How does pH affect the Nernst equation?

1 Answer
Sep 25, 2014

pH 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