Problem 69 - Chemistry The Central Science In Si Units

Question

A voltaic cell is constructed with two \(\mathrm{Cu}^{2+}-\mathrm{Cu}\) electrodes. The two half-cells have \(\left[\mathrm{Cu}^{2+}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \times 10^{-4} \mathrm{M},\) respectively. \((\mathbf{a})\) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [ \(\left.\mathrm{Cu}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

Step-by-Step Solution

Answer

(a) The cathode is at 0.100 M \\(\mathrm{Cu}^{2+}\\). (b) Standard emf is 0.00 V. (c) Cell emf is 0.118 V. (d) \\text{At the cathode, \\[\mathrm{Cu}^{2+}\] decreases; at the anode, it increases.}\\]

Step 1: Identify the Cell Components

In this voltaic cell, there are two copper half-cells with copper electrodes. The concentrations of the copper ions \([\mathrm{Cu}^{2+}]\) are 0.100 M and 1.00 \times 10^{-4} M.

Step 2: Determine the Cathode and Anode

According to the work principle of voltaic cells, the cathode is where reduction takes place, and it's determined by the higher concentration of \([\mathrm{Cu}^{2+}]\) because it will gain electrons. Thus, the half-cell with 0.100 M \([\mathrm{Cu}^{2+}]\) is the cathode.

Step 3: Calculate the Standard EMF

The standard EMF (electromotive force) is calculated using the standard reduction potentials of both electrodes, which are the same in this scenario (\(E^\circ = 0 \text{ V}\) for copper under standard conditions). Therefore, the standard emf of this cell is \mathbf{0.00 \, V}\.

Step 4: Apply the Nernst Equation to Find the Cell EMF

Use the Nernst equation: \[E = E^\circ - \frac{RT}{nF}\ln Q\] where \(E^\circ = 0\, \text{V}\), \(R\) is the gas constant (8.314 J/mol·K), \(T\) is the temperature in Kelvin (assume 298 K, unless specified), \(n\) is the number of electrons transferred (2 for copper), and \(F\) is Faraday's constant (96485 C/mol). \The reaction quotient \(Q\) is \\[Q = \frac{[\text{Cu}^{2+}]_{anode}}{[\text{Cu}^{2+}]_{cathode}} = \frac{1.00 \times 10^{-4} \text{ M}}{0.100 \text{ M}}\]Substituting these into the Nernst equation gives: \[E = 0 - \frac{(8.314)(298)}{2(96485)}\ln \left(\frac{1.00 \times 10^{-4}}{0.100}\right)\]Simplifying yields approximately \mathbf{0.118 \, \text{V}}\.

Step 5: Predict Concentration Changes as the Cell Operates

In the cathode compartment, the \[\mathrm{Cu}^{2+}\] concentration will decrease as it is reduced to \[\mathrm{Cu}\], while in the anode compartment, the \[\mathrm{Cu}^{2+}\] concentration will increase as \[\mathrm{Cu}\] is oxidized to \[\mathrm{Cu}^{2+}\].

Key Concepts

Voltaic Cell

A voltaic cell, also known as a galvanic cell, is a type of electrochemical cell that converts chemical energy into electrical energy through a spontaneous redox reaction. The cell is comprised of two separate half-cells, each containing an electrode and an electrolyte. In these half-cells, oxidation and reduction reactions occur, creating an electric current that can be harnessed for electrical power.

• The anode is where oxidation occurs.
• The cathode is where reduction occurs.

The cell has a salt bridge that allows ions to flow between the two solutions, thus maintaining charge balance as the reaction progresses. In the case of copper-copper (Cu-Cu) voltaic cells as discussed in the exercise, the entire setup involves copper electrodes immersed in copper ion solutions with different concentrations. The difference in concentration drives the movement of electrons from one half-cell to the other, generating electric current.

Cathode and Anode

Understanding the cathode and anode is essential in electrochemistry. In a voltaic cell, these are the electrodes where half-reactions occur. The anode is the electrode where oxidation occurs, meaning it loses electrons, and it is negatively charged. Conversely, the cathode is the electrode where reduction occurs, gaining electrons, and it is positively charged.

• To find which electrode is the anode or cathode, identify the direction of electron flow and concentration of ions.
• The cathode will generally be the half-cell with the higher ion concentration because it attracts electrons to reduce the ions.

In the exercise's context, since the cathode is where reduction takes place, the half-cell with the concentration of \([\text{Cu}^{2+}]\) at 0.100 M is the cathode due to the higher ion availability compared to the 1.00 x 10^-4 M concentration in the other half-cell.

Nernst Equation

The Nernst equation is crucial for calculating the electrochemical cell's potential under non-standard conditions. It refines our understanding of how concentration affects cell potential. The equation is:\[E = E^\circ - \frac{RT}{nF}\ln Q\]where:

• \(E^\circ\) is the standard electrode potential (0 V for a copper-copper cell with identical electrodes).
• \(R\) is the universal gas constant, 8.314 J/mol·K.
• \(T\) is the temperature in Kelvin (usually 298 K, unless otherwise noted).
• \(n\) is the number of moles of electrons exchanged (2 for copper).
• \(F\) is Faraday's constant, 96485 C/mol.
• \(Q\) is the reaction quotient, \(\frac{1.00 \times 10^{-4} \text{ M}}{0.100 \text{ M}}\).

By substituting known values into the equation, the cell's potential in real conditions is calculated effectively. In our copper example, this calculates to approximately 0.118 V.

Cell EMF

The cell emf (electromotive force), also known as cell potential, is the measure of the voltage or electrical potential difference between two half-cells in a voltaic cell. It indicates the cell's ability to drive electric current through a circuit.

• An emf of zero implies no net energy transfer; the system is in equilibrium.
• A positive emf suggests a spontaneous reaction, driving electrons from the anode to the cathode.

In the voltaic cell from the exercise, the standard emf is calculated as 0.00 V because the electrodes and reaction are identical under standard conditions. However, using the Nernst equation, we find that the cell emf under the given conditions is approximately 0.118 V, reflecting the influence of ionic concentration disparities on driving electron flow.