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

Two metals `A` and `B` have `E_(RP)^(@) = +0.76V` and `-0.80V` respect

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

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

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