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. Based on the given reduction potentials, which of the following would lead to a reaction? (A) Placing some \(\operatorname{Cr}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions (B) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (C) Placing some \(\mathrm{Cr}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (D) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions

Step by step solution

01

Understand Reduction Potentials

Here we are given the standard reduction potentials of two half-cell reactions. A standard reduction potential indicates the tendency of a species to be reduced. A species with a higher (less negative) reduction potential will tend to gain electrons and be reduced, while the species with a lower (more negative) reduction potential will lose electrons and be oxidized.

02

Identify the Half-Cell Reactions

We have two half-cell reactions: \(\mathrm{Cr}^{3+} + 3e \rightarrow \mathrm{Cr}(s)\) with a reduction potential of -0.41 V and \(\mathrm{Pb}^{2+} + 2e \rightarrow \mathrm{Pb}(s)\) with a reduction potential of -0.12 V. Given the standard reduction potential values, \(\mathrm{Pb}^{2+}\) has a higher tendency to gain electrons than \(\mathrm{Cr}^{3+}\).

03

Analyze the Reaction Scenarios

Looking at the scenarios A to D, we need to find out where a reaction would occur. This means that the solid metal would dissolve into the solution and its corresponding cation would be reduced. Given the reduction potentials, the reaction can only occur if the solid metal has a lower reduction potential (more negative) than the cation in the solution.

04

Determine the Possible Reaction

We consider the proposed scenarios. In scenario (A), \(\mathrm{Cr}(s)\) (reduction potential -0.41V) is placed in a solution of \(\mathrm{Pb}^{2+}\) (reduction potential -0.12V). Since \(\mathrm{Cr}\) has a lower reduction potential than \(\mathrm{Pb}^{2+}\), this scenario is possible and a reaction would occur. Scenario (B) putting \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) won't cause a reaction because \(\mathrm{Pb}^{2+}\) has a higher reduction potential. Scenarios (C) and (D) do not make sense either because they involve placing a metal in a solution with its own ions, which cannot lead to a reaction.

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