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

Options

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

Answer

D

Explanation

(d) From the question, we have an equation
`2H^(+)+2e^(-) to H_(2)(g)`
According to Nernst equation,
`E=E^(@)-(0.059)/(2)log""(p_(Hg))/([H^(+)]^(2))`
`=0-(0.059)/(2) log""(p_(Hg))/((10^(-7))^(2))" " [:.[H^(+)]=10^(-7)]`
`:.` For potential of `H_(2)` electrode to be zero, `p_(H_(2))` should be equal to `[H^(+)]^(2)`,i.e `10^(-14)` atm.
`:." " log""(10^(-14))/((10^(-7))^(2))=0`