Problem 29 - General Chemistry: Principles and Modern Applications

Question

For the reaction \(2 \mathrm{Cu}^{+}(\mathrm{aq})+\mathrm{Sn}^{4+}(\mathrm{aq}) \longrightarrow\) \(2 \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq}), E_{\mathrm{cell}}^{\circ}=-0.0050 \mathrm{V}\) (a) can a solution be prepared at \(298 \mathrm{K}\) that is \(0.500 \mathrm{M}\) in each of the four ions? (b) If not, in which direction will a reaction occur?

Step-by-Step Solution

Answer

No, a solution with 0.500 M concentration of each ion cannot be prepared, as the calculated negative cell potential suggests the reaction will not occur in the given direction. The reverse reaction will take place instead.

Step 1: Identify the redox processes

In the equation, \(2 \mathrm{Cu}^{+}(\mathrm{aq})+\mathrm{Sn}^{4+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})\), copper ions are being oxidized (losing electrons) and tin ions are being reduced (gaining electrons).

Step 2: Use the Nernst equation to calculate the cell potential under given conditions

The Nernst equation, which relates the cell potential under non-standard conditions to the standard cell potential, is given by: \(E = E_{\mathrm{cell}}^\circ - \frac{0.0592}{n} \log Q\), where Q is the reaction quotient. \nFor the given reaction, \(Q\) can be calculated with the equation: \(Q=[Cu^{2+}]^2[Sn^{2+}]/[Cu^+]^2[Sn^{4+}]\). Given that concentrations of all ions are 0.500 M, \(Q=1\). Therefore, \(E = E_{\mathrm{cell}}^\circ - \frac{0.0592}{2} \log 1 = E_{\mathrm{cell}}^\circ = -0.0050\;V\).

Step 3: Analyze the calculated E to determine the direction of reaction

Since \(E\) is negative, the reaction is not spontaneous in the forward direction under these conditions. Therefore, the reverse reaction will occur, which means it's impossible to prepare a solution that is 0.500 M in each of the four ions. Copper ions Cu2+ will be reduced to Cu+, and Tin ions Sn2+ will be oxidized to Sn4+.

Key Concepts

Nernst Equation

The Nernst equation is a fundamental equation in electrochemistry that enables us to calculate the cell potential of an electrochemical cell under non-standard conditions. The formula is expressed as:
\begin{align*}E = E_{\mathrm{cell}}^\circ - \frac{0.0592}{n} \log Qt{align*}Here, \(E\) represents the cell potential under non-standard conditions, \(E_{\mathrm{cell}}^\circ\) is the standard cell potential, \(n\) is the number of moles of electrons transferred in the redox reaction, and \(Q\) is the reaction quotient, which is a ratio of the concentrations of the products to the concentrations of the reactants, raised to the power of their coefficients in the balanced reaction. When \(Q = 1\), the log term becomes zero, and the cell potential under the given conditions is equal to the standard cell potential. The Nernst equation is invaluable for predicting the behavior of cells in a variety of chemical conditions and plays an essential role in understanding electrochemical processes.

Reaction Spontaneity

Reaction spontaneity in electrochemistry is determined by the sign of the cell potential (\(E\)). A positive cell potential indicates a spontaneous reaction, while a negative potential suggests the reaction is non-spontaneous.
For the given exercise, we have calculated \(E\) to be negative, which means the forward reaction will not occur spontaneously under standard conditions. As a rule of thumb, a spontaneous reaction will have a tendency to proceed without the need for additional energy. This concept is closely linked with the Gibbs free energy change (\(\Delta G\)), where a negative \(\Delta G\) corresponds to a spontaneous process. In electrochemistry, the relationship between \(\Delta G\) and the cell potential is given by \(\Delta G = -nFE\), where \(F\) is the Faraday constant.

Standard Cell Potential

The standard cell potential (\(E_{\mathrm{cell}}^\circ\)) is the potential difference between two half cells under standard conditions, which are 25°C (298 K), 1 atm pressure, and 1 M concentrations for all solutions. This value is crucial as it serves as a benchmark for determining the spontaneity of the reaction.
In the exercise, the given standard cell potential is \(-0.0050 \mathrm{V}\), a negative value implying that the reaction, if left on its own, prefers to proceed in the reverse direction under standard conditions. It's important to note that standard cell potentials are additive; they can be combined to determine the overall cell potential for a given electrochemical reaction, which is what has been done to arrive at the provided standard cell potential. By assessing the standard cell potential, chemists can predict whether a redox reaction will proceed without external energy input.

Oxidation-Reduction Processes

Oxidation-reduction (redox) processes are chemical reactions that involve the transfer of electrons between substances. Oxidation refers to the loss of electrons and reduction refers to the gain of electrons. These processes are essential to the functioning of electrochemical cells.
In our textbook exercise, the redox reaction pairs \(\mathrm{Cu}^{+}\) with \(\mathrm{Cu}^{2+}\) (oxidation) and \(\mathrm{Sn}^{4+}\) with \(\mathrm{Sn}^{2+}\) (reduction). To fully grasp these changes, students must remember that the species undergoing oxidation will provide electrons to the one undergoing reduction. In practical terms, recognizing these changes helps to identify which species acts as the anode or cathode within a cell. The relative strengths of the oxidizing and reducing agents involved are also reflected in their standard reduction potentials, which integrate into the standard cell potential and influence reaction spontaneity.