Read the passage carefully and answer the questions.
Nearly all the transition elements display typical metallic properties such as high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre. The transition metals (with the exception of Zn, Cd and Hg) are very hard and have low volatility. Their melting and boiling points are high. They have high enthalpies of atomization. Ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. However, the variation within a series is quite small. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction which essentially compensates for the expected increase in atomic size with increasing atomic number. The net result of the lanthanoid contraction is that the second and the third d series exhibit similar radii (e.g., Zr 160 pm, Hf 159 pm) and have very similar physical and chemical properties. There is an increase in ionisation enthalpy along each series of the transition elements from left to right. The first ionisation enthalpy, in general, increases, but the magnitude of the increase in the second and third ionisation enthalpies for the successive elements, is much higher along a series. One of the notable features of a transition elements is the great variety of oxidation states these may show in their compounds.
Atomic number of V, Mn, Cu and Zn is 23, 25, 29 and 30 respectively. Which of them shows a positive value for the standard electrode potential \( E°_{(M^{2+}/M)} \)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Cu
The standard electrode potential \( E^\circ(\ce{M^2+}/\ce{M}) \) measures how easily the solid metal changes into \( \ce{M^2+} \) ions in water, compared with hydrogen. The change needs energy for atomisation of the metal and for the first and second ionisation of the atoms, and it gives back the hydration enthalpy of \( \ce{M^2+} \). If the energy given back is less than the energy needed, the metal does not form the ion easily and its \( E^\circ \) is positive.
The values for the four metals are:
\( \displaystyle \ce{V}: -1.18\,\mathrm{V} \qquad \ce{Mn}: -1.18\,\mathrm{V} \qquad \ce{Cu}: +0.34\,\mathrm{V} \qquad \ce{Zn}: -0.76\,\mathrm{V} \)
Copper is the only one with a positive value. Cu (Z = 29) is \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \); its high enthalpy of atomisation and its high second ionisation enthalpy (the second electron comes from the full 3d10 set) are not balanced by the hydration enthalpy of \( \ce{Cu^2+} \). This is why copper does not liberate hydrogen from dilute acids such as HCl; only oxidising acids such as nitric acid attack it.
So the metal with a positive \( E^\circ(\ce{M^2+}/\ce{M}) \) is Cu.
Why the other options are incorrect
Option (1) Zn — Incorrect. Zn has \( E^\circ = -0.76 \) V; it liberates hydrogen from dilute acids, and the full 3d10 set of \( \ce{Zn^2+} \) makes the ion stable.
Option (3) V — Incorrect. V has \( E^\circ = -1.18 \) V; vanadium readily forms \( \ce{V^2+} \).
Option (4) Mn — Incorrect. Mn has \( E^\circ = -1.18 \) V; \( \ce{Mn^2+} \) has the stable half-filled 3d5 configuration, which makes the value more negative than the trend suggests.
NCERT (2026-27 edition):
Trends in the M2+/M Standard Electrode Potentials – “… The unique behaviour of Cu, having a positive Eo, accounts for its inability to liberate H2 from acids. Only oxidising acids (nitric and hot concentrated sulphuric) react with Cu, the acids being reduced. The high energy to transform Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy. The general trend towards less negative Eo values across the series is related to the general increase in the sum of the first and second ionisation enthalpies. It is interesting to note that the value of Eo for Mn, Ni and Zn are more negative than expected from the trend.”