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

`a.` Now constructing a cell of metal `M_(1)` with `H_(2)SO_(4)`,
`2M_(1)+nH_(2)SO_(4) rarr (M_(1))_(2)(SO_(4))_(n)+nH_(2). ,,,,,(i)`
`[M_(1)` undergoes oxidation to `M_(1)^(2+)` and `2h^(o+)` undergoes reduction to `H_(2)]`
`:. E^(c-)._(cell)=E^(c-)._(red)+E^(c-)._(o x i d )`
`=E^(c-)._(2H^(o+)|H_(2))+E^(c-)._(M_(1)|M_(1)^(2+))`
`=0+0.76=0.76V`
Since `E^(c-)._(cell)` is positive and reaction `(i)` is spontaneous, therefore, `M_(1)` will liberate `H_(2)` from `H_(2)SO_(4)`.
`b.` Now constructing a cell of metal `M_(2)` with `H_(2)SO_(4)`.
`:. E^(c-)._(cell)=E^(c-)._(red)+E^(c-)._(o x i d)`
`=E^(c-)._(2H^(o+)|H_(2))+E^(c-).M_(2)|M_(2)^(2+)`
`=0-0.80=-0.80V`
Since `E^(c-)._(cell)` is negative and reaction `(ii)` is non`-` spontaneous, thus, `M_(2)` will not liberate `H_(2)` from `H_(2)SO_(4)`.