Question

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

(a) 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

Copper has an exceptionally positive \( E^\circ_{\text{M}^{2+}/\text{M}} \) value because copper's \( 3d^{10} 4s^1 \) electron configuration makes its \( Cu^{2+} \) ion highly stable compared to the other transition metal ions. When copper undergoes oxidation from \( Cu \) to \( Cu^{2+} \), it loses electrons, and the reaction becomes more favorable due to the relatively low tendency of copper to lose electrons. Additionally, the stable \( 3d^{10} \) configuration of \( Cu^+ \) gives it a strong tendency to stay in the +1 oxidation state, making copper less likely to oxidize further to +2. Thus, the positive \( E^\circ_{\text{M}^{2+}/\text{M}} \) value reflects the fact that the copper ion is more stable in its lower oxidation state, giving it a tendency to resist further oxidation. \vspace{10pt}