Chapter 17

Given the following two standard reduction potentials, $$\begin{array}{ll} \mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.10 \mathrm{V} \\ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.50 \mathrm{V} \end{array}$$ solve for the standard reduction potential of the half-reaction $$ \mathbf{M}^{3+}+\mathbf{e}^{-} \longrightarrow \mathbf{M}^{2+} $$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

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 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} .\) The \(8^{\circ}\) 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 \(\mathscr{C}_{\text {cell }}\) 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 \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} \mathrm{M},\) 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 \(17-1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

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 \(\mathrm{NH}_{3}\)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned} &3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\ & \mathscr{E}^{\circ}=0.957 \mathrm{V} \end{aligned}$$ $$\begin{aligned} &\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) &\mathscr{E}^{\circ}=0.775 \mathrm{V} \end{aligned}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

Three electrochemical cells were connected in series so that the same quantity of electrical current passes through all three cells. In the first cell, 1.15 g chromium metal was deposited from a chromium(III) nitrate solution. In the second cell, 3.15 g osmium was deposited from a solution made of \(\mathrm{Os}^{n+}\) and nitrate ions. What is the name of the salt? In the third cell, the electrical charge passed through a solution containing \(\mathrm{X}^{2+}\) ions caused deposition of \(2.11 \mathrm{g}\) metallic \(\mathrm{X}\). What is the electron configuration of X?

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \\ \mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ}=-1.20 \mathrm{V} \end{array}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \mathrm{M},\) and the vanadium compartment contains a vanadium electrode and \(V^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with \(0.0800 M \space \mathrm{H}_{2} \mathrm{EDTA}^{2-},\) resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \space \mathrm{K=?}$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{mL} \space \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{V}\). The solution was buffered at a pH of \(10.00 .\) a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K,\) for the titration reaction. c. Calculate \(\mathscr{E}_{\text {cell }}\) at the halfway point in the titration.

The table below lists the cell potentials for the 10 possible galvanic cells assembled from the metals \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E},\) and their respective \(1.00 \space \mathrm{M} \space 2+\) ions in solution. Using the data in the table, establish a standard reduction potential table similar to Table \(17-1\) in the text. Assign a reduction potential of \(0.00 \mathrm{V}\) to the half-reaction that falls in the middle of the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct. $$\begin{array}{|lcccc|} \hline & \begin{array}{c} \mathrm{A}(s) \text { in } \\ \mathrm{A}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{B}(s) \text { in } \\ \mathrm{B}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{c}(s) \text { in } \\ \mathrm{c}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{D}(s) \text { in } \\ \mathrm{D}^{2+}(a q) \end{array} \\ \hline \mathrm{E}(s) \text { in } \mathrm{E}^{2+}(a q) & 0.28 \mathrm{V} & 0.81 \mathrm{V} & 0.13 \mathrm{V} & 1.00 \mathrm{V} \\ \mathrm{D}(s) \text { in } \mathrm{D}^{2+}(a q) & 0.72 \mathrm{V} & 0.19 \mathrm{V} & 1.13 \mathrm{V} & \- \\ \mathrm{C}(s) \text { in } \mathrm{C}^{2+}(a q) & 0.41 \mathrm{V} & 0.94 \mathrm{V} & \- & \- \\ \mathrm{B}(s) \text { in } \mathrm{B}^{2+}(a q) & 0.53 \mathrm{V} & \- & \- & \- \\ \hline \end{array}$$