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Chapter 18: Problem 75

From the following information, calculate the solubility product of AgBr: \(\mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \operatorname{Ag}(s) \quad E^{\circ}=0.80 \mathrm{~V}\) \(\operatorname{AgBr}(s)+e^{-} \longrightarrow \operatorname{Ag}(s)+\operatorname{Br}^{-}(a q) \quad E^{\circ}=0.07 \mathrm{~V}\)

Short Answer

Expert verified
The solubility product of AgBr is approximately \(2.14 × 10^{12}\).

Step by step solution

01

Find the full cell reaction

To find the overall reaction, add the half-reactions together. The silver ion (Ag+) is on the product side in the first reaction and the reactant side in the second reaction. Thus Ag+ can be cancelled out to obtain the full cell reaction: \[\operatorname{AgBr}(s) \longrightarrow \operatorname{Ag}(s) + \operatorname{Br}^{-}(a q)\] The balanced reaction above leads to no influx of electrons, indicating it is a balanced reaction. The overall reaction represents the dissociation of AgBr solid into its ions.
02

Determining the cell potential (E°)

The overall cell potential (E°) can be calculated by subtracting the second E° value from the first. That is, \(E° = E°_1 - E°_2 = 0.80V - 0.07V = 0.73V\) where \(E°_1\) and \(E°_2\) are given in the problem statement.
03

Calculation of the equilibrium constant (Ksp)

According to the Nernst Equation, the relationship between the cell potential and the equilibrium constant is: \(E° = \frac{0.0592}{n} \log Ksp \) Where n is the number of electrons transferred in the reaction, which is 1 in this case. Rearranging for \( Ksp \): \(Ksp = 10^{n*(E°/0.0592)} = 10^{(1*0.73V)/0.0592} = 10^{12.33} \approx 2.14 * 10^{12}\)
04

Calculating the solubility product

In this case, the solubility product (Ksp) of AgBr is equivalent to the equilibrium constant due to the stoichiometric ratio in the balanced equation. Therefore, the solubility product of AgBr is \(2.14 × 10^{12}\).

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Most popular questions from this chapter

Describe an experiment that would enable you to determine which is the cathode and which is the anode in a galvanic cell using copper and zinc electrodes.

A \(9.00 \times 10^{2}-\mathrm{mL} 0.200 \mathrm{M} \mathrm{MgI}_{2}\) was electrolyzed. As a result, hydrogen gas was generated at the cathode and iodine was formed at the anode. The volume of hydrogen collected at \(26^{\circ} \mathrm{C}\) and \(779 \mathrm{mmHg}\) was \(1.22 \times 10^{3} \mathrm{~mL}\). (a) Calculate the charge in coulombs consumed in the process. (b) How long (in min) did the electrolysis last if a current of \(7.55 \mathrm{~A}\) was used? (c) A white precipitate was formed in the process. What was it and what was its mass in grams? Assume the volume of the solution was constant.

The \(E^{\circ}\) value of one cell reaction is positive and that of another cell reaction is negative. Which cell reaction will proceed toward the formation of more products at equilibrium?

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