Question

Determine the half-cell reactions and the overall cell reaction, calculate the cell potential, and determine the equilibrium constant at \(298.15 \mathrm{K}\) for the cell \\[\begin{array}{c}\operatorname{Zn}(s)\left|\mathrm{Zn}^{2+}\left(a q, a_{\pm}=0.0120\right)\right| \\ \left|\mathrm{Mn}^{3+}\left(a q, a_{\pm}=0.200\right), \mathrm{Mn}^{2+}\left(a q, a_{\pm}=0.0250\right)\right| \mathrm{Pt}(s) \end{array}\\] Is the cell reaction spontaneous as written?

Short answer

The overall cell reaction for the given cell is: \(\mathrm{Zn (s)} + \mathrm{Mn^{3+} (aq)} \to \mathrm{Zn^{2+} (aq)} + \mathrm{Mn^{2+} (aq)}\). The cell potential is calculated as \(2.171\thinspace V\), and the equilibrium constant is \(3.56 \times 10^{20}\). Since the cell potential is positive, the reaction is spontaneous as written.

Step-by-step solution

Identify Half-Cell Reactions and Overall Cell Reaction

Based on the cell diagram, we can identify the half-cell reactions as follows: 1. Anode half-cell: $$\mathrm{Zn (s)} \to \mathrm{Zn^{2+} (aq)} + 2 \mathrm{e^-} $$ 2. Cathode half-cell: $$\mathrm{Mn^{3+} (aq)} + \mathrm{e^-} \to \mathrm{Mn^{2+} (aq)} $$ Combining both half-cell reactions to form the overall cell reaction: $$\mathrm{Zn (s)} + \mathrm{Mn^{3+} (aq)} \to \mathrm{Zn^{2+} (aq)} + \mathrm{Mn^{2+} (aq)}$$

Calculate Cell Potential using the Nernst Equation

To calculate the cell potential, we will use the Nernst equation: $$E_{cell} = E_{cell}^{0} - \frac{RT}{nF} \ln{Q}$$ Where: - \(E_{cell}^{0}\) is the standard cell potential - \(R\) is the gas constant - \(T\) is the temperature - \(n\) is the number of electrons transferred - \(F\) is the Faraday constant - \(Q\) is the reaction quotient Using the standard reduction potentials: $$E_{Zn^{2+}/Zn}= -0.76\thinspace V$$ $$E_{Mn^{3+}/Mn^{2+}}= 1.51\thinspace V$$ Calculate the standard cell potential: $$E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0} = 1.51\thinspace V - (-0.76\thinspace V) = 2.27\thinspace V$$ The reaction quotient Q is: $$Q = \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Mn^{3+}}]} = \frac{(0.0120)}{(0.200)}$$ Calculate the cell potential: $$E_{cell} = E_{cell}^{0} - \frac{RT}{nF} \ln{Q} = 2.27\thinspace V - \frac{8.314 \times 298.15}{(1 \times 96485)} \ln{\frac{0.0120}{0.200}} = 2.27\thinspace V - 0.0592\thinspace V \times \ln{0.060} = 2.27\thinspace V - 0.0987\thinspace V = 2.171\thinspace V$$

Calculate Equilibrium Constant

To calculate the equilibrium constant, we'll use the relationship between cell potential and the equilibrium constant: $$E_{cell} = \frac{RT}{nF} \ln{K}$$ Rearrange and solve for K: $$K = e^{\frac{nFE_{cell}}{RT}} = e^{\frac{1 \times 96485 \times 2.171\thinspace V}{8.314 \times 298.15}} = 3.56 \times 10^{20}$$

Determine if Reaction is Spontaneous

The cell potential at these concentrations is \(E_{cell} = 2.171\thinspace V\). Since the cell potential is positive, the reaction is spontaneous as written. In conclusion, the overall cell reaction is: $$\mathrm{Zn (s)} + \mathrm{Mn^{3+} (aq)} \to \mathrm{Zn^{2+} (aq)} + \mathrm{Mn^{2+} (aq)}$$ The cell potential is \(2.171\thinspace V\), the equilibrium constant is \(3.56 \times 10^{20}\), and the reaction is spontaneous as written.

