Skip to main content
Chemistry

Return to Question

Improved formatting and clarity. Added few tag.
Source Link
Mathew Mahindaratne
Mathew Mahindaratne
  • 42.1k
  • 29
  • 56
  • 111

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

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

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

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

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 $$\ce{E_{Ag^+|Ag} = E^\circ_{Ag^+|Ag} - \dfrac{\ce{RT}}{\ce{F}}\ln\dfrac{[\ce{Ag}]}{[\ce{Ag^+}]}}$$

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

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

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

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 $E_\ce{Ag^+|Ag} = E^\circ_\ce{Cl^-|AgCl,Ag}$?

Improved formatting and clarity.
Source Link
Mathew Mahindaratne
Mathew Mahindaratne
  • 42.1k
  • 29
  • 56
  • 111

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

I was given this proof:

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

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

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

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

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

I was given this proof:

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

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

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

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

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 $$\ce{E_{Ag^+|Ag} = E^\circ_{Ag^+|Ag} - \dfrac{\ce{RT}}{\ce{F}}\ln\dfrac{[\ce{Ag}]}{[\ce{Ag^+}]}}$$

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

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

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

Source Link
Ray Bradbury
Ray Bradbury
  • 760
  • 8
  • 24

Relation between standard silver-silver chloride electrode potential and solubility product of silver chloride

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

I was given this proof:

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

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

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

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