Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following best describes the activity in the salt bridge as the reaction progresses? (A) Electrons flow through the salt bridge from the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell. (B) \(\quad \mathrm{Pb}^{2+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell. (C) \(\mathrm{Na}^{+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half- cell. (D) \(\quad \mathrm{Na}^{+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2}\) half- cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half- cell.

Step by step solution

01

- Determining the Anode and Cathode

First, identify the anode and cathode. The metal with the more negative \(E^{\circ}\) undergoes oxidation and becomes the anode, and the metal with the less negative \(E^{\circ}\) undergoes reduction and becomes the cathode. So, here Cr/Cr3+ is the anode and Pb/Pb2+ is the cathode.

02

- Understanding the Role of the Salt Bridge

The salt bridge allows ions to flow and maintains electrical neutrality. In this case, the salt bridge likely contains \(NaCl\), which dissociates into \(Na^{+}\) and \(Cl^{-}\) ions.

03

- Determining the Direction of Ions

Anode is where oxidation is happening, so Cr will be losing electrons and the solution is gaining Cr3+ ions. Hence the salt bridge will provide \(Cl^{-}\) ions to the anode half-cell. If we look at the cathode where reduction is happening, Pb2+ ions are being reduced to Pb, thus the solution will be losing Pb2+ ions. To balance this, the salt bridge will provide \(Na^{+}\) ions to the cathode half cell.

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