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

Options

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

Answer

A

Explanation

`2H^(+)(aq) + 2e^(-) rarr H_(2)(g)`
`therefore E = E^(0)- (0.0591)/(2) "log" (P_(H_(2)))/([H^(+)]^(2))`
`0= 0- 0.0295 "log" (p_(H_(2)))/((10^(-7))^(2))`
`(p_(H_(2)))/((10^(-7))^(2))= 1`
`p_(H_(2))= 10^(-14)` atm