calculate the ph of half cell Pt,H2|H2SO4, if its electrode potential is 0.03V

Answer 1

ou can read much more here.
https://en.wikipedia.org/wiki/Nernst_equation

The reaction is 2H^+ + 2e ==> H2(g)
Ecell = Eocell - (RT/nF)log(red/ox)
You know Ecell from the problem. Eo is zero. RT/nF is 0.05916/n and n = 2. H2 as a gas and at 1 atm pressure is 1. Solve for H^+ and convert to pH.

Answer 2

With pH there is a much quicker way of doing this.
Ecell = Eo + (0.05916/1)*pH

Answer 3

I don't have my solution

Answer 4

No clear solution