The high atomization enthalpy (\( \Delta H^0_{\text{atom}} \)) and low hydration enthalpy (\( \Delta H^0_{\text{hydr}} \)) of copper make its standard reduction potential (\( E^0 \)) positive.
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.
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.
If E$_{cell}$ of the following reaction is x $\times$ 10$^{-1}$. Find x
\(\text{Pt/ HSnO$_2$ / Sn(OH)$_6^{2-}$, OH$^-$ / Bi$_2$O$_3$ / Bi / Pt}\)
\(\text{[Reaction Quotient, Q = 10$^6$]}\)
Given \( E^o_{\text{[Sn(OH)$_3$]}} \) = -0.90 V, \( E^o_{\text{Bi$_2$O$_3$ / Bi}} \) = -0.44 V

From the following information, calculate opening and closing inventory :