Two metals `M_(1)` and `M_(2)` have `E^(c-)._(red)=-0.76 V` and `0.80V`, respectively. Which will liberate `H_(2)(g)` from `H_(2)SO_(4)` ?
Given `:`
`{:(M_(1)^(n+)+n e^(-)rarr M_(1),,,,E^(c-)._(red)=-0.76V),(M_(2)^(n+)+n e^(-)rarr M_(2),,,,E^(c-)._(red)=0.80V),(2H^(o+)+2e^(-) rarr H_(2)(g),,,,E^(c-)._(red)=0.0V):}`

Explanation

To determine which metal, \( M_1 \) or \( M_2 \), will liberate \( H_2(g) \) from \( H_2SO_4 \), we need to analyze the standard reduction potentials given for both metals and compare them with the standard reduction potential of hydrogen.

### Step-by-Step Solution:

1. **Identify the Standard Reduction Potentials**:
- For metal \( M_1 \): \( E^{\circ}_{\text{red}} = -0.76 \, \text{V} \)
- For metal \( M_2 \): \( E^{\circ}_{\text{red}} = 0.80 \, \text{V} \)
- For hydrogen: \( E^{\circ}_{\text{red}} = 0.00 \, \text{V} \)
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