1.

What will be the emf for the given cell Pt|H_(2)(P_(1))|H_((aq))^(+)||H_(2)(P_(2))|Pt

Answer»

`(RT)/(F)ln(P_(1))/(P_(2))`
`(RT)/(2f)ln(P_(1))/(P_(2))`
`(RT)/(f)ln(P_(2))/(P_(1))`
None of these

Solution :Anodic reaction: `H_(2)(P_(1))to2H^(+)`
Cathodic reaction: `2H^(+)toH_(2)(P_(2))`
`E_("cathode")=-(RT)/(2F)ln (P_(2))/([H^(+)]^(2)),E_("anode")=-(RT)/(2F)ln([H^(+)]^(2))/(P_(1))`
`E_("INF")=E_("anode")+E_("cathode")` ltBrgt `=-(RT)/(2F)ln((H^(+))^(2))/(P_(1))-(RT)/(2F)ln(P_(2))/((H^(+))^(2))`
`=-(RT)/(2F)ln(P_(2))/(P_(1))=(RT)/(2F)ln(P_(1))/(P_(2))`


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