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Chapter 18: Problem 152

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 $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$.

Short Answer

Expert verified
The standard cell potential is 0.204V and the standard free energy change is -39.38 kJ/mol. The Nernst equation for the cell is \(\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{RT}{2F} \ln \frac{1}{[CrO^{2-}_{4}]}\). The cell potential when \([CrO^{2-}_{4}] = 1.00 * 10^{-5} M\) is 0.505 V. The unknown concentration of \(\text{CrO}^{2-}_{4}\) in the solution is 3.55 * 10^{-6} M. The solubility product \(K_{sp}\) for \(Ag_{2}CrO_{4}\) is 1.79 * 10^{-16}.

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 (ΔG°) 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·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, \(\mathscr{E}_{\text{cell}} = 0.204 V - \frac{(8.314 J/(mol·K)*298K)}{2*96485 C/mol} * \ln \frac{1}{1.00 * 10^{-5} M} = 0.505 V\).
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, \[[\text{CrO}^{2-}_{4}] = e^{\frac{2F(\mathscr{E}^{\circ}_{\text{cell}} - \mathscr{E}_{\text{cell}})}{RT}} = e^{\frac{2*96485 C/mol*(0.204 V-0.504 V)}{8.314 J/(mol∙K)*298 K}} = 3.55 * 10^{-6} M\].
06

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

For the reaction \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s) \longrightleftharpoons с\mathrm{CrO}^{2-}_{4}(aq)+2\mathrm{Ag}^+(aq)\), 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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