Question

A voltaic cell is constructed with two \(\mathrm{Zn}^{2+}-\) Zn electrodes. The two cell compartments have \(\left[\mathrm{Zn}^{2+}\right]=1.8 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=1.00 \times 10^{-2} \mathrm{M}\), respectively. (a) Which electrode is the anode 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 \(\left[\mathrm{Zn}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

Short answer

(a) The anode is the electrode with a concentration of \(\mathrm{Zn}^{2+}\) ions of 1.8 M. (b) The standard emf of a cell with the same type of electrodes is 0 V. (c) The cell emf for the given concentrations is approximately 0.043 V. (d) As the cell operates, the \(\mathrm{Zn}^{2+}\) concentration will decrease at the anode and increase at the cathode.

Step-by-step solution

(a) Identifying the anode

Since both electrodes are Zn²⁺-Zn, the anode will be the electrode where the Zn²⁺ ion concentration is higher, as it will generate more spontaneous oxidation. In this case, the anode is the one with a concentration of \(\mathrm{Zn}^{2+}\) ions of 1.8 M.

(b) Calculating the standard emf

The standard emf (E°) of a cell with the same type of electrodes on both sides is 0 V. This is because the oxidation and reduction potentials will be the same, so their difference (emf) will be zero.

(c) Calculating the cell emf for the given concentrations

We can use the Nernst equation to find the cell emf for the given concentrations. The Nernst equation is: \[E = E° - \frac{RT}{nF} \ln Q\] Where: - E is the cell emf - E° is the standard emf (which we found to be 0 V in part b) - R is the universal gas constant (8.314 J/mol∙K) - T is the temperature (assume standard conditions, 298 K) - n is the number of moles of electrons transferred (for Zn²⁺-Zn, it is 2) - F is the Faraday's constant (96485 C/mol) - Q is the reaction quotient, which is the ratio of the concentrations of products and reactants Since E° = 0, we can simplify the Nernst equation to: \[E = - \frac{RT}{nF} \ln Q\] Now, calculate Q for the given concentrations by dividing the lower concentration by the higher concentration: \[Q = \frac{1.00 \times 10^{-2} M}{1.8 M}\] Now, plug in the values into the Nernst equation and solve for E: \[E = - \frac{(8.314 J/mol∙K)(298 K)}{(2)(96485 C/mol)} \ln \left(\frac{1.00 \times 10^{-2} M}{1.8 M}\right)\] \[E = -0.0296 V \ln \left(\frac{1.00 \times 10^{-2} M}{1.8 M}\right)\] After calculation, we get: \[E \approx 0.043 V\] So, the cell emf for the given concentrations is approximately 0.043 V.

(d) Predicting the change in Zn²⁺ concentration for each electrode

As the cell operates, at the anode (1.8 M concentration), the \(\mathrm{Zn}^{2+}\) concentration will decrease due to the oxidation of Zn to \(\mathrm{Zn}^{2+}\) ions. On the other hand, at the cathode (1.00 x 10⁻² M concentration), the \(\mathrm{Zn}^{2+}\) concentration will increase due to the reduction of the \(\mathrm{Zn}^{2+}\) ions to Zn metal.

Key concepts

Nernst Equation

The Nernst equation is a fundamental concept in electrochemistry, allowing us to determine the electromotive force (emf) of a cell under non-standard conditions. At its core, the Nernst equation correlates the observable cell potential to the temperature, the number of moles of electrons transferred in the reaction (n), the universal gas constant (R), the Faraday constant (F), and the reaction quotient (Q).

The form of the Nernst equation used in the solution to the exercise is simplified since the standard emf (\(E^\text{o}\)) is zero for cells with identical electrodes. In this form, the equation is \[E = -\frac{RT}{nF} \ln Q\] which helps to calculate the cell's emf based on the concentrations of the zinc ions (\(\mathrm{Zn}^{2+}\)) at both electrodes.

The Nernst equation empowers students to calculate the actual emf of a battery in a real-world scenario where concentrations are often not at standard conditions. By understanding this equation, students can predict how the cell potential will change as the reacting species' concentrations vary over time.

Standard Electromotive Force (emf)

The standard electromotive force, denoted as \(E^\text{o}\), is the voltage difference between two half-cells when all reactants and products are at standard conditions. These conditions typically mean that the reactive species are at 1 mole per liter concentrations, the gas pressures are at 1 atmosphere, and the temperature is at 298 K (25°C).

In the given exercise, we found that the standard emf of the cell with zinc electrodes on both sides is 0 V. This happens because each zinc half-cell has the same potential, and when combined, the differences in potentials cancel each other out. Students often misunderstand this concept and expect a non-zero voltage. Emphasizing that the standard emf reflects reversible, equilibrium conditions helps clarify this point and is critical for understanding cell potentials in electrochemical scenarios.

Zinc Electrodes Reaction

In the context of the exercise, zinc electrodes undergo redox reactions during the operation of a voltaic cell. The anode is where oxidation occurs, while the cathode is the site of reduction. For a zinc electrode, the oxidation reaction can be written as \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^-\), and reduction as \(\mathrm{Zn}^{2+} + 2e^- \rightarrow \mathrm{Zn}\).

Identifying the anode in the exercise is a key step that hinges on understanding the directional nature of electron flow and the spontaneous reactions that occur at each electrode. In our case, the anode is where the zinc concentration is higher (1.8 M), as this will facilitate the oxidation process. This insight is crucial since the cell's operation alters the concentrations of \(\mathrm{Zn}^{2+}\) ions at both electrodes, influencing the cell's voltage and direction of the electron flow over time.

Concentration Cells

A concentration cell is a type of electrochemical cell where both electrodes are made of the same material, but they are immersed in solutions with different concentrations. The difference in concentration creates a potential difference, driving the flow of electrons from the higher to the lower concentration electrode.

In the given exercise, despite both compartments containing \(\mathrm{Zn/Zn}^{2+}\) electrodes, a potential difference arises due to the different concentrations of \(\mathrm{Zn}^{2+}\) ions. As the cell operates, the concentration at the anode (\(1.8 \mathrm{M}\)) will decrease while it will increase at the cathode (\(1.00 \times 10^{-2} \mathrm{M}\)). This is because zinc oxidizes at the anode, releasing \(\mathrm{Zn}^{2+}\) ions into the solution, and reduces at the cathode, consuming \(\mathrm{Zn}^{2+}\) ions to form metallic zinc. Understanding concentration cells is beneficial for comprehending how natural and man-made electrochemical processes function even in the absence of different electrode materials.