Question

How does pH affect the Nernst equation?

Answer 1

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