Two metals `A` and `B` have `E_(RP)^(@) = +0.76V` and `-0.80V` respectively. Which will liberated `H_(2)` from `H_(2)SO_(4)` ?

Explanation

Given , For `A, A^(n+)+n e rarr A, E_(RP)^(@) = +0.76V`
For `B, B^(n+)+n e rarr B, E_(RP)^(@) = -0.80V`
We have, for `H, H^(+)+e rarr 1//2H_(2), E_(RP)^(@) = 0`
Now coupling `A` with `H_(2)SO_(4)` :
`2A + nH_(2)SO_(4) rarr A_(2)(SO_(4))_(n)+nH_(2)`
`E_(Cell)^(@) = E_(OP_(A))^(@)+E_(RP_(H))^(@) = -0.76 + 0.0 = -0.76V`
Since `E^(@)` is `+ve`,
`2A + nH_(2)SO_(4) rarr A_(2)(SO_(4))_(n)+nH_(2)`
Reaction is non-spontaneous.
A will not liberate `H_(2)` from `H_(2)SO_(4)`.
Now coupling `B` with `H_(2)SO_(4)`
`2B+nH_(2)SO_(4) rarr B_(2)(SO_(4)(n)+nH_(2)`
`E_(Cell)^(@) = E_(OP_(B))^(@)+E_(RP_(H))^(@) = +0.80 + 0 = +0.80`
Since `E^(@)` is `+ve`,
`:. 2B+nH_(2)SO_(4) rarr B_(2)(SO_(4))_(n)+nH_(2)`
Reaction will be spontaneous.
`B` will liberate `H_(2)` from `H_(2)SO_(4)`.