Problem 98 - Chemistry Principles and Practice

Question

Calculate the potential of the half-reaction $$ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} $$ when the concentrations in solution are \(\left[\mathrm{Fe}^{3+}\right]=0.033 M\) and \(\left[\mathrm{Fe}^{2+}\right]=0.0025 M,\) and the temperature is \(298 \mathrm{~K}\).

Step-by-Step Solution

Answer

The potential is approximately 0.836 V.

Step 1: Understand the Nernst Equation

The Nernst equation is used to calculate the potential of a half-reaction in electrochemistry. The equation is given by: \[E = E^0 - \frac{RT}{nF} \ln \frac{[\text{Products}]}{[\text{Reactants}]}\]where:- \(E\) is the electrode potential;- \(E^0\) is the standard electrode potential;- \(R\) is the universal gas constant \(8.314 \text{ J/(mol K)}\);- \(T\) is the temperature in Kelvin;- \(n\) is the number of moles of electrons transferred;- \(F\) is the Faraday's constant \(96485 \text{ C/mol}\);- The reaction quotient is \(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\).

Step 2: Standard Electrode Potential

For the half-reaction \(\text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+}\), we need the standard electrode potential \(E^0\). From standard tables, \(E^0 = +0.77 \text{ V}\).

Step 3: Calculate the Reaction Quotient

Calculate the reaction quotient \(Q\) using the concentrations given:\[Q = \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = \frac{0.0025}{0.033}\]After calculation, \(Q \approx 0.0758\).

Step 4: Plug Values into the Nernst Equation

Using the Nernst equation, the temperature \(T = 298 \text{ K}\), and \(n = 1\) (1 electron is transferred), substitute all known values:\[E = 0.77 - \frac{(8.314)(298)}{(1)(96485)} \ln(0.0758)\]Calculate the second term.

Step 5: Calculate the Second Term

Compute \(\frac{(8.314)(298)}{(1)(96485)} \approx 0.025693 \text{ V}\). Now multiply this by \(\ln(0.0758)\):\[0.025693 \times (-2.579) = -0.0664 \text{ V}\]Thus, \(E = 0.77 - (-0.0664)\).

Step 6: Calculate the Electrode Potential

Simplify the equation from Step 4:\[E \approx 0.77 + 0.0664 = 0.8364 \text{ V}\]

Key Concepts

Electrode Potential

Electrode potential is a measure of the ability of a chemical species within a half-cell of an electrochemical cell to gain or lose electrons. This is quantified through the electrode potential and expressed in volts (V). Often referenced as the half-cell potential, it is a pivotal aspect of understanding how different elements interact electrochemically.

A higher electrode potential implies a greater tendency to accept electrons (reduction), whereas a lower potential indicates a tendency to lose electrons (oxidation). This potential is calculated using the Nernst Equation, which adjusts the standard electrode potential, considering the specific conditions like temperature and concentration of reactants and products.

When calculating an electrode potential, it's important to determine the standard electrode potential, denoted as \(E^0\), typically found in electrochemical tables. For example, the Fe reaction provided has a standard potential of \(+0.77\, \text{V}\). This value along with the reaction quotient helps to find the actual electrode potential under non-standard conditions.

Electrochemistry

Electrochemistry involves studying the movement of electrons in chemical reactions, which is essential for various applications, including batteries, corrosion, and electroplating. At its core, electrochemistry examines how chemical reactions can generate electric current, or conversely, how electric current can facilitate chemical transformations.

Key concepts within electrochemistry include:

• Redox reactions: These are chemical processes involving the transfer of electrons between two substances.
• Galvanic cells: These involve spontaneous redox reactions that produce electricity.
• Electrolytic cells: These require an external power source to drive non-spontaneous reactions.

In the exercise, the calculation of the electrode potential using the Nernst Equation exemplifies a practical application of electrochemical principles. By considering the concentration of the \(\text{Fe}^{2+}\) and \(\text{Fe}^{3+}\) ions, we calculate how electron flow potential is modified from its standard state, offering insights into the specific electrochemical behavior of iron.

Reaction Quotient

The reaction quotient, often denoted \(Q\), is a ratio that compares the concentrations of products to reactants at any point during a reaction. It is a crucial tool in predicting the direction of a chemical reaction's shift towards equilibrium.

For the half-reaction of \(\mathrm{Fe}^{3+} + \mathrm{e}^- \rightarrow \mathrm{Fe}^{2+}\), the reaction quotient is calculated as \(Q = \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}\). By substituting the given concentrations, \(Q\) helps determine how far a system deviates from equilibrium at a given moment. In this exercise, \(Q \approx 0.0758\), showing that there is more of the \(\text{Fe}^{3+}\) reactant present than the \(\text{Fe}^{2+}\) product.

The Nernst Equation uses the reaction quotient to tailor the standard electrode potential value, adapting it to the current state of the chemical system. It reveals how non-equilibrium conditions affect the overall potential of the electrochemical reaction.