Question
Pacemakers are electronic devices that help regulate the heart rate. Currently, lithium-iodine cells are commonly used to power pacemakers and have replaced zinc-mercury cells. Table 1 provides the operating cell potential, E , for each cell. Table 2 provides the standard reduction potentials for several half-reactions related to zinc-mercury and zinc-air cells.
The use of zinc-mercury cells in hearing aids has been replaced by zinc-air cells that operate using the oxidation of Zn by \(O_2\) from the air, generating a potential of +1.60V . Table 2 provides the standard reduction potentials for the half-reactions used in zinc-mercury and zinc-air cells. Which of the following best explains the modification to the cell design that is mostly responsible for the difference in standard cell potentials for zinc-mercury and zinc-air cells?
A The greater standard cell potential of the Zn-air cell compared to that of the zinc-mercury cell most likely results from the thermodynamically more favorable reduction of \(O_2\) compared to \(HgO\).
B The greater standard cell potential of the Zn-air cell compared to that of the zinc-mercury cell most likely results from the greater number of moles of \(e^−\) required to reduce \(O_2\) compared to \(HgO\).
C The greater standard cell potential of the Zn-air cell compared to that of the zinc-mercury cell most likely results from the thermodynamically less favorable reduction of \(O_2\) compared to \(HgO\).
D The greater standard cell potential of the Zn-aircell compared to that of the zinc-mercury cell most likely results from the greater number of moles of hydroxide ions required to reduce \([Zn(OH)_4]^{2−}\) compared to \(Zn(OH)_2\).
▶️Answer/Explanation
Ans:A
Based on the information in table 2 and the cell potential given, the reaction that takes place in the zinc-air cell is \(2Zn+4 OH^−+O^2+2 H_2O→2[Zn(OH)_4]^{2−}\). The relatively small difference in the standard reduction potential for Zn(OH)2 compared to that for [Zn(OH)4]2− indicates that the oxidation of zinc under both of these conditions cannot account for the difference in E°cell between the zinc-mercury and zinc-air cells. Therefore, the larger, more positive standard reduction potential for \(O_2\) compared to that for \(HgO\) is mostly responsible for the difference in cell potentials. This is justified by ΔG\(=−nFE\) because a larger, more positive E° results in a larger, more negative ΔG and a reaction that is thermodynamically more favorable.
Question
▶️Answer/Explanation
Ans:C
Replacing the \(Pb-Pb(NO_3)_2\) half-cell with an \(Al-Al(NO_3)_3\) half-cell, not the \(Cu-Cu(NO_3)_2\) half-cell with a \(Ag-AgNO_3\) half-cell, causes the greatest increase in E°cell. The replacement means that the reaction taking place is \(2Al(s)+3Cu_^{2+}(aq)→2Al_3+(aq)+3Cu(s)\), and \(E^{\circ}_{cell}=+2.00V\).
Question
The galvanic cell illustrated above was constructed using a salt bridge containing \(KNO_3\). A second cell is constructed from identical half-cells but uses \(NaNO_3\) for the salt bridge. Which of the following best explains whether the initial potential of the second cell will be different from the initial potential of the first cell?
A The initial potential of the second cell will be lower than the initial potential of the first cell because Na has a lower molar mass than K has.
B The initial potential of the second cell will be higher than the initial potential of the first cell because Na is less electronegative than K is.
C The initial potential of the second cell will be the same as the first cell because the ions from the salt bridge are not oxidized or reduced during cell operation.
D The initial potential of the second cell will be the same as the first cell because the standard reduction potentials of Group 1 metals are very similar.
▶️Answer/Explanation
Ans:C
Based on the movement of ions in the salt bridge shown in the illustration provided for the first cell, the redox reaction that takes place is \(Cu_2^+(aq)+Pb(s)→Cu(s)+Pb_2^+(aq)\) . The ions from the salt bridge are not part of the redox process that generates the cell potential, and as a result, the initial E° for the two cells will be the same.
Question
Standard reduction potentials for the half-reactions associated with the electrochemical cell shown above are given in the table below.
What is the value of E° for the cell?
(A) 0.04 V
(B) 0.84 V
(C) 1.56 V
(D) 2.36 V
▶️Answer/Explanation
Ans:C