I learned that $E^\circ_\ce{Cl^-|Ag,AgCl}, E^\circ_\ce{Ag^+|Ag}$ and $K_\mathrm{sp}$ of $\ce{AgCl}$ are related as $$\boxed{E^\circ_\ce{Cl^-|Ag,AgCl} = E^\circ_\ce{Ag^+|Ag} + \frac{RT}{F}\ln K_\mathrm{sp}}$$

I was given this proof:

For an $\ce{Ag^+|Ag}$ electrode in non-standard conditions, using the Nernst equation $$E_\ce{Ag^+|Ag} = E^\circ_\ce{Ag^+|Ag} - \frac{RT}{F}\ln\frac{[\ce{Ag}]}{[\ce{Ag^+}]}$$

As $\ce{[Ag^+][Cl^-]}=K_\mathrm{sp}$ and $\ce{[Cl^-]} = \pu{1 M} \ \text{(standard state)}$, so $\ce{[Ag^+]} = K_\mathrm{sp}$. Also $\ce{[Ag] = 1}$ as pure solids have activity of one.

So, $$E_\ce{Ag^+|Ag} = E^\circ_\ce{Ag^+|Ag} - \frac{RT}{F}\ln\frac{1}{K_\mathrm{sp}} = E^\circ_\ce{Ag^+|Ag} + \frac{RT}{F}\ln K_\mathrm{sp}$$

What I don't understand is the next step. In the proof, $E_\ce{Ag^+|Ag} = E^\circ_\ce{Cl^-|AgCl,Ag}$ and thus the boxed equation above is proved. But why is $E_\ce{Ag^+|Ag} = E^\circ_\ce{Cl^-|AgCl,Ag}$?

  • $\begingroup$ Because both the reactions involve reduction of $\ce{Ag+}$ $\endgroup$ Aug 3, 2021 at 9:35
  • $\begingroup$ @napstablook Then why is the non-standard potential one reaction equal to standard potential of the other? $\endgroup$ Aug 3, 2021 at 9:39

1 Answer 1


First of all you made a slight mistake one of the equations is wrong

First equilibrium reaction,

$$\ce{Ag(aq)+ + e- <=> Ag(s)}$$

Writing the Nernst equation for this reaction

$$\ce{E_{Ag|Ag^+} = E_{Ag|Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{[Ag]}}{\ce[Ag^+]}}$$

Second equilibrium reaction,

$$\ce{AgCl(s) + e- <=> Ag(s) + Cl-(aq)}$$

Writing the Nernst equation for the second reaction

$$\ce{E_{AgCl|Ag,Ag^+} = E_{AgCl|Ag,Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{[Ag][Cl^-]}}{\ce[AgCl]}}$$

Since the activity of solids are considered to be 1

$$\ce{E_{Ag|Ag^+} = E_{Ag|Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{1}}{\ce[Ag^+]}}$$

$$\ce{E_{AgCl|Ag,Ag^+} = E_{AgCl|Ag,Ag^+}^0 - \frac{RT}{F}ln\ce{[Cl^-]}}$$

When you short the wires of the two cells potential of the two cells equals

$$\ce{E_{Ag|Ag^+} = E_{AgCl|Ag,Ag^+}}$$


$$\ce{E_{Ag|Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{1}}{\ce[Ag^+]}} = E_{AgCl|Ag,Ag^+}^0 - \frac{RT}{F}ln\ce{[Cl^-]}$$

Simplifying the math here gives your equation,

$$\ce{E_{AgCl|Ag,Ag^+}^0 = E_{Ag|Ag^+}^0 + \frac{RT}{F}ln([Ag^+][Cl^-])}$$

Since we know

$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$

$$\ce{K_{sp} = [Ag^+][Cl^-]}$$

The equation is,

$$\ce{E_{AgCl|Ag,Ag^+}^0 = E_{Ag|Ag^+}^0 + \frac{RT}{F}ln(K_{sp})}$$

  • $\begingroup$ Thanks, this cleared up some things for me. However, I'm not sure what you mean by "shorting" here. How exactly do you short the wires attached to the two half-cells? $\endgroup$ Aug 3, 2021 at 12:34

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