Step 1: Explanation of \( E^0 \) Value - The electrode potential (\( E^0 \)) depends on: - Atomization enthalpy (\( \Delta H^0_{\text{atom}} \)): The energy required to convert solid Cu to Cu\(^{2+}\) is high. - Hydration enthalpy (\( \Delta H^0_{\text{hydr}} \)): Cu\(^{2+}\) has low hydration energy, making it less stable in aqueous solution.
Step 2: Effect on \( E^0 \) Value - Due to low hydration enthalpy, the reduction of Cu\(^{2+}\) to Cu is not highly favored. - Hence, Cu\(^{2+}/\)Cu has a positive \( E^0 \) value of \( +0.34 \) V, indicating that Cu is less reactive than expected.
Give explanation for each of the following observations:
(a) With the same d-orbital configuration (d4), Mn3+ ion is an oxidizing agent whereas Cr2+ ion is a reducing agent.
(b) Actinoid contraction is greater from element to element than that among lanthanoids.
(c) Transition metals form a large number of interstitial compounds with H, B, C, and N.
A transition metal (M) among Mn, Cr, Co, and Fe has the highest standard electrode potential $ M^{n}/M^{n+1} $. It forms a metal complex of the type $[M \text{CN}]^{n+}$. The number of electrons present in the $ e $-orbital of the complex is ... ...