Calculate the equilibrium constant for the possible reaction between \(\mathrm{Fe}^{2+}(\mathrm{aq})\) ions and \(\mathrm{Ce}^{4+}(\mathrm{aq})\) ions \(\mathrm{f} E_{\mathrm{Ce}^{4+} / \mathrm{ce}^{3+}}=1.44 \mathrm{~V}\) and \(E_{\mathrm{Fe}^{\circ} / \mathrm{Fe}^{2+}}^{\circ}=0.68 \mathrm{~V}\). Also comment on the spontaneity of the reaction.

Short Answer

Expert verified
The equilibrium constant (K) for the reaction is approximately \(10^{13}\), and the reaction is spontaneous.

Step by step solution

01

Calculate the Cell Potential

The first step would be to determine the cell potential (E) for the reaction using the given half-cell potentials. The cell potential for the reaction is calculated by subtracting the half-cell potential of the reduction half-reaction from that of the oxidation half-reaction. The corresponding half-reactions are: Oxidation: \(Fe^{2+}\) -> \(Fe\) + 2e⁻, E = 0.68 V Reduction: \(Ce^{4+}\) + e⁻ -> \(Ce^{3+}\), E = 1.44 V Hence, E_cell = E_(reduction) - E_(oxidation) = 1.44 V - 0.68 V = 0.76 V
02

Determine the Equilibrium Constant

The Nernst equation connects the equilibrium constant (K) and the cell potential (E). It is given by: \(E = E° - \frac{0.0592}{n} \log K\), where E° is the standard cell potential, n is the number of electrons exchanged (which is 1 in this case), and K is the equilibrium constant. Here, we are given E = E°. So, we can rearrange the equation to find K: \(\log K = \frac{nE°}{0.0592}\) Substituting for n and E°, we have: \(\log K = \frac{1 × 0.76 V}{0.0592} ≈ 13\) Finally, we find the equilibrium constant K by anti-logging the result, which gives \(K = 10^{13}\).
03

Comment on Spontaneity of Reaction

The spontaneity of a reaction can be assessed by the sign of the cell potential (E). If E is positive, the reaction is spontaneous; if it is negative, the reaction is non-spontaneous. Here, as calculated in step 1, E = 0.76 V, which is positive. Hence, the reaction is spontaneous.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Nernst Equation
The Nernst equation is a fundamental relation in electrochemistry that connects the cell potential of a reaction to its concentration ratio or, alternatively, to its equilibrium constant when the system reaches equilibrium. It is expressed as:
\[ E = E^\circ - \frac{0.0592}{n} \log Q \]
where \( E \) is the actual cell potential, \( E^\circ \) is the standard cell potential, \( n \) represents the number of electrons transferred in the electrochemical reaction, and \( Q \) is the reaction quotient, which is the ratio of the concentrations of the products raised to their stoichiometric coefficients over the concentrations of the reactants raised to theirs.
  • The equation helps predict the cell potential at any given set of concentrations.
  • At equilibrium, \( Q \) becomes the equilibrium constant, \( K \), which allows for the calculation of \( K \) from the standard cell potentials.
  • The term \(\frac{0.0592}{n}\) is derived from constants (the gas constant, the Faraday constant, and the temperature), assuming standard conditions of 298 K.

In the case of the reaction between \(\mathrm{Fe}^{2+}(\mathrm{aq})\) and \(\mathrm{Ce}^{4+}(\mathrm{aq})\) ions, the Nernst equation is simplified since the measured cell potential is equal to the standard cell potential (given the standard conditions), eliminating the need to know the concentrations for calculating the equilibrium constant.
Cell Potential
Cell potential, often denoted by \( E \), is a measure of the driving force behind an electrochemical reaction. It represents the capability of an electrochemical cell to produce an electric current when no external work is being done on it. For a galvanic (voltaic) cell, this potential arises due to the chemical reactions occurring at the electrodes.
  • A positive cell potential suggests that a reaction can release electrical energy, performing work.
  • The standard cell potential \( E^\circ \) is calculated when all reactants and products are at unit activity, which typically means 1M concentration for solutions or 1 atm pressure for gases at 298 K.

In the exercise, the cell potential is calculated by subtracting the oxidation potential from the reduction potential. The oxidation involves \(\mathrm{Fe}^{2+}\) losing two electrons to form \(\mathrm{Fe}\) and has an assigned potential of 0.68 V. The reduction involves \(\mathrm{Ce}^{4+}\) accepting an electron to form \(\mathrm{Ce}^{3+}\) and has a potential of 1.44 V. Subtracting the former from the latter gives the overall cell potential of 0.76 V, which indicates that the reaction can proceed spontaneously under standard conditions.
Reaction Spontaneity
Reaction spontaneity in an electrochemical context references whether a chemical reaction can proceed on its own without the need for additional energy. This concept is closely linked to the thermodynamic properties of a system, specifically Gibbs free energy.
  • If a reaction has a positive cell potential (\( E \) is greater than zero), it is spontaneous as it has a negative change in Gibbs free energy (\( \Delta G \) is less than zero).
  • A negative cell potential indicates a non-spontaneous reaction, which cannot occur without external energy.

The standard cell potential is a key indicator of spontaneity under standard conditions. For the reaction in the exercise, the cell potential calculated from the half-cell potentials is positive, which confirms that the reaction is indeed spontaneous. More precisely, the reaction between iron (II) and cerium (IV) ions spontaneously produces iron and cerium (III) ions under standard conditions, as described by the positive cell potential of 0.76 V calculated in the problem. This aligns with the thermodynamic expectation that spontaneous reactions occur when there is a potential for work to be done by the system.

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