Question

Assertion (A): Cu cannot liberate \( H_2 \) on reaction with dilute mineral acids. 
Reason (R): Cu has positive electrode potential.

• A. Assertion (A) is false, but Reason (R) is true.
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Hint:

Metals like copper, with a positive electrode potential, are less reactive and do not readily release hydrogen when reacting with dilute acids.

Answer 1

The Correct Option is D

To address the issue, we will assess the Assertion (A) and Reason (R) concerning the interaction of copper (Cu) with dilute mineral acids, verifying their accuracy and interrelation.

1. Analysis of Assertion (A):
Assertion (A) posits that Cu cannot produce H\(_2\) when reacting with dilute mineral acids. Dilute mineral acids, such as HCl or H\(_2\)SO\(_4\), supply H\(^+\) ions, which can be reduced to H\(_2\) gas by a metal. For a metal to liberate H\(_2\), it must displace hydrogen by reducing H\(^+\) (via the reaction 2H\(^+\) + 2e\(^-\) → H\(_2\)). This capability is determined by the metal's position in the electrochemical series. Copper (Cu) is situated below hydrogen in this series, indicating lower reactivity. Consequently, it cannot reduce H\(^+\) to H\(_2\) in dilute acids under standard conditions. Copper does not react with dilute HCl or H\(_2\)SO\(_4\) to generate H\(_2\). However, it can react with oxidizing acids like HNO\(_3\) or hot, concentrated H\(_2\)SO\(_4\) to yield different products (e.g., NO or SO\(_2\)). Therefore, the assertion holds true for dilute, non-oxidizing mineral acids.

2. Analysis of Reason (R):
Reason (R) states that Cu exhibits a positive electrode potential. The standard electrode potential (E\(^\circ\)) of a metal quantifies its propensity for reduction. The standard reduction potential for the reaction Cu\(^{2+}\) + 2e\(^-\) → Cu is approximately +0.34 V, which is a positive value. For hydrogen, the reduction potential is 0 V (2H\(^+\) + 2e\(^-\) → H\(_2\)). A metal with a positive reduction potential (like Cu) is more prone to reduction than H\(^+\). This implies that Cu cannot reduce H\(^+\) to H\(_2\), as the oxidation of Cu (Cu → Cu\(^{2+}\) + 2e\(^-\)) is less favorable than the reduction of hydrogen. Consequently, the reason is valid, as Cu's positive electrode potential signifies its inability to displace hydrogen.

3. Evaluation of the Relationship:
The reason provides a valid explanation for the assertion. Copper's positive electrode potential (+0.34 V) indicates a lower tendency for oxidation (electron loss) compared to hydrogen (0 V). For Cu to liberate H\(_2\), it would need to undergo oxidation (Cu → Cu\(^{2+}\) + 2e\(^-\)), while H\(^+\) is reduced. Given that Cu's reduction potential is higher than that of H\(^+\), the reaction is not spontaneous, and Cu does not liberate H\(_2\). Hence, R accurately elucidates why A is true.

4. Conclusion:
- Assertion (A) is accurate, as Cu cannot liberate H\(_2\) when reacting with dilute mineral acids.
- Reason (R) is accurate, as Cu possesses a positive electrode potential.
- The reason correctly explains the assertion, attributing Cu's inability to reduce H\(^+\) to H\(_2\) to its positive electrode potential.

Final Answer:
Assertion (A) is true, Reason (R) is true, and R correctly explains A.