³⁺ (aq) and Br^–(aq)
(iv) Ag(s) and Fe³⁺(aq)
(v) Br₂(aq) and Fe²⁺(aq)

From the table, standard electrode potents at 298 k are:
(E°_Fe3+/VF = 0.77 V, E°_I2/I– = 0.54 V)
(E°_Ag+/Ag = E°_Cu²⁺/Cu = 0.34)
(E°_Fe³⁺/Fe= 0.77 V, E°_Br2/Br- = 1.08 V)
(E°_Ag+/Ag= 0.8 V E°_{Fe³⁺/Fe²⁺} = 0.77 V)
(E°_{Fe³⁺/Fe²⁺}= 0.77 V,E°_Br2/Br- = 1.08 V)

(a) ${\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right) + {\mathrm{I}}^{-}\left(\mathrm{aq}\right) \to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right) + \frac{1}{2} {\mathrm{I}}_2 \left(\mathrm{g}\right)$${\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right) + {\mathrm{I}}^{-}\left(\mathrm{aq}\right) \to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+\frac{1}{2}{\mathrm{I}}_2\left(\mathrm{g}\right)$
In this reaction, Fe³⁺ is reduced to Fe²⁺ and I^– is oxidised to I₂. The cell giving above reaction will be

$$\begin{gathered} {\mathrm{I}}_2\left(\mathrm{s}\right) | {\mathrm{I}}^{-} (\mathrm{aq}) || {\mathrm{Fe}}^{3+} (\mathrm{aq}) | {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right) \\ \therefore {{\mathrm{E}}^0}_{\mathrm{cell}} = {{\mathrm{E}}^0}_{\mathrm{cathode}} - {{\mathrm{E}}^0}_{\mathrm{anode}} = {{\mathrm{E}}^0}_{\mathrm{right}} - {{\mathrm{E}}^0}_{\mathrm{left}} \\ = 0.77 \mathrm{V} - 0.54 \mathrm{V} = + 0.23 \mathrm{V} \end{gathered}$$

As E⁰ is positive, the reaction between Fe³⁺ (aq) and I^– (aq) occurs as indicated by possible reaction given above.


(b) $2 {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right) + \mathrm{Cu}\left(\mathrm{s}\right) \to 2 \mathrm{Ag}\left(\mathrm{s}\right) + {\mathrm{Cu}}^{2+} \left(\mathrm{aq}\right)$
Here, in this reaction, Ag⁺ is reduced to Ag (i.e., it should be cathode) and Cu(s) is oxidised to Cu²⁺(aq) (i.e., it should be anode).
The cell can be represented as
           
$$\begin{gathered} \mathrm{Cu}\left(\mathrm{s}\right) | {\mathrm{Cu}}^{2+}(\mathrm{aq}\left)\right|| {\mathrm{Ag}}^{+} (\mathrm{aq}) | \mathrm{Ag}\left(\mathrm{s}\right) \\ \mathrm{anode} \mathrm{cathode} \\ {{\mathrm{E}}^0}_{\mathrm{cell}} = {{\mathrm{E}}^0}_{\mathrm{cathode}} - {{\mathrm{E}}^0}_{\mathrm{anode}} \\ = 0.80 \mathrm{V} - 0.34 \mathrm{V} = + 0.46 \mathrm{V} \end{gathered}$$

As E°_cell is positive, the reaction between (Ag⁺ (aq) and Cu(s) occurs as indicated by possible reaction given above.


(c) ${\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right) + {\mathrm{Br}}^{-}\left(\mathrm{aq}\right) \to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right) + \frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{aq}\right)$
 In this reaction Fe³⁺ is reduced to Fe²⁺ (i.e., Fe³/Fe²⁺ electrode should be cathode) and Br is oxidised to Br₂ (i.e., Br₂/Br^– electrode should be anode.
The cell can be represented as:

$$\begin{gathered} {\mathrm{Br}}_2\left(\mathrm{aq}\right)\left|{\mathrm{Br}}^{-}\right(\mathrm{aq}\left)\right|\left|{\mathrm{Fe}}^{2+}\right(\mathrm{aq}) | {\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right) \\ {{\mathrm{E}}^0}_{\mathrm{cell}} = {{\mathrm{E}}^0}_{\mathrm{cathode}} - {{\mathrm{E}}^0}_{\mathrm{anode}} = {{\mathrm{E}}^0}_{\mathrm{right} }-{{\mathrm{E}}^0}_{\mathrm{left}} \\ = 0.77 \mathrm{V} - 1.08 \mathrm{V} = -0.31 \mathrm{V} \end{gathered}$$

As E°_cell is negative, no reaction will occur between Fe³⁺ (aq) and Br^–(aq).


(d) $\mathrm{Ag}\left(\mathrm{s}\right) + {\mathrm{Fe}}^{3+} \left(\mathrm{aq}\right) \to {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right) + {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)$     
Two half-cell reactions can be expressed as:

$$\begin{gathered} \mathrm{Ag}\left(\mathrm{s}\right) \to {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right) + {\mathrm{e}}^{-} \\ (\mathrm{oxidation} \mathrm{anode}) \\ {\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right) + {\mathrm{e}}^{-} \to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right) \\ (\mathrm{reduction}, \mathrm{cathode}) \\ {{\mathrm{E}}^0}_{\mathrm{cell}} = {{\mathrm{E}}^0}_{\mathrm{cathode}} - {{\mathrm{E}}^0}_{\mathrm{anode}} \\ = 0.77 \mathrm{V} - 0.80 \mathrm{V} = -0.3 \mathrm{V} \end{gathered}$$

As E°_cell is negative, no reaction occurs between Fe³⁺(aq) and Ag(s).


(e) $\frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{aq}\right) + {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right) \to {\mathrm{Fe}}^{3+} \left(\mathrm{aq}\right) + {\mathrm{Br}}^{-} \left(\mathrm{aq}\right)$
The two half-cell reactions are

   $$\begin{gathered} \frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{aq}\right) + {\mathrm{e}}^{-} \to {\mathrm{Br}}^{-} \left(\mathrm{aq}\right) \\ (\mathrm{reduction}, \mathrm{cathode}) \\ {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right) \to {\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right) + {\mathrm{e}}^{-} \\ (\mathrm{oxidation} \mathrm{anode}) \\ {{\mathrm{E}}^0}_{\mathrm{cell}} = {{\mathrm{E}}^0}_{\mathrm{cathode}} - {{\mathrm{E}}^0}_{\mathrm{anode}} \\ = 1.80 \mathrm{V} - 0.77 \mathrm{V} = +0.31 \mathrm{V} \end{gathered}$$


As E°_cell is positive, the reaction is feasible, i.e., reaction between Br₂(aq) and Fe²⁺ (aq) occurs as indicated by possible reaction given above.