1.

The pressure of `H_(2)` required to make the potential of `H_(2)-`electrode zero in pure water at 289K is :A. `10^(-4)atm`B. `10^(-14)atm`C. `10^(-12)atm`D. `10^(-10)atm`

Answer» Correct Answer - B
`2H_(aq.)^(+) = 2_(e-) rarr H_(2)` (reduction reaction)
as pure water has `(H^+) = (-OH)=10^(-7)`
`E = E^(@)-(0.059)/(2) log(P_(H_2))/([H_(aq)^(+)]^(2))`
`0-0.059/2 log(P_(H_2))/([10^(-7)]^(2))`
so as to make `E = 0, log(P_(H_2))/([H_(2)^(+)]^(2)) = 0`
which is only possible when `P_(H_2) = 10^(-14)`
so log`(P_(H_2))/([H^(+)]^(2)) = log(10^(-14))/(10^(-14)) = log 1 = 0`.


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