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 solve the problem, we need to evaluate the Assertion (A) and Reason (R) regarding the reaction of copper (Cu) with dilute mineral acids and determine their validity and relationship.

1. Analyzing the Assertion (A):
Assertion (A) states that Cu cannot liberate H\(_2\) on reaction with dilute mineral acids. Dilute mineral acids, such as dilute HCl or H\(_2\)SO\(_4\), provide H\(^+\) ions that could potentially be reduced to H\(_2\) gas by a metal. For a metal to liberate H\(_2\), it must displace hydrogen by reducing H\(^+\) (2H\(^+\) + 2e\(^-\) → H\(_2\)). This depends on the metal’s position in the electrochemical series. Copper (Cu) is below hydrogen in the series, meaning it is less reactive and 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 produce H\(_2\), though it can react with oxidizing acids like HNO\(_3\) or hot, concentrated H\(_2\)SO\(_4\) to form other products (e.g., NO or SO\(_2\)). Thus, the assertion is true for dilute, non-oxidizing mineral acids.

2. Analyzing the Reason (R):
Reason (R) states that Cu has a positive electrode potential. The standard electrode potential (E\(^\circ\)) of a metal indicates its tendency to be reduced. The standard reduction potential for Cu\(^{2+}\) + 2e\(^-\) → Cu is approximately +0.34 V, which is positive. For hydrogen, the reduction potential is 0 V (2H\(^+\) + 2e\(^-\) → H\(_2\)). A metal with a positive reduction potential (like Cu) is more likely to be reduced than H\(^+\), meaning Cu cannot reduce H\(^+\) to H\(_2\), as its oxidation (Cu → Cu\(^{2+}\) + 2e\(^-\)) is not favored compared to hydrogen’s reduction. Thus, the reason is true, as Cu’s positive electrode potential indicates its inability to displace hydrogen.

3. Evaluating the Relationship:
The reason explains the assertion. Copper’s positive electrode potential (+0.34 V) means it is less likely to oxidize (lose electrons) compared to hydrogen (0 V). For Cu to liberate H\(_2\), it would need to oxidize (Cu → Cu\(^{2+}\) + 2e\(^-\)) while H\(^+\) is reduced. Since Cu’s reduction potential is higher than that of H\(^+\), the reaction is not spontaneous, and Cu does not liberate H\(_2\). Thus, R directly explains why A is true.

4. Conclusion:
- Assertion (A) is true because Cu cannot liberate H\(_2\) when reacted with dilute mineral acids.
- Reason (R) is true because Cu has a positive electrode potential.
- The reason correctly explains the assertion, as the positive electrode potential of Cu prevents it from reducing H\(^+\) to H\(_2\).

Final Answer:
Assertion (A) is true, Reason (R) is true, and R is the correct explanation for A.