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Chapter 17: Problem 84

You have a concentration cell in which the cathode has a silver electrode with 0.10 \(M\) Ag \(^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \space \mathrm{M}\space \mathrm{S}_{2} \mathrm{O}_{3}^{2-},\) and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-} .\) You read the voltage to be 0.76 \(\mathrm{V}\) a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}\) $$\mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?$$

Short Answer

Expert verified
The concentration of Ag+ at the anode is \(1.51\times10^{-6} M\), and the equilibrium constant K for the formation of \(Ag(S_2O_3)_2^{3-}\) is 8.82.

Step by step solution

01

Write down the Nernst Equation

The Nernst Equation is given by: \(E_{cell}=E_{0}-\frac{RT}{nF}\ln Q\) where: - \(E_{cell}\) is the cell voltage - \(E_{0}\) is the standard cell voltage (since both electrodes are Ag, \(E_{0} = 0\)) - R is the gas constant (\(R = 8.314\space J\space mol^{-1}\space K^{-1}\)) - T is the temperature (assume standard temperature, 298 K) - n is the number of electrons involved in the redox reaction (n = 1) - F is the Faraday's constant (\(F = 9.648\times10^4\space C\space mol^{-1}\)) - Q is the reaction quotient
02

Determine the value of Q

The reaction quotient Q is given by the ratio of concentrations: \(Q = \frac{[Ag+]_{anode}}{[Ag+]_{cathode}}\) Since we know the concentration of Ag+ at the cathode, we have: \(Q = \frac{[Ag+]_{anode}}{0.10}\)
03

Use the Nernst Equation to find Ag+ concentration at the anode

Plug in the known values into the Nernst Equation and solve for [Ag+] at the anode: \(0.76 V = 0-\frac{8.314 J\space mol^{-1}\space K^{-1}\times 298 K}{1\times9.648\times10^4\space C\space mol^{-1}}\times\ln\frac{[Ag+]_{anode}}{0.10}\) Now, isolate [Ag+] and solve: \([Ag+]_{anode} = 0.10\times e^{\frac{0.76\times9.648\times10^4}{8.314\times298}}\) \([Ag+]_{anode} = 1.51\times10^{-6} M\) So the concentration of Ag+ at the anode is \(1.51\times10^{-6} M\).
04

Write the equilibrium expression for the reaction

The equilibrium expression for the reaction is given as: \(K = \frac{[Ag(S_2O_3)_2^{3-}]}{[Ag^+][S_2O_3^{2-}]^2}\)
05

Calculate the value of the equilibrium constant

Plug the known concentration values into the equilibrium expression, and solve for K: \(K = \frac{1.0\times10^{-3}}{(1.51\times10^{-6})(0.050)^2}\) \(K = 8.82\) So the equilibrium constant K for the formation of \(Ag(S_2O_3)_2^{3-}\) is 8.82.

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Most popular questions from this chapter

The ultimate electron acceptor in the respiration process is molecular oxygen. Electron transfer through the respiratory chain takes place through a complex series of oxidationreduction reactions. Some of the electron transport steps use iron-containing proteins called cytochromes. All cytochromes transport electrons by converting the iron in the cytochromes from the +3 to the +2 oxidation state. Consider the following reduction potentials for three different cytochromes used in the transfer process of electrons to oxygen (the potentials have been corrected for \(\mathrm{pH}\) and for temperature): $$\begin{aligned} &\text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{2+}\right)\ &\mathscr{E}^{\circ}=0.385 \mathrm{V}\\\ &\text { cytochrome } \mathbf{b}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{b}\left(\mathrm{Fe}^{2+}\right)\ &\mathscr{E}^{\circ}=0.030 \mathrm{V}\\\ &\text { cytochrome } c\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{c}\left(\mathrm{Fe}^{2+}\right)\ &\mathscr{E}^{\circ}=0.254 \mathrm{V} \end{aligned}$$ In the electron transfer series, electrons are transferred from one cytochrome to another. Using this information, determine the cytochrome order necessary for spontaneous transport of electrons from one cytochrome to another, which eventually will lead to electron transfer to \(\mathrm{O}_{2}\)

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of 0.10 \(M\) NaOH that is saturated with \(\mathrm{Cu}(\mathrm{OH})_{2},\) what is the cell potential at \(25^{\circ} \mathrm{C} ?\left[\mathrm{For} \mathrm{Cu}(\mathrm{OH})_{2}\right.\) \(K_{\mathrm{sp}}=1.6 \times 10^{-19} \cdot \mathrm{J}\)

Gold is produced electrochemically from an aqueous solution of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) containing an excess of \(\mathrm{CN}^{-} .\) Gold metal and oxygen gas are produced at the electrodes. What amount (moles) of \(\mathbf{O}_{2}\) will be produced during the production of 1.00 mole of gold?

Consider the following galvanic cell: A 15.0 -mole sample of \(\mathrm{NH}_{3}\) is added to the Ag compartment (assume \(1.00 \mathrm{L}\) of total solution after the addition). The silver ion reacts with ammonia to form complex ions as shown: $$\begin{aligned} \mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \\\& K_{1}=2.1 \times 10^{3} \\ \mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \\\\\ & K_{2}=8.2 \times 10^{3} \end{aligned}$$ Calculate the cell potential after the addition of 15.0 moles of \(\mathrm{NH}_{3}\)

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