The potential of hydrogen electrode is -118MV.thr H^+ concentration of the solution is

Question

a)0.01M
b)2M
C)10^4M
d)1M

Answer 1

I assume you mean -118 millivolts and that pH2 = 1 atmosphere..
-0.118 v = Eo - (0.06/n)log (H2/(H^+)^2 for the equation
2H^+ 2e ==> H2
Post your work if you get stuck.

Answer 2

Taking E =118millivolt...
2H+. +2e ...H2
E= E*-0.059/2log H2/[H+]2
-118=0-0.059/2log 1/[H+]2
Solving it by using antilog we will get 0.01