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A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 111\()\) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) . The \(\mathscr{E}^{\circ}\) value for the following half-reaction is 0.446 \(\mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{E}_{\text { cell } \text { and }} \Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) . b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C} )\) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} M,\) what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is 0.504 \(\mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\).

Step by step solution

01

Calculate the standard cell potential.

The standard cell potential \(\mathscr{E}^{\circ}_{\text{cell}}\) is the difference between the standard potentials of the two half cells. In this case, we can write this equation \(\mathscr{E}^{\circ}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{Ag}_{2}\text{CrO}_{4}/\text{Ag}} - \mathscr{E}^{\circ}_{\text{SCE}}\). However, the SCE is oftentimes used as a reference electrode in which the potential is defined to be exactly 0.242V. Therefore, \(\mathscr{E}^{\circ}_{\text{cell}} = 0.446V - 0.242V = 0.204V\).

02

Calculate the standard free energy change.

There is a relationship between the standard cell potential and the standard free-energy change (螖G掳) given by the equation \(\Delta G^{\circ} = -nF\mathscr{E}^{\circ}_{\text{cell}}\), where F is the Faraday constant (96485 C/mol) and n is the number of moles of electrons transferred in the balanced equation for the cell reaction, which is 2 in this case. So, we can substitute these values into the equation to find \(\Delta G^{\circ} = -2 * 96485 C/mol * 0.204 V = -39383.72 J/mol = -39.38 kJ/mol\).

03

Write the Nernst Equation for the cell.

The Nernst equation describes the relationship between the cell potential and the concentrations of the reactants and products. It is given by: \[\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q\] where R is the universal gas constant (8.314 J/(mol路K)), T is the temperature in Kelvin, n is the number of moles of electrons transferred in the balanced equation for the cell reaction, F is the Faraday constant, and Q is the reaction quotient. For this cell, the Nernst Equation is \(\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{RT}{2F} \ln \frac{1}{[CrO^{2-}_{4}]}\).

04

Calculate the cell potential for different \([CrO^{2-}_{4}]\).

Now we can use the Nernst equation from step 3 to find the cell potential when \([CrO^{2-}_{4}] = 1.00 * 10^{-5} M\). First, convert the temperature to Kelvin (T=298K). Then substitute all the values into the equation, \(\mathscr{E}_{\text{cell}} = 0.204 V - \frac{(8.314 J/(mol路K)*298K)}{2*96485 C/mol} * \ln \frac{1}{1.00 * 10^{-5} M} = 0.505 V\).

05

Calculate the unknown [CrO^{2-}_{4}].

We can rearrange the Nernst equation to calculate the unknown [\(\text{CrO}^{2-}_{4}\)]. It can be written as, \[[\text{CrO}^{2-}_{4}] = e^{\frac{2F(\mathscr{E}^{\circ}_{\text{cell}} - \mathscr{E}_{\text{cell}})}{RT}} = e^{\frac{2*96485 C/mol*(0.204 V-0.504 V)}{8.314 J/(mol鈭橩)*298 K}} = 3.55 * 10^{-6} M\].

06

Calculate the solubility product \(K_{sp}\).

For the reaction \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s) \longrightleftharpoons 褋\mathrm{CrO}^{2-}_{4}(aq)+2\mathrm{Ag}^+(aq)\), the solubility product \(K_{sp}\) is given by the equation \(K_{sp} = [\mathrm{Ag}^+]^2 [\mathrm{CrO}^{2-}_{4}]\). And since the concentration of \(\mathrm{Ag}^+\) can also be calculated using \( [\mathrm{CrO}^{2-}_{4}] \), we can substitute this into the above equation to find \(K_{sp} = [\mathrm{CrO}^{2-}_{4}]*[\mathrm{Ag}^+]^2 = (3.55 * 10^{-6} M) * (7.10 * 10^{-6} M)^2 = 1.79 * 10^{-16}\).

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Key Concepts

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

Nernst Equation

The Nernst Equation is fundamental in electrochemistry. It relates the cell potential of an electrochemical cell to the concentrations of reactants and products involved. This equation allows chemists to determine the cell potential under non-standard conditions.
The basic format of the Nernst Equation is:

• \[ \mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q \]

In this equation:

• \( \mathscr{E} \) is the cell potential at non-standard conditions.
• \( \mathscr{E}^{\circ} \) is the standard cell potential.
• \( R \) is the universal gas constant, which is \( 8.314 \text{ J/(mol路K)} \).
• \( T \) is the temperature in Kelvin.
• \( n \) is the number of moles of electrons transferred in the cell reaction.
• \( F \) is the Faraday constant, \( 96485 \text{ C/mol} \).
• \( Q \) is the reaction quotient, which is calculated from the concentrations of the products over reactants raised to the power of their stoichiometric coefficients.

Thus, the Nernst Equation helps in predicting how the cell potential changes with varying ion concentrations in the reaction.

Cell Potential

Cell potential, or electromotive force (emf), is a measure of the voltage across an electrochemical cell. This value indicates the energy per charge available from the chemical reaction taking place in the cell.
To determine cell potential, it is important to understand that each half-cell in the electrochemical reaction contributes to the overall potential.

• The standard cell potential (\( \mathscr{E}^{\circ}_{\text{cell}} \)) is calculated using the standard potentials of the individual half reactions.
• For a standard cell potential, conditions are 1 atm, 1 M concentration for all reactants and products, and a temperature of 25掳C (298 K).

The cell potential is found using:

• \( \mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q \)

Here, \( Q \) stands for the reaction quotient, taking into account the actual concentrations, which allows calculating non-standard cell potentials. By assessing the change in \( \mathscr{E}_{\text{cell}} \), chemists can infer the spontaneity and feasibility of electrochemical reactions.

Solubility Product

The solubility product constant, denoted as \( K_{sp} \), helps assess the solubility of slightly soluble ionic compounds. For a dissolution equilibrium, this value reflects the maximum concentration of ions in solution.
In any precipitation or dissolution reaction, you're dealing with equilibrium:

• The reaction for \( \text{Ag}_2\text{CrO}_4 \) can be written as: \[ \text{Ag}_2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq) \]
• \( K_{sp} \) is expressed as \( [\text{Ag}^+]^2[\text{CrO}_4^{2-}] \).

Calculating \( K_{sp} \), once ionic concentrations are known, tells how much solute can dissolve before a solution becomes saturated:

• The smaller the \( K_{sp} \), the less soluble the compound.
• \( K_{sp} \) will aid in predicting precipitation in mixed solutions.

This constant is valuable in electrochemistry for analyzing precipitation reactions and ensuring the equations for \( \Delta G^{\circ} \) and cell potentials remain true under real experimental conditions.

Standard Free Energy Change

The standard free energy change, denoted \( \Delta G^{\circ} \), is a thermodynamic quantity that indicates the amount of energy available to do work from a chemical reaction under standard conditions.
The relationship between \( \Delta G^{\circ} \) and the standard cell potential is given by:

• \[ \Delta G^{\circ} = -nF\mathscr{E}^{\circ}_{\text{cell}} \]

Here:

• \( n \) is the number of moles of electrons transferred during the reaction.
• \( F \) is Faraday's constant, \( 96485 \text{ C/mol} \).

The significance of \( \Delta G^{\circ} \) is:

• Negative \( \Delta G^{\circ} \) indicates a spontaneous reaction.
• Positive \( \Delta G^{\circ} \) means the reaction is non-spontaneous under standard conditions.
• Zero \( \Delta G^{\circ} \) represents a system in equilibrium.

By analyzing \( \Delta G^{\circ} \), you understand the driving force behind a reaction and its capability to perform work.