Question

A galvanic cell is based on the following half-reactions at \(25^{\circ} \mathrm{C} :\) $$\begin{array}{c}{\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}} \\\ {\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}}\end{array}$$ Predict whether \(\mathscr{E}_{\text{cell}}\) is larger or smaller than \(\mathscr{E}^{\circ}_{\text{cell}}\) for the following cases. a. [Ag1] 5 1.0 a. \(\left[\mathrm{Ag}^{+}\right]=1.0 M,\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=2.0 M,\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) b. \(\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=1.0 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} \mathrm{M}\)

Short answer

In scenario a, the cell potential \(\mathscr{E}_{\text{cell}}\) is larger than the standard cell potential \(\mathscr{E}^{\circ}_{\text{cell}}\), while in scenario b, the cell potential \(\mathscr{E}_{\text{cell}}\) is smaller than the standard cell potential \(\mathscr{E}^{\circ}_{\text{cell}}\).

Step-by-step solution

Write the balanced cell reaction

First, we need to combine the two given half-reactions into one balanced cell reaction. To do this, we need to add the two half-reactions together, making sure to multiply them by appropriate coefficients to balance the electrons. The balanced cell reaction is: $$2\mathrm{Ag}^{+}+\mathrm{H}_{2}\mathrm{O}_{2}+2\mathrm{H}^{+}\longrightarrow 2\mathrm{Ag}+2\mathrm{H}_{2}\mathrm{O}$$

Determine the standard cell potential

To find the standard cell potential, we will need to subtract the standard reduction potential of the reduction half-reaction (the first reaction) from the standard reduction potential of the oxidation half-reaction (the second reaction). This information is normally provided or can be looked up in tables. Because it is not given in this exercise, we will denote the standard cell potential \(\mathscr{E}^{\circ}_{\text{cell}}\) as a variable.

Apply the Nernst equation

Now that we have the balanced cell reaction, we can apply the Nernst equation to calculate \(\mathscr{E}_{\text{cell}}\) for both scenarios (a and b). The Nernst equation is given as: $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln\left(\frac{\text{products}}{\text{reactants}}\right)$$ where \(R\) is the universal gas constant (\(8.314\ \text{J/mol}\cdot\text{K}\)), \(T\) is the temperature in Kelvin, \(n\) is the number of transferred electrons, and \(F\) is the Faraday constant (\(96485\ \text{C/mol}\)). For both scenarios (a and b), the temperature is given as \(25^{\circ}\mathrm{C}\) which equals to \(298.15\ \text{K}\), and the number of transferred electrons is 2 (from the balanced cell reaction). Scenario a:

Step 4a: Plug in given concentrations into Nernst equation for scenario a

In scenario a, we have following concentrations: \(\left[\mathrm{Ag}^{+}\right]=1.0 \mathrm{M},\left[\mathrm{H}_{2}\mathrm{O}_{2}\right]=2.0 \mathrm{M},\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) Using these concentrations, we can plug in the values into the Nernst equation: $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{(8.314)(298.15)}{(2)(96485)} \ln\left(\frac{\left[\mathrm{H}_{2}\mathrm{O}\right]^2}{\left[\mathrm{Ag}^{+}\right]^2 \left[\mathrm{H}_{2}\mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]^2}\right)$$ $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{(8.314)(298.15)}{(2)(96485)} \ln\left(\frac{1}{(1.0)^2(2.0)(2.0)^2}\right)$$ $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - 0.0257\ \ln(0.25)$$ Scenario b:

Step 4b: Plug in given concentrations into Nernst equation for scenario b

In scenario b, we have following concentrations: \(\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2}\mathrm{O}_{2}\right]=1.0 \mathrm{M},\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7}\ \mathrm{M}\) Using these concentrations, we can plug in the values into the Nernst equation: $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{(8.314)(298.15)}{(2)(96485)} \ln\left(\frac{\left[\mathrm{H}_{2}\mathrm{O}\right]^2}{\left[\mathrm{Ag}^{+}\right]^2 \left[\mathrm{H}_{2}\mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]^2}\right)$$ $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{(8.314)(298.15)}{(2)(96485)} \ln\left(\frac{1}{(2.0)^2(1.0)(1.0 \times 10^{-7})^2}\right)$$ $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - 0.0257\ \ln(5\times10^{14})$$

