(i)(I) WHY $E^\circ_{M^{2+}/M}$ SHOWS AN IRREGULAR TREND

Step 1 - Break the electrode process into steps.
The overall change $M(s) \rightarrow M^{2+}(aq) + 2e^-$ can be imagined as a
Born-Haber type cycle of three steps:
$$M(s) \xrightarrow{\Delta_a H} M(g) \xrightarrow{IE_1 + IE_2} M^{2+}(g) \xrightarrow{\Delta_{hyd} H} M^{2+}(aq)$$

Step 2 - Identify the three controlling enthalpies.
(1) ENTHALPY OF ATOMISATION (sublimation), $\Delta_a H$ - energy to convert the solid
metal into gaseous atoms.
(2) SUM OF THE FIRST AND SECOND IONISATION ENTHALPIES, $IE_1 + IE_2$ - energy to remove
two electrons.
(3) HYDRATION ENTHALPY, $\Delta_{hyd} H$ - energy RELEASED when the gaseous ion is
hydrated.

Step 3 - Explain the irregularity.
$E^\circ_{M^{2+}/M}$ depends on the ALGEBRAIC SUM of all three terms. Each of these three
quantities varies IRREGULARLY across the 3d series (because of changes in nuclear
charge, in the stability of half-filled and fully-filled $d$ configurations, and in
ionic size). Since none of them changes smoothly, and since they do not vary in step
with one another, their SUM shows NO REGULAR TREND.

Hence $E^\circ_{M^{2+}/M}$ values show an irregular variation from V to Cu.

(i)(II) WHY $E^\circ_{Cu^{2+}/Cu}$ IS EXCEPTIONALLY POSITIVE $(+0.34\ V)$

Step 1 - Look at the energy input for copper.
Copper has an unusually HIGH ENTHALPY OF ATOMISATION and a HIGH SUM OF IONISATION
ENTHALPIES $(IE_1 + IE_2)$. A large amount of energy is therefore needed to convert
$Cu(s)$ into $Cu^{2+}(g)$.

Step 2 - Look at the energy released.
The hydration enthalpy of $Cu^{2+}$, though large, is NOT LARGE ENOUGH TO COMPENSATE
for this high input.

Step 3 - Draw the conclusion.
The overall energy change for
$$Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$$
is therefore POSITIVE, i.e. oxidation of copper is NOT favourable. The REVERSE process
(reduction, $Cu^{2+} + 2e^- \rightarrow Cu$) is favoured, which means $E^\circ$ is POSITIVE.

Step 4 - Chemical consequence.
Copper is the ONLY first-row transition metal that does NOT liberate $H_2$ from dilute
acids, because $E^\circ_{Cu^{2+}/Cu}$ is positive (it lies below hydrogen in the
electrochemical series).

(i)(III) WHY $E^\circ_{Mn^{2+}/Mn}$ IS HIGHLY NEGATIVE $(-1.18\ V)$

Step 1 - Write the configuration of the product ion.
$$Mn\,(Z=25) = [Ar]3d^5 4s^2 \qquad \Longrightarrow \qquad Mn^{2+} = [Ar]3d^5$$

Step 2 - Note the special stability.
$Mn^{2+}$ has an EXACTLY HALF-FILLED $3d^5$ configuration, which has EXTRA STABILITY
(maximum exchange energy, spherically symmetrical charge distribution).

Step 3 - Consequence.
Because the product ion is unusually stable, its formation is energetically easy - the
second ionisation enthalpy of Mn is abnormally LOW. Manganese is therefore OXIDISED
VERY READILY to $Mn^{2+}$.

Step 4 - Relate to $E^\circ$.
Easy oxidation means a strong tendency to lose electrons, i.e. a very NEGATIVE reduction
potential. Hence $E^\circ_{Mn^{2+}/Mn} = -1.18$ V, which is much more negative than the
expected trend value.
(For the same kind of reason, $E^\circ_{Zn^{2+}/Zn}$ is also very negative, because
$Zn^{2+}$ has the stable $3d^{10}$ configuration.)

(ii) IONIC EQUATIONS FOR THE OXIDATION OF $I^-$ BY $KMnO_4$

CASE 1 : IN ACIDIC SOLUTION
Here $MnO_4^-$ (Mn in $+7$) is reduced all the way to $Mn^{2+}$ (a 5-electron change),
and iodide is oxidised to iodine.

Reduction half reaction $(\times 2)$:
$$MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O$$
Oxidation half reaction $(\times 5)$:
$$2I^- \longrightarrow I_2 + 2e^-$$

OVERALL IONIC EQUATION:
$$\boxed{2MnO_4^- + 16H^+ + 10I^- \longrightarrow 2Mn^{2+} + 8H_2O + 5I_2}$$
(The purple colour of $MnO_4^-$ disappears and a brown solution of iodine is formed.)

CASE 2 : IN ALKALINE (or neutral/faintly basic) SOLUTION
Here $MnO_4^-$ is reduced only to $MnO_2$ (Mn in $+4$, a 3-electron change), and iodide
is oxidised further, to IODATE $IO_3^-$ (I goes from $-1$ to $+5$, a 6-electron change).

Reduction half reaction $(\times 2)$:
$$MnO_4^- + 2H_2O + 3e^- \longrightarrow MnO_2 + 4OH^-$$
Oxidation half reaction $(\times 1)$:
$$I^- + 6OH^- \longrightarrow IO_3^- + 3H_2O + 6e^-$$

OVERALL IONIC EQUATION:
$$\boxed{2MnO_4^- + H_2O + I^- \longrightarrow 2MnO_2 + 2OH^- + IO_3^-}$$
(A brown precipitate of $MnO_2$ is formed.)