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a concentration cell is set up using silver chloride and 1.0m silver nitrate using a 1.0m sodium nitrate salt bridge at 298 k. ag has a standard reduction potential of 0.80v. the voltmeter measured a voltage of 0.29v. what is the ksp of the silver chloride? use 298 k and f

User Willow
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Final answer:

To find the Ksp of AgCl using the measured cell potential, we use the Nernst equation and the fact that [Ag+] equals Ksp when [Cl-] is at 1.0M, leading to the conclusion that Ksp will be equal to the concentration of Ag+ in the saturated solution.

Step-by-step explanation:

To calculate the Ksp of silver chloride using the given cell potential, we can start with Nernst equation:

E = E° - (RT/nF) * lnQ

Where:


  • E = cell potential under non-standard conditions

  • E° = standard reduction potential of the electrodes

  • R = universal gas constant (8.314 J/(mol*K))

  • T = temperature in Kelvin (298 K)

  • n = number of moles of electrons transferred (1 for Ag+/Ag)

  • F = Faraday's constant (96485 C/mol)

  • Q = reaction quotient ([Ag+] in the saturated AgCl solution)

The given cell potential (0.29V) is already adjusted for the Ag/AgCl reference electrode, so E° for Ag/AgCl is not required. Substituting the known values and isolating Q gives us the concentration of Ag+ ions in equilibrium with solid AgCl, which equals the Ksp when [Cl-] is at 1.0M, as is the case with a saturated solution.

After finding [Ag+] (which equals Ksp), we can use the relation:

Ksp = [Ag+][Cl-]

Since the concentration of Cl- is 1.0 M in a saturated solution of AgCl, Ksp will be equal to the concentration of Ag+.

Given the typical solubility of AgCl at 25 °C is 1.67 × 10-5 M, which gives us the Ksp under standard conditions.

User Dasi
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