Question

The elements of the 3d transition series are given as: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn. Answer the following:

Copper has an exceptionally positive \( E^\circ_{\text{M}^{2+}/\text{M}} \) value, why?

Hint:

Copper€™s positive standard electrode potential indicates that it is less likely to undergo oxidation to \( Cu^{2+} \) due to its stable \( 3d^{10} \) electron configuration.

Answer 1

Reasons for Copper's Highly Positive Standard Electrode Potential

Copper possesses a notably positive standard electrode potential for the half-cell reaction:

\( E^\circ_{\text{Cu}^{2+}/\text{Cu}} \)

This is attributed to its distinct electronic configuration:

\( \text{Cu}: [\text{Ar}]\,3d^{10}\,4s^1 \)

The oxidation of copper from \( \text{Cu} \) to \( \text{Cu}^{2+} \) involves the loss of two electrons. This process is less favored due to the following:

• Electron loss disrupts the highly stable, completely filled \( 3d^{10} \) configuration.
• Copper exhibits a limited propensity to lose electrons, thus reducing the likelihood of oxidation.
• The \( \text{Cu}^+ \) ion benefits from a stable \( 3d^{10} \) configuration, which enhances the stability of the +1 oxidation state and impedes further oxidation to the +2 state.

Consequently, the half-cell potential:

\( E^\circ_{\text{Cu}^{2+}/\text{Cu}} \)

is significantly positive, indicating copper's low reactivity and considerable resistance to oxidation. The inherent stability of the copper ion promotes reduction over oxidation, explaining the substantially positive standard electrode potential.