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If the potential of a hydrogen electrode based on

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Answer #1

Nernest eq is

E cell = Eo cell - ( 0.059/n) log Q

where Q = 1 /[H+]^2             , given pH = 6.75 , [H+] = 10^ -6.75 = 1.78 x 10^ - 7 M

Q = ( 1/ 1.78x10^-7)^2 = 3.156 x 10^ 13

n = number of electrons involved per reduction = 2

Eo cell = 0   for H+/H2 reduction

E cell = 0 - ( 0.059/2) log ( 3.156x10^13)

   = -0.37 V

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