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Chapter 17: Problem 117

Consider the standard galvanic cell based on the following half-reactions: $$\begin{array}{c} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{array}$$ The electrodes in this cell are \(\mathrm{Ag}(s)\) and \(\mathrm{Cu}(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\mathrm{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\left.\mathrm{Cu}^{2+} \text { reacts with } \mathrm{NH}_{3} \text { to form } \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: \(\left.\mathrm{Ag}^{+} \text {reacts with } \mathrm{Cl}^{-} \text {to form } \mathrm{AgCl}(s) .\right]\) d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$ \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V} $$

Short Answer

Expert verified
In summary: a) When CuSO4(s) is added to the copper half-cell, the cell potential increases. b) When NH3(aq) is added to the copper half-cell, the cell potential decreases. c) When NaCl(s) is added to the silver half-cell, the cell potential increases. d) When water is added to both half-cell compartments, the cell potential remains the same. e) When the silver electrode is replaced with a platinum electrode, the cell potential remains the same.

Step by step solution

01

Write the overall cell reaction

First, we will write the overall cell reaction by combining the copper and silver half-reactions: \[ Cu^{2+}(aq) + 2Ag^+(aq) \longrightarrow Cu(s) + 2Ag(s) \] Now, we can analyze each change in the exercise and determine its effect on the cell potential.
02

a. Adding CuSO4(s) to the copper half-cell

When CuSO4 is added to the copper half-cell compartment, it will dissolve and increase the concentration of Cu²⁺ ions. According to Le Chatelier's principle, the reaction will shift in the direction that reduces the stress caused by this change, namely, towards the reduction of Cu²⁺ ions. Therefore, the cell potential will increase.
03

b. Adding NH3(aq) to the copper half-cell

When NH3 is added to the copper half-cell, it will react with Cu²⁺ ions to form Cu(NH3)₄²⁺ complex ions. This reaction will decrease the concentration of free Cu²⁺ ions in the solution. According to Le Chatelier's principle, the reaction will shift to restore the concentration of Cu²⁺ ions, favoring the oxidation of Cu(s). Therefore, the cell potential will decrease.
04

c. Adding NaCl(s) to the silver half-cell

When NaCl(s) is added to the silver half-cell compartment, it will dissolve and increase the concentration of Cl⁻ ions. Cl⁻ ions react with Ag⁺ ions to form AgCl(s), decreasing the concentration of free Ag⁺ ions in the solution. According to Le Chatelier's principle, the reaction will shift to restore the concentration of Ag⁺ ions, favoring the reduction of Ag⁺ ions. Therefore, the cell potential will increase.
05

d. Doubling the volume of the solution in both half-cells

When water is added to both half-cell compartments, diluting the solution, the concentrations of Cu²⁺ and Ag⁺ ions will decrease. According to the Nernst equation, the cell potential depends on the concentrations of the ions involved in the reaction. However, since both half-cells are diluted by the same factor, the ratio of the concentrations of Cu²⁺ and Ag⁺ ions will remain the same, and the cell potential will remain the same.
06

e. Replacing the silver electrode with a platinum electrode

Changing the electrode material does not affect the overall cell reaction or the redox processes taking place at the electrodes, as long as the new electrode (in this case, platinum) is a chemically inert conductor. In this situation, the platinum electrode will only serve as an electron conductor and will not participate in the redox reactions directly. Therefore, the cell potential will remain the same. In summary: a) Cell potential increases b) Cell potential decreases c) Cell potential increases d) Cell potential remains the same e) Cell potential remains the same

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Most popular questions from this chapter

Which of the following statements concerning corrosion is(are) true? For the false statements, correct them. a. Corrosion is an example of an electrolytic process. b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen. c. Steel rusts more easily in the dry (arid) Southwest states than in the humid Midwest states. d. Salting roads in the winter has the added benefit of hindering the corrosion of steel. e. The key to cathodic protection is to connect via a wire a metal more easily oxidized than iron to the steel surface to be protected.

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of 4.00 A. How long will it take to plate out \(99 \%\) of the platinum in 0.50 L of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+} ?\)

What volume of \(\mathrm{F}_{2}\) gas, at \(25^{\circ} \mathrm{C}\) and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for \(2.00 \mathrm{h} ?\) What mass of potassium metal is produced? At which electrode does each reaction occur?

Consider the following electrochemical cell: a. If silver metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. b. If copper metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. c. If the above cell is a galvanic cell, determine the standard cell potential. d. If the above cell is an electrolytic cell, determine the minimum external potential that must be applied to cause the reaction to occur.

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.
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