1.

The pressure of H_(2) required to make the potential of H_(2)-electrode zero in pure water at 298K is

Answer»

<P>`10^(-14)atm`
`10^(-12)atm`
`10^(-10)atm`
`10^(-4)atm`

Solution :`2H^(+)(aq)=2e^(-)toH_(2)(g)`
`thereforeE=E^(0)-(0.0591)/(2)"log"(P_(H_(2)))/([H^(+)]^(2))`
`0=0-0.295"log"(P_(H_(2)))/((10^(-7))^(2))`
`(P_(H_(2)))/((10^(-7))^(2))=1`
`P_(H_(2))=10^(-14)atm`


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