`Pt(H_(2))(p_(1))|H^(o+)(1M)|(H_(2))(p_(2)),Pt` cell reaction will be exergonic ifA. `p_(1)=p_(2)`B. `p_(1)gtp_(2)`C. `p_(2)gtp_(1)`D. `p_(1)=1 atm`

`Pt(H_(2))(p_(1))|H^(o+)(1M)|(H_(2))(p_(2)),Pt` cell reaction will be

1.	`Pt(H_(2))(p_(1))\|H^(o+)(1M)\|(H_(2))(p_(2)),Pt` cell reaction will be exergonic ifA. `p_(1)=p_(2)`B. `p_(1)gtp_(2)`C. `p_(2)gtp_(1)`D. `p_(1)=1 atm`
Answer» Correct Answer - b For a cell to be spontaneous `(E_(cell)=+ve or Delta G =-ve)(` exergonic `)p_(1)p_(2) or c_(2)gtc_(1)`. Factual statement.

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`Pt(H_(2))(p_(1))|H^(o+)(1M)|(H_(2))(p_(2)),Pt` cell reaction will be exergonic ifA. `p_(1)=p_(2)`B. `p_(1)gtp_(2)`C. `p_(2)gtp_(1)`D. `p_(1)=1 atm`

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