SEM = 171,1mV = 0,1711V E⁰(Ag⁺|Ag) = 0,80V E⁰(AgCl|Ag) = 0,197V zał. T = 298K R = 8,314 J/mol*K F = 96500 C/mol [latex]SEM = E_P - E_L = E_{Ag^+|Ag} - E_{AgCl|Ag}\ \ E_{Ag^+|Ag} =E^0_{Ag^+|Ag}+frac{RT}{zF}lnfrac{[Ag^+]}{[Ag]}\ \ E_{AgCl|Ag} = E^0_{AgCl|Ag} +frac{RT}{zF}lnfrac{1}{[Cl^-]}[/latex] [Ag] = 1 i z w obu równaniach również jest równe 1. Wtedy: [latex]SEM = E^0_{Ag^+|Ag}+frac{RT}{F}ln[Ag^+] - E^0_{AgCl|Ag} -frac{RT}{F}lnfrac{1}{[Cl^-]}\ \ 0,1711 = 0,80 + 0,02567ln0,01 - 0,197 - 0,02567lnfrac{1}{[Cl^-]}\ \ -0,4319 = -0,1182 - 0,02567lnfrac{1}{[Cl^-]}\ \ -0,3137 = -0,02567lnfrac{1}{[Cl^-]}\ \ lnfrac{1}{[Cl^-]} = 12,22\ \ frac{1}{[Cl^-]} = 202805\ \ [[Cl^-] =4,93cdot 10^-^6[/latex] Ir = [Ag⁺][Cl⁻] = [Cl⁻]² = (4,93*10⁻⁶)² = 2,43*10⁻¹¹
