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

Options

A. `10^(-4)` atm
B. `10^(-14)` atm
C. `10^(-32)` atm
D. `10^(-10)` atm

Answer

B

Explanation

`2H^(+) + 2e^(-) to H_(2)`
`E = E^(@) - 0.059/2 log P_(H_(2))/(H^(+))^(2)`
`log 1=0`
`P_(H_(2)) = (10^(-7))^(2) = 10^(-14)` atm.