Question

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

Answer 1

Why Copper Has an Exceptionally Positive \( E^\circ_{\text{M}^{2+}/\text{M}} \) Value:

Explanation:
The standard electrode potential (\( E^\circ \)) of a metal represents its tendency to lose electrons and undergo oxidation. A positive \( E^\circ \) value indicates that the metal has a strong tendency to gain electrons (i.e., it is easily reduced) and exist in its elemental form. Copper (Cu) has a very positive \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} \) value, which means it is relatively less likely to oxidize and lose electrons compared to many other metals.

Reason for the Positive \( E^\circ \) Value of Copper:
- Electron Configuration and Stability: Copper has a unique electron configuration. In its ground state, copper has a relatively stable \( 3d^{10} 4s^1 \) electron configuration. The \( \text{Cu}^{2+} \) ion, with a \( 3d^9 \) configuration, is less stable due to the loss of an electron from the filled \( 3d \) orbital. This makes the reduction of \( \text{Cu}^{2+} \) to copper metal more favorable, leading to a higher \( E^\circ \). - Lattice Energy and Solvation Energy: The energy involved in the solvation of \( \text{Cu}^{2+} \) ions is significant. The \( \text{Cu}^{2+} \) ion is highly charged and is stabilized by the solvation process in aqueous solutions, but this does not outweigh the stabilizing effect of copper's metallic state, making copper less likely to lose electrons. - Comparative Inertness: Copper is less reactive than many other metals (such as sodium or iron), which means it resists corrosion and oxidation. This inertness is reflected in its positive electrode potential value, indicating that copper is more stable in its elemental form than as \( \text{Cu}^{2+} \).

Conclusion:
The high positive \( E^\circ \) value of copper reflects its tendency to be reduced and stay in its elemental form, as well as the stability of its atomic structure and its resistance to oxidation. This is why copper has a relatively positive electrode potential compared to many other metals.

Answer 2

The element that is a strong reducing agent in the +2 oxidation state is Zinc (Zn). Zinc has a relatively low \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} \) value of \(-0.76 \, \text{V}\), which means it readily loses electrons to form \( Zn^{2+} \). The more negative the \( E^\circ \) value, the stronger the reducing agent, because it indicates a greater tendency to donate electrons.
In the +2 oxidation state, zinc acts as a strong reducing agent, easily donating electrons to reduce other species.

Answer 3

Why Zn\(^{2+}\) Salts are Colourless:

Explanation:
Zn\(^{2+}\) salts are colourless due to the electronic configuration of the zinc ion and the nature of the d-orbitals involved in its chemistry.

1. Electronic Configuration of Zn\(^{2+}\):
In its elemental state, zinc has the electron configuration [Ar] 3d\(^{10}\) 4s\(^{2}\). When zinc loses two electrons to form the Zn\(^{2+}\) ion, the electron configuration becomes [Ar] 3d\(^{10}\), meaning that all of the d-orbitals are fully occupied with 10 electrons.

2. Lack of d-orbital Excitation:
For a compound to show colour, there must be electronic transitions involving the absorption of visible light. These transitions usually occur when electrons are excited from one d-orbital to another. However, in Zn\(^{2+}\), the 3d orbitals are completely filled, meaning there are no available electron transitions within the d-orbital to absorb visible light. As a result, Zn\(^{2+}\) salts do not absorb light in the visible spectrum and hence appear colourless.

3. Absence of Ligand-to-Metal Charge Transfer:
Unlike other transition metal ions (such as Cu\(^{2+}\) or Fe\(^{3+}\)), which have partially filled d-orbitals that allow for ligand-to-metal charge transfer (a process that can lead to coloured solutions), Zn\(^{2+}\) does not exhibit such behaviour. The filled d-orbitals in Zn\(^{2+}\) result in the absence of the charge transfer interactions that would give rise to colour.

Conclusion:
Zn\(^{2+}\) salts are colourless because the 3d orbitals of the Zn\(^{2+}\) ion are fully occupied, preventing any electronic transitions that could absorb visible light and produce colour. Therefore, no colour is observed in Zn\(^{2+}\) salts.