How Does Electrochemistry Calculate the Solubility of AgI?

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SUMMARY

The discussion focuses on calculating the solubility of silver iodide (AgI) using the Nernst equation. The standard potential of the cell is given as +0.9509 V at 25°C, while the standard reduction potential for Ag+ is 0.7991 V. The user incorrectly interprets the result of -18.51 as the natural logarithm of the solubility, when it is actually the decimal logarithm. The correct approach involves using the Nernst equation to derive the natural log of the solubility (ln s) accurately.

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  • Nernst equation application
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  • Knowledge of solubility product constants (Ksp)
  • Basic electrochemistry principles
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koomanchoo
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hey i seem to be having a lot of trouble qith this question may i get some help?
The standard potential of the cell below is + 0.9509 V at 25 oC. Calculate the solubility, normally called s, of AgI (Note: Enter the natural log of s (ln s) as your answer)

Ag|AgI(s)|AgI(aq)|Ag

i'm using nernst equation and found the standard potential for Ag+,Ag to be 0.7991V

i am using Eo as 0.7991 and E as 0.9509 and rearanging the equation to find the solubility as LnS.. the correct answer is suppose to be -18.51.. where i am going wrong? please help.
thanks

n.b standard reduction potential of AgI(s) is -.15V
 
Last edited:
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IIRC -18.51 is not a natural log of the AgI solubility product - it's decimal log.
 

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