How does pH affect the Nernst equation?

doesn’t affect the Nernst equation.
But the Nernst equation predicts the cell potential of reactions that depend on pH.
If H⁺ is involved in the cell reaction, then the value of ##E## will depend on the pH.
For the half-reaction, 2H⁺ + 2e⁻ → H₂, ##E^°## = 0
According to the Nernst equation,
##E_”H⁺/H₂” = E^° – (RT)/(zF)lnQ = -(RT)/(zF)ln((P_”H₂”)/(“[H⁺]”^2))##
If ##P_”H₂”## = 1 atm and ##T## = 25 °C,
##E_”H⁺/H₂” = -(RT)/(zF)ln((P_”H₂”)/(“[H⁺]”^2)) = -(“8.314 J·K”^-1 × “298.15 K”)/(“2 × 96 485 J·V”^-1 )× ln(1/”[H⁺]”^2)## = 0.012 85 V × 2ln[H⁺] = 0.02569 V × 2.303log [H⁺]
##E_”H⁺/H₂” = “-0.059 16 V × pH”##
EXAMPLE
Calculate the cell potential for the following cell as a function of pH.
Cu²⁺(1 mol/L) + H₂(1 atm) → Cu(s) + 2H⁺(aq)
Solution
Cu²⁺ + 2e⁻ → Cu; ##E^°## = +0.34 V
H₂ → 2H⁺ + 2e⁻; ##E^°## = 0
Cu²⁺ + H₂ → Cu + 2H⁺; ##E^°## = +0.34 V
##E_”Cu²⁺/Cu”^°## = E^° – (RT)/(zF)ln(1/[Cu²⁺]) = E^° + 0 = +0.34 V
##E_”H₂/H⁺” = -E_”H⁺/H₂”## = 0.059 16 V × pH
##E_”cell”## = (0.034 + 0.059 16 × pH) V