How to Calculate Ksp for Ferrous Oxalate in a Galvanic Cell?

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SUMMARY

The discussion focuses on calculating the solubility product constant (Ksp) for ferrous oxalate in a galvanic cell setup. The cell consists of one half-cell with 0.11M ferrous nitrate (Fe(NO3)2) and another half-cell with 0.15M oxalic acid (C2H2O4) and 1.0ml of 0.11M ferrous nitrate. The measured voltage between the half-cells is 100mV. The Nernst Equation is utilized to determine the Ksp based on the concentration of ions in the solution.

PREREQUISITES
  • Understanding of galvanic cells and their components
  • Familiarity with the Nernst Equation
  • Knowledge of solubility product constants (Ksp)
  • Basic chemistry concepts regarding ion concentrations
NEXT STEPS
  • Study the Nernst Equation in detail, focusing on its application in electrochemistry
  • Research the solubility product constant (Ksp) and its significance in chemical equilibria
  • Explore the properties and reactions of ferrous oxalate in aqueous solutions
  • Investigate concentration cells and their role in galvanic cell voltage generation
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Chemistry students, electrochemists, and anyone interested in understanding the principles of galvanic cells and solubility product calculations.

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Homework Statement



A student constructs a galvanic cell with one half-cell containing 0.11M ferrous nitrate, Fe(NO3)2, and the other half-cell made from 50.0ml of 0.15M oxalic acid and 1.0ml of the 0.11M ferrousnitrate. Ferrous oxalate is only slightly soluble. The half-cells contain iron electrodes and they are connected with a salt bridge. The student measures a voltage of 100mV between the two half-cells. Calculate the experimental Ksp for ferrous oxalate based on this data.

Homework Equations



Nernst Equation: E = E(standard) - (.05971/n)log Q
Ferrous Nitrate: Fe(NO3)2
Oxalic Acid: C2H2O4

The Attempt at a Solution



I'm not sure what to do because one of the half cells contains multiple chemicals. In addition, I'm not sure what the equation would look like (meaning, I'm having trouble finding Q).
 
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This is basically a concentration cell - the only source of voltage is the difference between concentrations of Fe2+ in both half cells.
 

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