Question

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

Answer 1

Rationale for Copper's Notably Positive Standard Electrode Potential (\( E^\circ_{\text{M}^{2+}/\text{M}} \)):

Explanation:
The standard electrode potential (\( E^\circ \)) quantifies a metal's propensity to undergo oxidation (lose electrons). A positive \( E^\circ \) signifies a strong tendency for the metal to gain electrons, i.e., to be reduced and exist in its elemental form. Copper (Cu) exhibits a highly positive \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} \) value, indicating a comparatively lower likelihood of oxidation compared to many other metals.

Factors Contributing to Copper's Positive \( E^\circ \) Value:
- Electron Configuration and Stability: Copper's ground-state electron configuration is a notably stable \( 3d^{10} 4s^1 \). The \( \text{Cu}^{2+} \) ion, with a \( 3d^9 \) configuration, is less stable due to the electron removal from a filled \( 3d \) orbital. This instability favors the reduction of \( \text{Cu}^{2+} \) to copper metal, resulting in a higher \( E^\circ \). - Lattice Energy and Solvation Energy: While \( \text{Cu}^{2+} \) ions are significantly stabilized by solvation in aqueous solutions due to their high charge, this effect is less dominant than the inherent stability of copper in its metallic state, thereby reducing its tendency to lose electrons. - Relative Inertness: Copper is less reactive than many metals (e.g., sodium, iron), exhibiting resistance to corrosion and oxidation. This inertness is reflected in its positive electrode potential, signifying greater stability in its elemental form than as \( \text{Cu}^{2+} \).

Summary:
Copper's high positive \( E^\circ \) value underscores its inclination towards reduction and its preference for existing in its elemental state, alongside the stability of its atomic structure and its resistance to oxidation. Consequently, copper possesses a relatively positive electrode potential compared to numerous other metals.

Answer 2

Zinc (Zn) is a potent reducing agent in its +2 oxidation state. This is evidenced by its low standard reduction potential, \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V} \), signifying a high propensity to lose electrons and form \( Zn^{2+} \). A more negative \( E^\circ \) value correlates with a stronger reducing agent due to its increased electron-donating capability. Consequently, in the +2 oxidation state, zinc efficiently reduces other substances by donating electrons.

Answer 3

Reason for Colourless Zn\(^{2+}\) Salts:

Explanation:
The absence of colour in Zn\(^{2+}\) salts stems from its electronic configuration and the nature of its d-orbitals.

1. Zn\(^{2+}\) Electronic Configuration:
Zinc's elemental electron configuration is [Ar] 3d\(^{10}\) 4s\(^{2}\). Upon forming the Zn\(^{2+}\) ion, it loses two electrons, resulting in a [Ar] 3d\(^{10}\) configuration, where all 3d orbitals are completely filled with 10 electrons.

2. Inability for d-orbital Excitation:
Visible colour in compounds arises from electronic transitions that absorb specific wavelengths of visible light, typically involving d-orbital excitations. Since Zn\(^{2+}\) possesses fully occupied 3d orbitals, there are no available d-d transitions to absorb visible light. Consequently, Zn\(^{2+}\) salts do not absorb in the visible spectrum and appear colourless.

3. Lack of Ligand-to-Metal Charge Transfer:
Unlike transition metal ions with partially filled d-orbitals (e.g., Cu\(^{2+}\), Fe\(^{3+}\)) that can undergo ligand-to-metal charge transfer, Zn\(^{2+}\) does not exhibit this phenomenon due to its filled d-orbitals, which precludes colour-generating charge transfer interactions.

Conclusion:
The complete filling of the 3d orbitals in the Zn\(^{2+}\) ion prevents any electronic transitions capable of absorbing visible light. Thus, Zn\(^{2+}\) salts are colourless.