Question

From the given data of standard electrode potential E°_M²⁺/M values (in volts): Cr = -1.18, Mn = -1.18, Fe = -0.44, Co = -0.28, Ni = -0.25, Cu = +0.34, answer the following: (I) Why does E°_M²⁺/M show an irregular trend in the above values? (II) Why is E°_Cu²⁺/Cu value exceptionally positive? (III) Why is E°_Mn²⁺/Mn value highly negative?

Hint:

The standard electrode potential E° M 2+ /M depends on three energy terms acting together: the enthalpy of atomisation (energy to make gaseous atoms), the ionisation enthalpy (energy to remove the two electrons), and

Answer 1

Concept: The standard electrode potential E°_M²⁺/M depends on three energy terms acting together: the enthalpy of atomisation (energy to make gaseous atoms), the ionisation enthalpy (energy to remove the two electrons), and the hydration enthalpy (energy released when the M²⁺ ion is surrounded by water). The overall potential reflects how these balance.
Answer:
(I) The trend is irregular because the three terms (atomisation enthalpy, sum of first and second ionisation enthalpies, and hydration enthalpy) do not change smoothly across the series. Their irregular combination, linked to the changing stability of the d-electron configurations, makes E° rise and fall irregularly instead of changing steadily.
(II) E°_Cu²⁺/Cu is exceptionally positive (+0.34 V) because copper has a high (and unfavourable) sum of atomisation plus ionisation enthalpies, which is not balanced by its hydration enthalpy. This makes the formation of Cu²⁺(aq) from Cu(s) energetically unfavourable, so the reverse (Cu²⁺ being reduced to Cu) is favoured, giving a positive potential.
(III) E°_Mn²⁺/Mn is highly negative (-1.18 V) because the Mn²⁺ ion has a stable half-filled d⁵ configuration. The extra stability of this half-filled set makes Mn give up its two electrons easily to form Mn²⁺, so Mn is readily oxidised, which corresponds to a strongly negative electrode potential.