Question

The reduction potential of hydrogen electrode at ph = 7 will be

Answer 1

2H+ + 2e----->H2

E°=0

E=E°-0.059/2*log(1/[H+]²)

[H+]=10⁻⁷

E=-0.059/2*log(1/10⁻¹⁴)

=-0.059/2*(14)

=-0.413