Question 1Level 2 · ExamSingle correct
Standard reduction potentials at 298 K for the couples M3+(aq)/M2+(aq) of some 3d metals are (in V): Ti −0.37, V −0.26, Cr −0.41, Mn +1.57, Fe +0.77, Co +1.97. Among the six M2+(aq) ions the strongest reducing agent, and among the six M3+(aq) ions the strongest oxidising agent, are respectively ______. Exactly one option is correct.
ATi2+ and Co3+
BCr2+ and Mn3+
CCr2+ and Co3+
DV2+ and Mn3+
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Answer: C
For M3++e−→M2+, the most negative E∘ means M2+ is most easily oxidised (strongest reductant): Cr, −0.41 V (more negative than Ti, −0.37 V, and V, −0.26 V). The most positive E∘ means M3+ is most easily reduced (strongest oxidant): Co, +1.97 V (above Mn, +1.57 V).\nChemically: Cr2+ (d4) is oxidised to Cr3+ (d3, half-filled t2g), while Co3+ is reduced to Co2+, which is the stable aqueous state of cobalt.\nAnswer: (C).
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
Standard reduction potentials E∘(M2+/M) at 298 K for the 3d metals are (in V): Ti −1.63, V −1.18, Cr −0.90, Mn −1.18, Fe −0.44, Co −0.28, Ni −0.25, Cu +0.34, Zn −0.76. Which statement about these values is correct? Exactly one option is correct.
AThe value for Mn is more negative than the smooth trend (more negative than Cr) mainly because of the extra stability of the half-filled d5 configuration of Mn2+.
BCopper has a positive value because Cu2+ has the least negative hydration enthalpy of all the 3d M2+ ions.
CThe general shift towards less negative values across the series arises because the sum of the first and second ionisation enthalpies generally decreases from Ti to Ni.
DZinc has a more negative value than nickel because forming Zn2+ requires removal of electrons from the filled 3d subshell.
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Answer: A
(A) Correct. The stability of the half-filled d5 configuration of Mn2+ (aided by the low enthalpy of atomisation of Mn) makes the Mn→Mn2+ change unusually favourable, so E∘ is more negative than expected.\n(B) Wrong. Cu2+ has one of the most negative hydration enthalpies of the series. Cu has a positive E∘ because its high enthalpy of atomisation plus ΔiH1+ΔiH2 is not balanced by the hydration enthalpy.\n(C) Wrong. The trend towards less negative values is linked to a general increase in ΔiH1+ΔiH2 across the series.\n(D) Wrong. Zn loses its two 4s electrons; Zn2+ keeps a filled 3d10 subshell, and that stability is why its E∘ is more negative than the trend.\nAnswer: (A).
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
At 298 K, E∘(Cu2+/Cu)=+0.34 V and E∘(Cu+/Cu)=+0.52 V. Take 2.303RT/F=0.059 V. For the disproportionation 2Cu+(aq)⇌Cu2+(aq)+Cu(s), the equilibrium constant K=[Cu2+]/[Cu+]2 (in L mol−1) is closest to ______. Exactly one option is correct.
A1.1×103
B1.3×106
C1.6×1012
D7.9×10−7
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Answer: B
Potentials are combined through free energies, not added directly. Cu2++2e−→Cu: ΔG∘=−2F(0.34). Cu++e−→Cu: ΔG∘=−F(0.52). Subtracting: Cu2++e−→Cu+ has −FE∘=−2F(0.34)+F(0.52), so E∘(Cu2+/Cu+)=0.68−0.52=0.16 V.\nDisproportionation (one electron transferred): Ecell∘=E∘(Cu+/Cu)−E∘(Cu2+/Cu+)=0.52−0.16=0.36 V.\nlogK=nE∘/0.059=0.36/0.059=6.10, so K=1.3×106.\nDistractors: 1.1×103 uses 0.52−0.34; 1.6×1012 uses n=2; 7.9×10−7 is the reverse reaction.\nThis large K is why Cu+ is unstable in water: the much more negative hydration enthalpy of Cu2+ more than compensates for the second ionisation enthalpy.\nAnswer: (B).
Difficulty: Very hard