Problem 112 - Chemistry The Molecular Nature of Matter

Question

A galvanic cell was constructed with a nickel electrode that was dipped into \(1.20 \mathrm{M} \mathrm{NiSO}_{4}\) solution and a chromium electrode that was immersed into a solution containing \(\mathrm{Cr}^{3+}\) at an unknown concentration. The potential of the cell was measured to be \(0.552 \mathrm{~V}\), with the chromium serving as the anode. The standard cell potential for this system was determined to be \(0.487 \mathrm{~V}\). What was the concentration of \(\mathrm{Cr}^{3+}\) in the solution of unknown concentration?

Step-by-Step Solution

Answer

The concentration of \(\text{Cr}^{3+}\) in the solution is \(\frac{1}{10^{\frac{3(0.552 - 0.487)}{0.0592}}}\) M.

Step 1: Write Down the Half-Reactions

Identify the oxidation and reduction half-reactions that are occurring at the anode and cathode, respectively. For chromium, which is the anode, the reaction is: \(\text{Cr} \rightarrow \text{Cr}^{3+} + 3e^-\). For nickel, which is the cathode, the reaction is: \(\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}\).

Step 2: Use the Nernst Equation

Apply the Nernst equation to relate cell potential with ion concentration. The Nernst equation is: \(E = E^0 - \frac{0.0592}{n}\log\frac{[\text{Red}]}{[\text{Ox}]}\), where \(E\) is the cell potential, \(E^0\) is the standard cell potential, \(n\) is the number of moles of electrons exchanged, \([\text{Red}]\) is the concentration of the reduced form, and \([\text{Ox}]\) is the concentration of the oxidized form.

Step 3: Calculate the Number of Moles of Electrons Transferred

From the half-reactions, determine the number of electrons transferred in the cell reaction. Chromium releases 3 moles of electrons, while nickel accepts 2 moles of electrons. The lowest common multiple of the electron transfers is 6, so the reaction is: \(2\text{Cr} + 3\text{Ni}^{2+} \rightarrow 2\text{Cr}^{3+} + 3\text{Ni}\), where 6 moles of electrons are transferred.

Step 4: Insert Values into the Nernst Equation

Since chromium is the anode, we focus on the oxidation reaction. The Nernst equation for the half-cell reaction at the chromium electrode is: \(0.552 \mathrm{V} = 0.487 \mathrm{V} - \frac{0.0592}{3}\log\frac{1}{[\text{Cr}^{3+}]}\).

Step 5: Solve for the Concentration of \(\text{Cr}^{3+}\)

Rearrange the Nernst equation to solve for \([\text{Cr}^{3+}]\): \(\log\frac{1}{[\text{Cr}^{3+}]} = \frac{3(0.552 - 0.487)}{0.0592}\). Calculate the final concentration of \(\text{Cr}^{3+}\) by taking antilogarithm (inverse of the logarithm) on both sides of the equation.

Step 6: Compute the Antilogarithm

After calculating the right side of the equation in the previous step, take the antilog to find the concentration \([\text{Cr}^{3+}]\) in molarity, as follows: \(\frac{1}{[\text{Cr}^{3+}]} = 10^{\frac{3(0.552 - 0.487)}{0.0592}}\) and then \([\text{Cr}^{3+}] = \frac{1}{10^{\frac{3(0.552 - 0.487)}{0.0592}}}\).

Key Concepts

Nernst Equation

The Nernst Equation is a fundamental formula in electrochemistry that links the electrochemical cell potential to the concentration of the reactants and products involved in the reaction. It offers a deeper understanding of how the potential changes as the reaction proceeds. The equation is given by:
\[\begin{equation}E = E^0 - \frac{0.0592}{n} \log\frac{[\text{Red}]}{[\text{Ox}]}\end{equation}\]
Where:

• \(E\) is the measured cell potential
• \(E^0\) is the standard cell potential
• \(n\) is the number of moles of electrons exchanged in the reaction
• \([\text{Red}]\) is the concentration of the reduced species
• \([\text{Ox}]\) is the concentration of the oxidized species

When applying this to a practical situation, like calculating the concentration of an ionic species in an electrochemical cell, the Nernst equation allows for the precise determination of unknown concentrations, as seen in the textbook exercise.

Electrochemical Cell Potential

Electrochemical cell potential, often symbolized as \(E\), is a measure of the voltage produced by an electrochemical cell, which can consist of two different metals submerged in their respective electrolytes and connected by a salt bridge. This potential is driven by the tendancy for electrons to be transferred from the anode to the cathode through an external circuit.
As seen in the given problem, the standard cell potential \(E^0\) represents the voltage when the concentrations of all the reactants and products are at standard conditions, typically 1 molar concentration and 1 atm of pressure. The cell potential is affected by the concentrations of the species in the cell—something the Nernst equation adjusts for, allowing us to calculate instantaneous cell potentials at any given condition.

Half-Reactions in Electrochemistry

Half-reactions in electrochemistry are a way of breaking down the overall redox reaction into two separate parts - the oxidation part, where electrons are lost, and the reduction part, where electrons are gained. The importance of half-reactions is that they clearly show which species is being oxidized and which is being reduced in an electrochemical cell.
For instance, in the provided exercise, the chromium electrode undergoes oxidation, represented by the half-reaction:\[\begin{equation}\text{Cr} \rightarrow \text{Cr}^{3+} + 3e^- \end{equation}\]While the nickel electrode undergoes reduction, represented by the half-reaction:\[\begin{equation}\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}\end{equation}\]
Understanding these individual reactions is crucial when using the Nernst equation because it informs us of the number of electrons involved, \(n\), which is necessary to calculate the cell potential properly.

Electron Transfer in Redox Reactions

A redox (reduction-oxidation) reaction involves the transfer of electrons between two species. In such reactions, one substance loses electrons (oxidation) and another gains electrons (reduction). These electron transfer reactions are the basis for the functioning of galvanic cells where chemical energy is converted into electrical energy.
In our textbook problem, the chromium electrode acts as the anode where oxidation occurs (loss of electrons), while the nickel electrode functions as the cathode where reduction takes place (gain of electrons). Through the cell's external circuit, electrons flow from the anode to the cathode which generates an electric current. The number of electrons transferred is pivotal as it relates directly to the cell potential and is used in the Nernst Equation to calculate concentrations in an electrochemical system.