Key concepts

Cell Potential

The cell potential, also known as electromotive force (EMF), of an electrochemical cell is a measure of the propensity of the cell to drive an electric current through an external circuit. It represents the energy per unit charge which is available from the redox reactions occurring in the cell and is expressed in volts (V). In the context of our exercise, calculating the cell potential involves understanding the half-cell potentials for each electrode and then combining them to find the overall potential of the cell. These half-cell potentials are determined under standard conditions (concentration of 1 M, pressure of 1 atm, temperature of 25°C), and are adjusted for non-standard conditions using the Nernst equation. The positive cell potential calculated in our exercise indicates that the reaction is capable of doing work if it were set up in a galvanic cell.

Students often struggle with interpreting the meaning of cell potential. It is crucial to understand that a positive cell potential signifies a spontaneous reaction, while a negative value would suggest a non-spontaneous process. Moreover, the cell potential is related to the free energy change for the reaction, thereby connecting electrochemical processes with thermodynamics.

Nernst Equation

The Nernst equation is a formula used to calculate the actual cell potential when conditions are non-standard. It provides a link between the concentrations of reactants and products at any given moment (via the reaction quotient, Q) and the electrical potential of the cell. The formula inherently includes variables such as the number of moles of electrons transferred in the reaction (n), the universal gas constant (R), the temperature (T), and Faraday's constant (F). One common mistake students make is not using the correct unit for temperature, which should always be in Kelvin. In our exercise, the Nernst equation allows us to incorporate the actual concentrations (or activities) of zinc and manganese ions present in the half-cells to compute the cell potential under specific conditions. Thus, the final cell potential can differ from the standard potential, reflecting the real-life scenario where concentrations do not always align with standard states.

Equilibrium Constant

The equilibrium constant (K) is a dimensionless value that gives us information about the position of equilibrium for a chemical reaction. For reactions occurring in a cell, K can be calculated from the cell potential. An important point for students to note is that a large equilibrium constant indicates that the products are favored at equilibrium, while a small constant means the reactants are favored. In our provided exercise, the very large equilibrium constant (on the order of 10^20) strongly suggests a reaction heavily skewed towards the formation of products, aligning with the spontaneous nature of the cell reaction under the given conditions. Understanding the relationship between the equilibrium constant and cell potential is crucial, as it bridges the gap between electrochemistry and chemical equilibrium concepts.

Half-Cell Reactions

Analyzing half-cell reactions is fundamental in understanding the mechanics of an electrochemical cell. Each half-cell reaction involves either the oxidation or reduction process, which occurs at the respective electrode. In a complete cell, we have one half-cell where oxidation occurs, known as the anode, and another where reduction happens, termed the cathode. In the exercise, zinc is being oxidized to zinc ions while manganese ions are being reduced to a lower oxidation state. These half-cell reactions are not just theoretically significant; they also guide the practical setup of an electrochemical cell in the laboratory.

To correctly identify the half-cell reactions, students must be able to assign oxidation states and determine the direction in which electrons flow. Proficiency in these areas helps paint a clear picture of the electrochemical processes at play and is highly important in predicting the behavior of the cells under different conditions.

Spontaneous Reaction

A spontaneous reaction is one that proceeds on its own without the need for external energy once it has been initiated. In electrochemistry, the sign of the cell potential is a direct indicator of whether a reaction is spontaneous: a positive value indicates a spontaneous reaction under standard conditions, while a negative value means it is not. In our exercise, the positive cell potential (\(2.171 \thinspace V\)) confirms that the reaction is spontaneous as written.

It's essential for students to grasp that 'spontaneity' does not imply that a reaction will occur quickly, only that the direction of change is energetically favorable. The rate at which a reaction proceeds toward equilibrium can be slow or fast and is governed by kinetics, not thermodynamics. Connecting spontaneous reactions to everyday batteries can be helpful, as it illustrates the practical applications of such reactions in providing electrical energy for devices.