Q36

5 marksLong AnswerSection E

(a) (i) From the given data of E° values, answer the following questions: (3 + 2)

$E^\circ(M^{2+}/M)$ values (V): $V = -1.18, Cr = -0.91, Mn = -1.18, Fe = -0.44, Co = -0.28, Ni = -0.25, Cu = +0.34$

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

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

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

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

OR

(b) Answer the following questions: (1 + 1 + 1 + 2)

(i) Name a member of the lanthanoid series (I) which exhibits +4 oxidation state (II) which exhibits +2 oxidation state.

(ii) Why transition metals act as good catalyst?

(iii) Why Cr has higher melting point than Mn?

(iv) What happens when acidic solution of potassium permanganate is allowed to stand for sometime? Give the equation involved. What is this type of reaction called?

Q: What does $KMnO_4$ oxidise iodide to in acidic versus alkaline medium?

In acidic medium, $I^-$ is oxidised to iodine ($I_2$) and $MnO_4^-$ is reduced to $Mn^{2+}$; in alkaline medium, $I^-$ is oxidised further to iodate ($IO_3^-$) while $MnO_4^-$ is reduced only to $MnO_2$.

Q: Why does Cr have a higher melting point than Mn?

Cr ($3d^5 4s^1$) has more unpaired d-electrons contributing to strong metallic bonding, whereas Mn ($3d^5 4s^2$) has a stable half-filled d configuration with weaker interatomic bonding, so Mn melts at a lower temperature.

The d- and f-Block Elements

Standard electrode potentials and oxidising action of KMnO₄ (d-block)

Official Answer

Primary answer — part (a)

(a)(i) Trends in $E^\circ(M^{2+}/M)$

(I) Irregular trend: $E^\circ(M^{2+}/M)$ depends on the sum of the enthalpy of atomisation and the ionisation enthalpies ( $IE_1 + IE_2$ ) offset by the hydration enthalpy of M²⁺. Since these three energy terms vary irregularly across the 3d series, the overall E° also varies irregularly.
(II) $E^\circ(Cu^{2+}/Cu)$ exceptionally positive (+0.34 V): The high sum of atomisation + ionisation enthalpies of copper is not compensated by its comparatively low hydration enthalpy. Converting $Cu(s)$ to $Cu^{2+}(aq)$ is therefore energetically unfavourable, making E° positive — Cu does not liberate $H_2$ from acids.
(III) $E^\circ(Mn^{2+}/Mn)$ highly negative (−1.18 V): $Mn^{2+}$ has a stable half-filled $3d^5$ configuration. This extra stability makes the formation of $Mn^{2+}$ easy, giving a strongly negative electrode potential.

(a)(ii) Oxidising action of $KMnO_4$ with $I^-$

Acidic medium ( $I^- \to I_2$ ):

$2MnO_4^- + 16H^+ + 10I^- \to 2Mn^{2+} + 8H_2O + 5I_2$

Alkaline / weakly basic medium ( $I^- \to IO_3^-$ ):

$2MnO_4^- + H_2O + I^- \to 2MnO_2 + 2OH^- + IO_3^-$

OR — part (b) (brief)

(i) +4 oxidation state: Ce (also Pr, Tb); +2 oxidation state: Eu (also Yb, Sm).
(ii) Transition metals are good catalysts because their variable oxidation states let them form intermediates/provide alternate low-energy pathways, and their partially filled d-orbitals adsorb reactant molecules on the surface.
(iii) Cr has a higher melting point than Mn because $Cr$ ( $3d^5 4s^1$ ) has more unpaired electrons available for strong metallic (interatomic) bonding, whereas $Mn$ ( $3d^5 4s^2$ ) has a stable half-filled 3d and weaker interatomic bonding.
(iv) On standing, acidified $KMnO_4$ slowly decomposes, evolving $O_2$ :

$4MnO_4^- + 4H^+ \to 4MnO_2 + 3O_2 + 2H_2O$

It is a self- (auto-) decomposition / redox decomposition reaction (accelerated by light). [Note: some textbooks label this simply as a decomposition reaction.]

standard electrode potential M2+/Mhydration enthalpy ionisation enthalpyCu exceptionally positive E°Mn half-filled d5 stabilityKMnO₄ oxidising I⁻acidic alkaline permanganatelanthanoid oxidation states Ce EuCr higher melting point

