Consider the standard galvanic cell based on the following half-reactions: $$\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}$$ $$\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}$$ The electrodes in this cell are \(A g(s)\) and \(C u(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\operatorname{CuSO}_{4}(s)\) 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: \(\mathrm{Cu}^{2+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) . ]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: Ag' reacts with Cl- to form AgCl(s). ] 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. $$\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V}$$

Step by step solution

01

(1) Calculate the standard cell potential

To begin, let's first establish the standard cell potential. Since this is galvanic cell with Cu²⁺/Cu and Ag⁺/Ag as electrodes, we can calculate the standard cell potential through: \(E_{cell}^o = E_{reduction, Ag}^o - E_{reduction, Cu}^o\). Given values for \(E_{reduction, Ag}^o = 0.80 V\), and \(E_{reduction, Cu}^o = 0.34 V\). Thus, \(E_{cell}^o = 0.80 - 0.34 = 0.46 V\). Now, let's analyze the effects of the changes in parts a through e. #a. Adding CuSO4 to Cu half-cell compartment#

02

(2a) Effect of CuSO4 on Cu²⁺ concentration

When we add CuSO4 in the copper half-cell compartment, it dissociates into Cu²⁺ and SO₄²⁻ ions increasing the concentration of Cu²⁺ ions. This would create more driving force for the reduction of Cu²⁺ ions.

03

(2b) Calculate the effect on potential

The increase in Cu²⁺ concentration shifts the Cu²⁺/Cu half-reaction according to the Nernst equation: \(E = E^o - \frac{RT}{nF} \ln{Q}\) Since the standard cell potential will remain the same, E_{reduction, Ag}^o, and \(E_{reduction, Cu}^o\) will also remain the same. As the concentration of Cu²⁺ increases, Q will decrease making the denominator in the equation more negative, thus increasing the cell potential. So, the cell potential will increase in this case. #b. Adding NH3 to Cu half-cell compartment#

04

(3a) Effect of NH3 on Cu²⁺ concentration

Cu²⁺ reacts with NH₃ to form Cu(NH₃)₄²⁺, which effectively reduces the concentration of Cu²⁺ in the solution by removing them to form the complex ion.

05

(3b) Calculate the effect on potential

The reduction in Cu²⁺ concentration as a consequence of the formation of Cu(NH₃)₄²⁺ will decrease Q value leading to a decrease in cell potential. #c. Adding NaCl to the silver half-cell compartment#

06

(4a) Effect of NaCl on Ag⁺ concentration

NaCl dissociates into Na⁺ and Cl⁻ ions; Cl⁻ ions would slowly remove Ag⁺ concentration by formation of silver chloride AgCl precipitate.

07

(4b) Calculate the effect on potential

The decrease in Ag⁺ concentration raises the log term in the Nernst equation, which will reduce the cell potential. #d. Doubling the volume of both half-cell compartments#

08

(5) Effect of dilution on the cell potential

Diluting both compartments by doubling the volume causes an equal decrease in concentrations of Cu²⁺ and Ag⁺ ions. Since the change is equal to both half-cell compartments, there will be no effect on the standard cell potential. #e. Replacing the silver electrode with a platinum electrode#

09

(6) Effect of replacing the electrode on potential

Changing the electrode from silver to platinum does not affect the half-reactions occurring in the cell. The platinum electrode does not change the concentrations of the species involved in the half-reactions and thus does not affect the standard cell potential.

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