Question

(i) From the given data of \( E^\circ \) values, answer the following questions:

M	V	Cr	Mn	Fe	Co	Ni	Cu
\( E^\circ_{\mathrm{M^{2+}/M}} \)	-1.18	-0.91	-1.18	-0.44	-0.28	-0.25	+0.34

(I) Why \( E^\circ_{\mathrm{M^{2+}/M}} \) show irregular trend in the above values?

(II) Why is \( E^\circ_{\mathrm{Cu^{2+}/Cu}} \) value exceptionally positive?

(III) Why \( E^\circ_{\mathrm{Mn^{2+}/Mn}} \) value is highly negative?

(ii) Write the ionic equations for the oxidising action of potassium permanganate for its reaction with \( \mathrm{I^-} \) in both acidic and alkaline solutions.

Hint:

The potential depends on atomisation, ionisation and hydration energies. Mn2+ is a stable d5 ion. In acid MnO4- goes to Mn2+, and in alkali it goes to MnO2.

Answer 1

Step 1: Understand what the data shows:
The value \( E^\circ_{\mathrm{M^{2+}/M}} \) tells how easily the metal M changes into \( \mathrm{M^{2+}} \) ions in water.
A more negative value means the metal loses electrons more easily.
This change takes place in three energy stages.
\[ \mathrm{M(s)} \xrightarrow{\Delta_a H} \mathrm{M(g)} \xrightarrow{\Delta_i H_1 + \Delta_i H_2} \mathrm{M^{2+}(g)} \xrightarrow{\Delta_{hyd} H} \mathrm{M^{2+}(aq)} \]
Here \( \Delta_a H \) is the enthalpy of atomisation, \( \Delta_i H \) is the ionisation enthalpy and \( \Delta_{hyd} H \) is the hydration enthalpy of the ion.

Step 2: Part (i)(I), why the trend is irregular:
Across the 3d series, the sum \( \Delta_i H_1 + \Delta_i H_2 \) does not rise smoothly. The stable half-filled and fully filled arrangements disturb it.
The atomisation enthalpies and hydration enthalpies also change unevenly from metal to metal.
These three terms do not move together, so the net \( E^\circ \) has no smooth pattern. For example, Cr is -0.91 volt but both V and Mn are -1.18 volt.

Step 3: Part (i)(II), why Cu has a positive value:
Copper has a high enthalpy of atomisation and a very high sum of the first two ionisation enthalpies.
The hydration enthalpy of \( \mathrm{Cu^{2+}} \) is not large enough to pay back this energy.
So the change \( \mathrm{Cu \rightarrow Cu^{2+}} \) is not favoured, and \( E^\circ_{\mathrm{Cu^{2+}/Cu}} = +0.34 \) volt. Copper is a poor reducing agent and cannot release hydrogen from dilute acids.

Step 4: Part (i)(III), why Mn has a highly negative value:
Manganese has the configuration \( 3d^5 4s^2 \). It loses the two 4s electrons to form \( \mathrm{Mn^{2+}} \).
The ion \( \mathrm{Mn^{2+}} \) has the stable half-filled \( 3d^5 \) configuration, so it forms very easily.
That is why \( E^\circ_{\mathrm{Mn^{2+}/Mn}} = -1.18 \) volt is highly negative, and Mn is a strong reducing agent.

Step 5: Part (ii), acidic medium:
In acid, \( \mathrm{MnO_4^-} \) (Mn is +7) changes to \( \mathrm{Mn^{2+}} \) (Mn is +2) by gaining 5 electrons. Iodide changes to iodine.
\[ \mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O} \]
\[ \mathrm{2I^- \rightarrow I_2 + 2e^-} \]
Multiply the first by 2 and the second by 5 so the electrons cancel.
\[ \mathrm{2MnO_4^- + 16H^+ + 10I^- \rightarrow 2Mn^{2+} + 8H_2O + 5I_2} \]

Step 6: Part (ii), alkaline medium:
In neutral or alkaline solution, \( \mathrm{MnO_4^-} \) changes to \( \mathrm{MnO_2} \) (Mn is +4) by gaining 3 electrons. Iodide is oxidised to iodate, \( \mathrm{IO_3^-} \).
\[ \mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-} \]
\[ \mathrm{I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-} \]
Multiply the first by 2 and add.
\[ \mathrm{2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-} \]

Final Answer:
The trend is irregular because \( \Delta_a H \), \( \Delta_i H \) and \( \Delta_{hyd} H \) vary unevenly. Cu is positive because its energy cost is not repaid by hydration. Mn is highly negative because \( \mathrm{Mn^{2+}} \) is a stable \( 3d^5 \) ion.
\[ \boxed{\mathrm{2MnO_4^- + 16H^+ + 10I^- \rightarrow 2Mn^{2+} + 8H_2O + 5I_2 \ (acidic)}} \]
\[ \boxed{\mathrm{2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^- \ (alkaline)}} \]