Marking Scheme

1(a)(i) 3 marks: 1 mark each — (I) irregular variation of atomisation + ionisation vs hydration enthalpies; (II) high atomisation+IE of Cu not offset by low hydration enthalpy; (III) extra stability of half-filled $d^5$ $Mn^{2+}$ .
2(a)(ii) 2 marks: 1 mark for balanced acidic equation ( $2MnO_4^- + 16H^+ + 10I^- \to 2Mn^{2+} + 8H_2O + 5I_2$ ); 1 mark for balanced alkaline equation ( $2MnO_4^- + H_2O + I^- \to 2MnO_2 + 2OH^- + IO_3^-$ ).
3OR (b): (i) 1 mark — Ce (+4) and Eu (+2); (ii) 1 mark — variable oxidation states/surface adsorption via d-orbitals; (iii) 1 mark — Cr has more unpaired d-electrons for metallic bonding; (iv) 2 marks — decomposition equation with O₂ evolution and naming it a self/auto-decomposition (redox) reaction.

Hint

E°(M²⁺/M) = balance of atomisation + ionisation enthalpies against hydration enthalpy. For KMnO₄ + I⁻, the product changes with medium: I₂ (acidic) vs IO₃⁻ + MnO₂ (alkaline).

Quick Oral Answer

The $E^\circ(M^{2+}/M)$ values wander because atomisation, ionisation and hydration enthalpies each vary irregularly; copper's value is positive since its low hydration enthalpy cannot offset the high energy to make $Cu^{2+}$ , and manganese's is very negative because $Mn^{2+}$ enjoys the stability of a half-filled $d^5$ shell; permanganate oxidises iodide to iodine in acid but to iodate in alkali.

Analysis & Explanation

Concept — the thermochemical cycle behind E°

The electrode potential $E^\circ(M^{2+}/M)$ is not a single quantity but the net result of three steps:

atomisation: $M(s) \to M(g)$ (ΔH_atom)
ionisation: $M(g) \to M^{2+}(g) + 2e^-$ (IE₁ + IE₂)
hydration: $M^{2+}(g) \to M^{2+}(aq)$ (ΔH_hyd, negative)

Because each of these varies irregularly across the series, E° does too. Copper's low ΔH_hyd fails to pay back its high atomisation + ionisation cost, so E° is positive; manganese's negative value reflects the special stability of the half-filled $d^5$ $Mn^{2+}$ .

Exam trap

For $KMnO_4$ , the product of $I^-$ oxidation depends on the medium: iodine ( $I_2$ ) in acid but iodate ( $IO_3^-$ ) with $MnO_2$ in alkaline solution. Writing $I_2$ for both loses marks.
In part (b)(iii), the melting-point argument must be about unpaired d-electrons in metallic bonding, not atomic size.

Real-world

Acidified $KMnO_4$ is a workhorse volumetric oxidant (permanganometry); its gradual self-decomposition is exactly why standard permanganate solutions must be freshly standardised and stored in dark bottles.

Common Mistakes

1Writing $I_2$ as the product of $MnO_4^- + I^-$ in alkaline medium — in alkali the product is iodate ( $IO_3^-$ ) with $MnO_2$ , not iodine.
2Explaining Cu's positive E° by 'inertness' rather than the unfavourable balance of high sublimation+ionisation enthalpy vs low hydration enthalpy.
3Leaving the $KMnO_4 + I^-$ equations unbalanced (wrong $H^+$ /electron count in acidic medium).

Interesting Facts

Copper's positive $E^\circ(Cu^{2+}/Cu)$ is the reason copper does not dissolve in dilute $HCl$ or $H_2SO_4$ to give $H_2$ — it needs an oxidising acid like $HNO_3$ .

Cerium(IV), as ceric ammonium sulphate/nitrate, is such a good one-electron oxidant that cerimetric titrations rival permanganometry in analytical chemistry.

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Frequently Asked Questions

Why is $E^\circ(Cu^{2+}/Cu)$ positive while most 3d metals are negative?

The high sum of copper's atomisation and ionisation enthalpies is not compensated by its comparatively low hydration enthalpy, so forming $Cu^{2+}(aq)$ is unfavourable and E° comes out positive — copper cannot displace $H_2$ from dilute acids.

What does $KMnO_4$ oxidise iodide to in acidic versus alkaline medium?

In acidic medium, $I^-$ is oxidised to iodine ( $I_2$ ) and $MnO_4^-$ is reduced to $Mn^{2+}$ ; in alkaline medium, $I^-$ is oxidised further to iodate ( $IO_3^-$ ) while $MnO_4^-$ is reduced only to $MnO_2$ .

Why does Cr have a higher melting point than Mn?

Cr ( $3d^5 4s^1$ ) has more unpaired d-electrons contributing to strong metallic bonding, whereas Mn ( $3d^5 4s^2$ ) has a stable half-filled d configuration with weaker interatomic bonding, so Mn melts at a lower temperature.