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Chapter 20: Problem 46

A voltaic cell, with \(E_{\text {cell }}=0.180 \mathrm{V},\) is constructed as follows: $$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}\left(\operatorname{satd} \mathrm{Ag}_{3} \mathrm{PO}_{4}\right) \| \mathrm{Ag}^{+}(0.140 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})$$ What is the \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{3} \mathrm{PO}_{4} ?\)

Short Answer

Expert verified
To solve this problem, first use the Nernst equation to determine the concentration of Ag+ in the solution. Then use this concentration to calculate the solubility product constant (\(K_{\mathrm{sp}}\)) of \(\mathrm{Ag}_3 \mathrm{PO}_4\).

Step by step solution

01

Identify the relevant equations

The Nernst equation is given by: \[E_{\text {cell }}=E^{\circ}-\left(\frac{0.0592}{n}\right) \log Q\]The value of \(Q\) for the reaction is \([Ag^{+}]\). Since there is no information about the standard electrode potential \(E^{\circ}\) the equation simplifies to:\[E_{\text {cell }}=-\left(\frac{0.0592}{n}\right) \log Q\]or \[E_{\text {cell }}=\left(\frac{-0.0592}{n}\right) \log \left(\frac{1}{[Ag^{+}]}\right)\]
02

Calculate the concentration of Ag+

In the voltaic cell, the reduction potential is given as 0.180 V. Substituting this into the Nernst equation, we get:\[0.180 = \left(\frac{-0.0592}{n}\right) \log \left(\frac{1}{[Ag^{+}]}\right)\]Assuming a single-electron process (n=1), we can solve for [Ag+]. This leads to:\[\log \left(\frac{1}{[Ag^{+}]}\right) = \frac{0.180}{-0.0592}\]Calculate [Ag+] to get:\[[Ag^+] = 0.140 \, M\]
03

Calculate the Ksp

The equilibrium constant for the solubility of \(\mathrm{Ag}_3 \mathrm{PO}_4\) is:\[K_{\mathrm{sp}} = [Ag^+]^3[PO_4^{3-}]\]For a saturated solution of \(\mathrm{Ag}_3 \mathrm{PO}_4\), [Ag+] is three times the [PO4^3-]. This leads to \[K_{\mathrm{sp}} = [Ag^+]^3 \times \left(\frac{[Ag^+]}{3}\right)\]Substitute [Ag+] with the concentration calculated from Step 2 and solve for \(K_{\mathrm{sp}}\).

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Most popular questions from this chapter

You prepare \(1.00 \mathrm{L}\) of a buffer solution that is \(1.00 \mathrm{M}\) \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) and \(1.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4} .\) The solution is divided in half between the two compartments of an electrolytic cell. Both electrodes used are Pt. Assume that the only electrolysis is that of water. If 1.25 A of current is passed for 212 min, what will be the \(\mathrm{pH}\) in each cell compartment at the end of the electrolysis?

In each of the following examples, sketch a voltaic cell that uses the given reaction. Label the anode and cathode; indicate the direction of electron flow; write a balanced equation for the cell reaction; and calculate \(E_{\mathrm{cell}}^{\circ}\). (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})\) (b) \(\mathrm{Pb}^{2+}(\mathrm{aq})\) is displaced from solution by \(\mathrm{Al}(\mathrm{s})\) (c) \(\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{aq})\) (d) \(\mathrm{Zn}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{Zn}^{2+}+\) \(\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}(\mathrm{g})\)

For the reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}(\mathrm{g}),\) describe the voltaic cell in which it occurs, label the anode and cathode,use a table of standard electrode potentials to evaluate \(E_{\text {cell }}^{\circ},\) and balance the equation for the cell reaction.

Refer to standard reduction potentials, and predict which metal in each of the following pairs is the stronger reducing agent: (a) sodium or potassium (b) magnesium or barium

Given that \(E_{\text {cell }}^{\circ}=3.20 \mathrm{V}\) for the reaction $$2 \mathrm{Na}(\mathrm{in} \mathrm{Hg})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})$$ What is \(E^{\circ}\) for the reduction \(2 \mathrm{Na}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow\) \(2 \mathrm{Na}(\text { in } \mathrm{Hg}) ?\)
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