Compare the cell potentials to the standard cell potential

Now that we have found the cell potentials (\(\mathscr{E}_{\text{cell}}\)) for both scenarios, we can compare them to the standard cell potential (\(\mathscr{E}^{\circ}_{\text{cell}}\)). Scenario a: If \(\mathscr{E}_{\text{cell}}>\mathscr{E}^{\circ}_{\text{cell}}\), then \(\mathscr{E}^{\circ}_{\text{cell}} - 0.0257\ \ln(0.25)>\mathscr{E}^{\circ}_{\text{cell}}\). As the \(\ln(0.25)=-1.386\), the term \(-0.0257\ \ln(0.25)=0.03565\) is positive. It is clear that \(\mathscr{E}_{\text{cell}} > \mathscr{E}^{\circ}_{\text{cell}}\) for scenario a. Scenario b: If \(\mathscr{E}_{\text{cell}}<\mathscr{E}^{\circ}_{\text{cell}}\), then \(\mathscr{E}^{\circ}_{\text{cell}} - 0.0257\ \ln(5\times10^{14})<\mathscr{E}^{\circ}_{\text{cell}}\). As the \(\ln(5\times10^{14})=33.835\), the term \(-0.0257\ \ln(5\times10^{14})=-0.86837\) is negative. It is clear that \(\mathscr{E}_{\text{cell}} < \mathscr{E}^{\circ}_{\text{cell}}\) for scenario b. So, the cell potential is larger than the standard cell potential for scenario a, while the cell potential is smaller than the standard cell potential for scenario b.

Key concepts

Nernst equation

The Nernst equation is a fundamental tool in electrochemistry used to calculate the cell potential of an electrochemical cell under non-standard conditions. It allows us to predict how the concentrations of the reactants and products in a galvanic cell will affect the actual cell potential (\( \mathscr{E}_{\text{cell}}\)).
The general form of the Nernst equation is:\[\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln\left(\frac{\text{products}}{\text{reactants}}\right)\]
In this equation:

• \( \mathscr{E}^{\circ}_{\text{cell}} \) is the standard cell potential, which is the cell potential under standard conditions (1M concentration for all solutions, 1 atm pressure for gases, and pure solids or liquids).
• \( R \) is the universal gas constant (8.314 J/mol·K).
• \( T \) is the absolute temperature in Kelvin (for room temperature, \( 25^{\circ} \text{C} \), \( T = 298.15 \text{K}\)).
• \( n \) is the number of moles of electrons transferred in the balanced overall cell reaction.
• \( F \) is Faraday's constant (96485 C/mol), representing the charge of one mole of electrons.

The Nernst equation shows that as the ratio of products to reactants in the cell changes, the cell potential deviates from the standard cell potential. In electrochemical terms, it provides a way to measure the effect of concentration on a cell's voltage, which is essential in predicting the feasibility and direction of chemical reactions under various conditions.

standard cell potential

The standard cell potential, denoted as \( \mathscr{E}^{\circ}_{\text{cell}} \), is a measure of the voltage produced by a galvanic cell when all species are at standard conditions. This potential is calculated from the standard reduction potentials of the two half-reactions occurring in the cell.
To find \( \mathscr{E}^{\circ}_{\text{cell}} \), we use the formula:\[\mathscr{E}^{\circ}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cathode}} - \mathscr{E}^{\circ}_{\text{anode}}\]
Here, \( \mathscr{E}^{\circ}_{\text{cathode}} \) and \( \mathscr{E}^{\circ}_{\text{anode}} \) represent the standard reduction potentials of the cathode and anode reactions, respectively. These values are typically found in electrochemical series tables.

• For the galvanic cell to work, \( \mathscr{E}^{\circ}_{\text{cell}} \) should be positive, indicating that the chemical reaction will proceed spontaneously under standard conditions.
• The standard cell potential serves as a benchmark for evaluating changes in the cell potential when the cell operates under non-standard conditions, using the Nernst equation.

Understanding \( \mathscr{E}^{\circ}_{\text{cell}} \) is crucial for predicting how alterations in conditions such as concentration and temperature influence electrochemical cell behavior and efficiency.

half-reactions

In electrochemical cells, the overall reaction is often split into two half-reactions, one occurring at the anode and the other at the cathode. Each half-reaction represents a specific oxidation or reduction reaction and is essential for balancing the overall cell equation.

• **Oxidation half-reaction:** This reaction occurs at the anode, where there is a loss of electrons. For instance, in the given galvanic cell, the oxidation half-reaction might involve the transformation of a species by losing electrons.
• **Reduction half-reaction:** This reaction takes place at the cathode, involving the gain of electrons. In the example, the reduction of \( \text{Ag}^{+} \) to \( \text{Ag} \) is indicative of this process.

When writing these half-reactions:

• Ensure the charges are balanced by adding electrons to the appropriate side.
• Combine the half-reactions to form a balanced full-cell reaction, making sure the number of electrons lost equals the number of electrons gained.

Half-reactions help not only in creating a clear picture of what transpires within an electrochemical cell but also in assisting with calculating theoretical cell potentials and efficiencies.