`Pt(H_(2))(p_(1))|H^(o+)(1M)|(H_(2))(p_(2)),Pt` cell reaction will be exergonic ifA. `P_(1) gt P_(2)`B. `P_(1) gt P_(2)`C. `P_(2) gt P_(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) gt P_(2)`B. `P_(1) gt P_(2)`C. `P_(2) gt P_(1)`D. `P_(1) = 1` atm
Answer» Correct Answer - B `E_("cell")` for the given change `E_("cell") = (0.059)/(2) "log" ([H^(+)]xxP_(1(H_(2))))/(P_(2H_(2))xx[H^(+)])` Thus For `+ve E_("cell") = P_(1H_(2)) gt P_(2H_(2))`

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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) gt P_(2)`B. `P_(1) gt P_(2)`C. `P_(2) gt P_(1)`D. `P_(1) = 1` atm

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