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

Why Copper Has an Exceptionally Positive Standard Electrode Potential 

Copper exhibits a highly positive standard electrode potential for the half-cell reaction:

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

This is due to its unique electronic configuration:

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

When copper oxidizes from \( \text{Cu} \) to \( \text{Cu}^{2+} \), it loses two electrons. However, this process is less favorable because:

• The loss of electrons disturbs the stable, fully filled \( 3d^{10} \) configuration.
• Copper has a low tendency to lose electrons, making oxidation less likely.
• The \( \text{Cu}^+ \) ion has a stable \( 3d^{10} \) configuration, which further stabilizes the +1 state and resists further oxidation to +2.

As a result, the half-cell potential:

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

is highly positive, reflecting copper’s low reactivity and high resistance to oxidation. This stability of the copper ion leads to a greater tendency for the reduction reaction (rather than oxidation), which is why the standard electrode potential is significantly positive.