# How does pH affect the Nernst equation?

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 ${P}_{\text{H₂}}$ = 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⁺]

${E}_{\text{H⁺/H₂" = "-0.059 16 V × pH}}$

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
${E}_{\text{H₂/H⁺" = -E_"H⁺/H₂}}$ = 0.059 16 V × pH
${E}_{\text{cell}}$ = (0.034 + 0.059 16 × pH) V