The pressure of `H_(2)` required to make the potential of `H_(2)-`electrode zero in pure water at 289K is :

Options

A. `10^(-10)m`
B. `10^(-4)atm`
C. `10^(-14)atm`
D. `10^(-12)atm`

Answer

C

Explanation

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`