Two metals A and B have `E_(RP)^(0) = – 0.76 V and + 0.80 V` respectively. Which of these will liberate `H_(2)` from `H_(2) SO _(4)` ?

Options

A. A
B. B
C. Both A and B
D. None of these

Answer

To determine which metal, A or B, will liberate hydrogen gas from sulfuric acid (H₂SO₄), we need to analyze their standard reduction potential values.

### Step-by-Step Solution:

1. **Identify the Standard Reduction Potentials**:
- Metal A has a standard reduction potential (E°) of -0.76 V.
- Metal B has a standard reduction potential (E°) of +0.80 V.

2. **Understand the Concept of Reactivity**:
- A metal can displace hydrogen from an acid if it is more reactive than hydrogen. This can be determined by comparing the reduction potentials.
- The standard reduction potential for hydrogen (H⁺ + e⁻ → H₂) is defined as 0 V.

3. **Determine the Oxidation Potential**:
- The oxidation potential is the negative of the reduction potential. Therefore:
- For Metal A: Oxidation potential = -(-0.76 V) = +0.76 V
- For Metal B: Oxidation potential = -(+0.80 V) = -0.80 V

4. **Compare with Hydrogen**:
- Since hydrogen has a reduction potential of 0 V, we need to compare the reduction potentials of A and B with that of hydrogen.
- Metal A (-0.76 V) has a lower (more negative) reduction potential than hydrogen (0 V), indicating it is more likely to be oxidized and can displace hydrogen from H₂SO₄.
- Metal B (+0.80 V) has a higher reduction potential than hydrogen, indicating it is less likely to be oxidized and cannot displace hydrogen.

5. **Conclusion**:
- Since Metal A has a lower reduction potential than hydrogen, it will liberate hydrogen gas from H₂SO₄. Therefore, the answer is Metal A.

### Final Answer:
Metal A will liberate hydrogen from H₂SO₄.