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दो धातुओं A व B के `E_(R.P.)^(@)` क्रमशः `-0.76V` तथा `+0.80V` है कौन सी धातु `H_(2)SO_(4)` से `H_(2)` मुक्त करेगी ?

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धातु A के लिए,
`A^(n+)+"ne "to A," "E_(R.P.)^(@)=-0.76V`
धातु B के लिए, `B^(n+)+"ne " to B," "E_(R.P.)^(@)=+0.80V`
यदि A की `H_(2)SO_(4)` के साथ अभिक्रिया इस प्रकार हो,
`2A+nH_(2)SO_(4)toA_(2)(SO_(4))_(n)+nH_(2)`
`E_("cell")^(@)=E_(O.P._(A))^(@)+E_(R.P._(H))^(@)=0.76+0.0=0.76`
चूँकि `E_("cell")^(@)` धनात्मक है अतः `A,H_(2)SO_(4)` से `H_(2)` मुक्त करेगी |
यदि B की `H_(2)SO_(4)` के साथ अभिक्रिया इस प्रकार हो
`2B+nH_(2)SO_(4)toB_(2)(SO_(4))_(n)+nH_(2)`
`E_("cell")^(@)=E_(O.P._(B))^(@)+E_(R.P._(H))^(@)=-0.80+0=-0.80V`
चूँकि `E^(@)` ऋणात्मक है अतः `B,H_(2)SO_(4)` से `H_(2)` मुक्त नहीं करेगी ।

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