Question

The half-cell potential for the reaction \\[\begin{array}{l}\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \text { is }+1.03 \mathrm{V} \text { at } \\ 298.15 \mathrm{K} \text { when } a_{\mathrm{O}_{2}}=1.00 . \text { Determine } a_{\mathrm{H}^{+}} \end{array}\\]

Short answer

The activity of hydrogen ions (\(a_{H^+}\)) is 1.

Step-by-step solution

Write down the Nernst equation for the given half-cell reaction

The Nernst equation is used to calculate the electrochemical potential of a half-cell reaction at non-standard conditions. It is written as: \(E = E^\circ - \frac{RT}{nF} \ln Q\) Here, E is the cell potential, \(E^\circ\) is the standard cell potential, R is the gas constant (8.314 J/molK), T is the temperature in kelvin, n is the number of moles of electrons transferred, F is the Faraday's constant (96485 C/mol), and Q is the reaction quotient.

Substitute the given values into the Nernst equation and solve for the activity of hydrogen ions

The half-cell potential is given as +1.03V, and the standard cell potential is the same since the oxygen gas activity is given as 1.00. The temperature is also given as 298.15K. For the given half-cell reaction, n = 4, since 4 electrons are transferred. Now, substituting the given values into the Nernst equation, we have: \(1.03 = 1.03 - \frac{8.314 \times 298.15}{4 \times 96485} \ln Q\) To simplify this equation, we will move the first term on the right side to the left to get: \(0 = - \frac{8.314 \times 298.15}{4 \times 96485} \ln Q\) Since we are trying to determine the activity of hydrogen ions, we need to find the reaction quotient, Q. In this case, Q = \({a_{H^+}}^4\). This means : \(0 = - \frac{8.314 \times 298.15}{4 \times 96485} \ln {a_{H^+}}^4 \) Next, we can simplify the equation by moving variables to the other side of the equation: \(\ln {a_{H^+}}^4 = 0\) Now, since \(\ln 1 = 0\), we can rewrite the equation as: \({a_{H^+}}^4 = 1\) Finally, we will solve for the activity of hydrogen ions, \(a_{H^+}\), by taking the fourth root of both sides of the equation: \(a_{H^+} = {\sqrt[4]{1}}\) Since the fourth root of 1 is still 1, the activity of hydrogen ions is: \(a_{H^+} = 1\)

Key concepts

Electrochemical Potential

Electrochemical potential is a crucial concept in understanding how batteries and electrochemical cells work. It represents the energy that drives electrons from one electrode to another. When we place a piece of zinc in a copper sulfate solution, we can watch the zinc dissolve and copper gets deposited on the zinc surface. This reaction occurs because electrons move from zinc to copper ions in the solution, and the energy that drives this electron transfer is related to differences in electrochemical potential between the zinc and copper ions.

In an electrochemical cell, we harness this potential to do work, such as powering a light bulb or starting a car. The Nernst equation, which we use to calculate this potential under various conditions, allows us to predict how the voltage of a cell will change when the concentration of reactants and products change.

Reaction Quotient

The reaction quotient, Q, plays a vital role in electrochemistry as it quantifies the relative amounts of reactants and products at any point during a chemical reaction, not just at equilibrium. To calculate Q, you use the same expression as the equilibrium constant (K), but with the current concentrations or pressures of the reactants and products instead of their equilibrium values.

For our half-cell reaction involving oxygen and hydrogen ions, the reaction quotient would look at the current activities of the reactants and products. In the Nernst equation, Q allows us to determine how far from equilibrium our cell is and, therefore, how much voltage or electrical push we can obtain from our half-cell. Understanding Q can help us predict how a cell's electrochemical potential will change as a reaction proceeds or as environmental conditions shift.

Half-Cell Reaction

A half-cell reaction is one part of a redox reaction, where oxidation or reduction occurs. It is called 'half-cell' because it represents only one of the two necessary reactions in a full electrochemical cell. A typical electrochemical cell has two half-cells: one for the oxidation reaction and one for the reduction reaction.

In our exercise, the half-cell reaction describes the process in which oxygen gas reacts with hydrogen ions and electrons to form water. This reaction is key to determining the overall voltage of the cell. By isolating and analyzing half-cell reactions, students can better understand how electrons and ions interact within the bigger context of an electrochemical cell.

Standard Cell Potential

Standard cell potential, denoted as \(E^\circ\), is a measure of how much voltage a cell can produce under standard conditions. These conditions generally include all reactants and products at a concentration of 1 mole per liter, a pressure of 1 atmosphere for any gases, and a temperature of 25 degrees Celsius (298 K).

The cell potential is essentially the driving force behind the electron transfer in the redox reaction. In our example, the standard cell potential for the half-cell reaction involving oxygen and hydrogen ions was +1.03V. By knowing \(E^\circ\), we can use the Nernst equation to find out the actual potential of a cell under different conditions, which is invaluable for real-world applications where conditions are rarely standard.

Electron Transfer

Electron transfer is the movement of electrons from one atom, ion, or molecule to another in a redox reaction. It is central to the operation of electrochemical cells, where one species loses electrons (oxidation) and the other gains electrons (reduction).

Understanding electron transfer helps us comprehend how batteries work or how corrosion occurs. For instance, when a battery powers a device, it's because of the flow of electrons through the circuit from the negative terminal to the positive one. The Nernst equation plays a critical role in our understanding of this process because it allows us to calculate the potential for these electron transfers to occur under a set of conditions.

Activity of Ions

The activity of ions, often denoted by the letter 'a', measures the effective concentration of those ions in a solution – how 'active' or 'available' they are to participate in a reaction. It is similar to concentration but also takes into account interactions between ions in the solution which can affect their reactivity.

Activity becomes significantly different from concentration in highly concentrated solutions where ion pairs and other interactions may occur. In dilute solutions, such as those under standard conditions (1 M concentration), the activity of ions is typically assumed to be equivalent to their molarity. In our exercise, by solving the Nernst equation, we found the activity of hydrogen ions \(a_{H^+}\) to be 1, meaning they are fully available for the reaction under the given conditions.