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What is the solubility product of saturated solution of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) in water at \(298 \mathrm{~K}\) if the EMF of the cell: Ag | Ag' (satd. \(\left.\mathrm{Ag}_{2} \mathrm{CrO}_{4}\right) \| \mathrm{Ag}^{+}(0.1 \mathrm{M}) \mid \mathrm{Ag}\) is \(0.162\) \(\mathrm{V}\) at \(298 \mathrm{~K} ?[2.303 R T / F=0.06, \log 2=0.3]\) (a) \(2.0 \times 10^{-4}\) (b) \(3.2 \times 10^{-11}\)

Short Answer

Expert verified
The solubility product, Ksp, of Ag2CrO4 at 298 K is approximately 3.2 x 10^-11.

Step by step solution

01

Write the Nernst Equation

The cell potential, E, can be related to the concentration of the ions using the Nernst equation. For the given Ag/Ag+ half-cell, the Nernst equation is: E = E^0 - (0.06/n) * log([Ag^+])
02

Calculate the Standard Cell Potential, E^0

The standard cell potential, E^0, is the potential of the cell under standard conditions. Since both electrodes are silver, E^0 for this cell is 0 because it is a concentration cell.
03

Calculate the Concentration of Silver Ions, [Ag+]

Using the EMF value of the cell, calculate the concentration of the silver ions using the Nernst equation: 0.162 = 0 - (0.06/1) * log([Ag+]). Since the concentration of Ag+ on one side is given (0.1 M), we need to find the concentration of Ag+ in saturated Ag2CrO4 solution.
04

Calculate the Solubility Product, Ksp

The solubility product, Ksp, for Ag2CrO4 can be calculated knowing that the dissolution of Ag2CrO4 in water produces 2Ag+ and CrO4^2-. Ksp = [Ag+]^2 * [CrO4^2-]. Using the concentration of Ag+ from the previous step, substitute into the expression for Ksp to find its value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nernst Equation
Understanding the Nernst equation is like getting a master key to unlocking the door of electrochemistry. It helps us predict the behavior of cells under non-standard conditions. The Nernst equation is a fundamental relation that bridges cell potential with the concentrations of the chemical species involved in the reaction.

In its general form, it expresses the potential of an electrochemical cell as a function of the standard cell potential (\(E^0\)) and the activities (approximately, concentrations) of the reactants and products. The equation is: \[ E = E^0 - \frac{RT}{nF} \ln(Q) \] where \(E\) is the cell potential, \(R\) is the universal gas constant, \(T\) is the temperature in kelvins, \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient, which is the ratio of the products' concentrations to reactants' concentrations, each raised to the power of their stoichiometric coefficients. In our context, \(E = 0.162 V \), and by converting the natural logarithm to log base 10 using the provided relationship \(2.303 RT/F = 0.06 V\), the Nernst equation simplifies to a more manageable form that can be used directly for calculations.
Concentration Cell
A concentration cell is another fascinating concept you will encounter in the world of electrochemistry. Imagine two half-cells that are composed of the same materials but submerged in solutions of their ions at different concentrations. That's essentially what a concentration cell is. The driving force for electrical potential in this type of cell is the concentration gradient between the two half-cells.

The key thing to note for a concentration cell is that the standard cell potential (\(E^0\)) is zero because the electrode materials are identical under standard conditions. However, as we have different ion concentrations in the two half-cells, there is an electric potential difference due to this imbalance. The Nernst equation becomes particularly handy here to calculate this potential difference, as seen in our exercise.

Relevance to the Exercise

In the given problem, the presence of an EMF (\(0.162 V\)) signifies that there are different concentrations of \(Ag^+\) ions across the cell. The concentration cell concept helps us recognize that the electrical potential arises due to this concentration difference, even when the electrodes are made of the same material.
Standard Cell Potential
The term standard cell potential (\(E^0\)), a cornerstone of electrochemistry, refers to the electrical potential of a cell when all components are at their standard states, typically 1 molar concentration and 1 atmospheric pressure for gases, at a specified temperature, often 25°C or 298 K. It's a measure of how much voltage a cell could produce under these idealized conditions.

