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

Options

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

Answer

B

Explanation

Electrode reaction:
`2H^(+)(aq.)+2e^(-) rarr H_(2(g))`
Nernst equation:
`E = E^(@) - (0.0591 V)/(2)"log"(p_(H_(2)))/(C_(H^(+))^(2))`
At `298 k, pH` of pure water is `7`, thus, `C_(H^(+)) = 10^(-7) molL^(-1)`
Thus,
`0 = 0-0.0296 V "log" (p_(H_(2)))/(10^(-7))^(2)`
Thus, `(p_(H_(2)))/(10^(-7))^(2) = 1`
`(p_(H_(2))) = 10^(-14) atm`