1.

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

Answer»

`10^(-10)m`
`10^(-4)atm`
`10^(-14)atm`
`10^(-12)atm`

SOLUTION :The reduction REACTION for `H_(2)`-electrode is
`2H^(+)(AQ)+2^(-)toH_(2)(g)`
`H_(H^(+)//H_(2))^(@)=E_(H^(+)//H_(2))^(@)-(0.0591)/(2)"log"(p_(H_(2)))/([H^(+)]^(2))`
In pure WATER at 298K, `[H^(+)]=10^(-7)M`
`therefore0=0-(0.0591)/(2)"log"(p_(H_(2)))/((10^(-7))^(2))`
or `"log"(p_(H_(2)))/(10^(-14))=0` or `(p_(H_(2)))/(10^(-14))=1""(becauselog1=0)`
or `p_(H_(2))=10^(-14)atm`


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