In a concentration cell, \(E^0\) is curiously zero, as we have seen in the exercise solution, because the electrodes are the same material, canceling out any potential difference if they were at equal concentrations. This zeros out any inherent 'preference' for electron movement, and any observed potential comes purely from the concentration differences of the ions involved.

Contextual Insight

With the exercise provided, recognizing that \(E^0\) equals zero for a concentration cell is a crucial step in applying the Nernst equation to find the unknown ion concentrations, which then allows us to decipher the solubility product for \(Ag_2CrO_4\).
Physical Chemistry Competitive Examinations
For students gearing up for competitive examinations, having a strong grip on concepts like the Nernst equation, concentration cells, and standard cell potentials is essential. Physical chemistry often acts as a critical segment in these exams, demanding a clear understanding and the ability to apply these principles to solve complex problems.

When you delve into physical chemistry, you will frequently navigate through calculations involving the interplay of various electrochemical concepts, just like our problem involving the solubility product of \(Ag_2CrO_4\). Being comfortable with these concepts can make a substantial difference in your performance in these competitive settings. It is advisable to practice these kinds of problems, understand the theoretical underpinnings, and learn the shortcuts and approximations that can save time during the exam.

Why Are These Concepts Core?

They form the fundamental building blocks for understanding the behaviors of chemical systems, predicting the feasibility of reactions, and harnessing the power of electrochemistry for practical applications. Students well-versed in these areas stand a better chance of excelling in examinations and further, in real-world applications of chemistry.

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Most popular questions from this chapter

The standard reduction potentials in acidic conditions are \(0.77 \mathrm{~V}\) and \(0.53 \mathrm{~V}\), respectively, for \(\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}\) and \(\mathrm{I}_{3}^{-} \mid \mathrm{I}^{-}\) couples. The equilibrium constant for the reaction: \(2 \mathrm{Fe}^{3+}+3 \mathrm{I}^{-} \rightleftharpoons 2 \mathrm{Fe}^{2+}+\mathrm{I}_{3}^{-}\), is \((2.303 R T / F=0.06)\) (a) \(2 \times 10^{8}\) (b) \(10^{8}\) (c) \(10^{4}\) (d) \(10^{-8}\)

Electrolysis of an acetate solution produces ethane according to the reaction: $$ 2 \mathrm{CH}_{3} \mathrm{COO}^{-} \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})+2 \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} $$ What total volume of ethane and \(\mathrm{CO}_{2}\) would be produced at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), if a current of \(0.5 \mathrm{~A}\) is passed through the solution for \(482.5\) min? Assume current efficiency \(80 \%\). (a) \(1.344 \mathrm{~L}\) (b) \(2.688 \mathrm{~L}\) (c) \(4.032 \mathrm{~L}\) (d) \(1.792 \mathrm{~L}\)

A certain current liberated \(0.50 \mathrm{~g}\) of hydrogen in \(2 \mathrm{~h}\). How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution? \((\mathrm{Cu}=63.5)\) (a) \(12.7 \mathrm{~g}\) (b) \(15.88 \mathrm{~g}\) (c) \(31.75 \mathrm{~g}\) (d) \(63.5 \mathrm{~g}\)

Which one of the following does not get oxidized by bromine water? (a) \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\) (b) \(\mathrm{Cu}^{+}\) to \(\mathrm{Cu}^{2+}\) (c) \(\mathrm{Mn}^{2+}\) to \(\mathrm{MnO}_{4}^{-}\) (d) \(\mathrm{Sn}^{2+}\) to \(\mathrm{Sn}^{4+}\)

For the cell \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\), if the concentration of both, \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cu}^{2+}\) ions are doubled, the EMF of the cell (a) doubles (b) reduces to half (c) remains same (d) becomes zero

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