Chapter 5: Problem 9
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
Step by step solution
Calculate the Cell Potential
Determine the Equilibrium Constant
Comment on Spontaneity of Reaction
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
The Nernst Equation
\[ E = E^\circ - \frac{0.0592}{n} \log Q \]
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
- 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
- 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.