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Avon97
Avon97
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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

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

So

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

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

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

Overall reaction is this, is obtained by summing the two half cell reactions. (Need to connect the two electrodes to obtain this)(As one reduction and one oxidation doesn't matter since its in equilibrium)

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

The equation is,

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

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

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

So

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

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

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

Overall reaction is this, is obtained by summing the two half cell reactions. (Need to connect the two electrodes to obtain this)(As one reduction and one oxidation doesn't matter since its in equilibrium)

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

The equation is,

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

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

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

So

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

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,

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

Improved the meaning of the answer
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Avon97
Avon97
  • 191
  • 2
  • 13

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

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

So

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

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

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

Overall reaction is this, is obtained by summing the two half cell reactions. (Need to connect the two electrodes to obtain this)(As one reduction and one oxidation doesn't matter since its in equilibrium)

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

The equation is,

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

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

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

So

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

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,

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

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

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

So

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

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

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

Overall reaction is this, is obtained by summing the two half cell reactions. (Need to connect the two electrodes to obtain this)(As one reduction and one oxidation doesn't matter since its in equilibrium)

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

The equation is,

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

fixed spelling mistake
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Avon97
Avon97
  • 191
  • 2
  • 13

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 nerstNernst 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 nerstNernst 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

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

So

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

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,

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

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 nerst 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 nerst 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

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

So

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

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,

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

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

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

So

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

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,

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

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Avon97
Avon97
  • 191
  • 2
  • 13