# 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 VH₂ → 2H⁺ + 2e⁻; ##E^°## = 0Cu²⁺ + 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

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