1.

Value of E_(H_(2)O//H_(2)(1atm)Pt) at 298 K would

Answer»

`-0.207V`
`+0.207V`
`-0.414V`
`+0.414V`

Solution :For water at 298K,`[H^(+)]=10^(-7)M`
Reduction reaction is: `H^(+)+e^(-)to(1)/(2)H_(2)`
`thereforeE=-(RT)/(F)"LN"(p_(H_(2))^(1//2))/([H^(+)])=-0.0591"LOG"(p_(H_(2))^(1//2))/([H^(+)])`
`=0.0591"log"(1)/(10^(-7))=-0.4137=-0.414V`


Discussion

No Comment Found

Related InterviewSolutions