Electrode potential of hydrogen electrode is 18 m V then `[H^(+)]` is -

Options

A. `0.2`
B. `1`
C. 2
D. 5

Answer

C

Explanation

Given `E_(H^(+) | H_(2)) = 18 mV = 18 xx 10^(-3) V , [H^(+)]= ?` and `H^(+) + e^(-) to (1)/(2) H_(2) , pH_(2) = 1 ` atm
AlC to Nernst equation
`E_(H^(+) | H_(2)) = E_(H^(+) | H_(2))^(@) - (0.0592)/(n) "log" (1)/([H^(+)])`
`18 xx 10^(-3) V = 0 + (0.0592)/(1) "log" [H^(+)]`
`therefore log [H^(+)] = (18 xx 10^(-3) V)/(0.0592) = 0 . 3046`
`[H^(+)] = AL (0.3046) = 2.02 = 2`