Question

Assertion (A): $E^\circ_{Cu^{2+}/Cu} = +0.34$ V.
Reason (R): It is due to low $\Delta H^\circ_{hyd}$ and high $\Delta H^\circ_a$.

• A. Both A and R are true and R is the correct explanation of A.
• B. Both A and R are true but R is not the correct explanation of A.
• C. A is true but R is false.
• D. A is false but R is true.

Hint:

Cu has positive $E^\circ$ because high $\Delta H^\circ_a$ and $\Delta H^\circ_{IE}$ are not compensated by (insufficient) $\Delta H^\circ_{hyd}$.

Answer 1

The Correct Option is C

Step 1: Check the Assertion.
The standard reduction potential $E^\circ_{Cu^{2+}/Cu} = +0.34$ V is a well-established value, meaning copper is reduced rather than oxidised under standard conditions. Assertion is TRUE.
Step 2: Thermodynamic factors governing $E^\circ$.
The electrode potential reflects the net thermodynamics of: sublimation enthalpy ($\Delta H^\circ_a$), ionisation enthalpy ($\Delta H^\circ_{IE}$), and hydration enthalpy ($\Delta H^\circ_{hyd}$). For copper, both $\Delta H^\circ_a$ and $\Delta H^\circ_{IE}$ are HIGH, while $\Delta H^\circ_{hyd}$ is relatively LOW, making dissolution of Cu thermodynamically unfavourable, hence positive $E^\circ$.
Step 3: Why the Reason is false.
The Reason incorrectly frames the cause. It is the HIGH atomisation and ionisation enthalpies combined with LOW hydration enthalpy that explain positive $E^\circ$, but the Reason as stated misrepresents this relationship. Reason is FALSE.
Step 4: Conclusion.
A is true but R is false.
\[ \boxed{\text{A true; R false}} \]