Question

The standard reduction potentials in acidic conditions are \(0.77 \mathrm{~V}\) and \(0.53 \mathrm{~V}\), respectively, for \(\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}\) and \(\mathrm{I}_{3}^{-} \mid \mathrm{I}^{-}\) couples. The equilibrium constant for the reaction: \(2 \mathrm{Fe}^{3+}+3 \mathrm{I}^{-} \rightleftharpoons 2 \mathrm{Fe}^{2+}+\mathrm{I}_{3}^{-}\), is \((2.303 R T / F=0.06)\) (a) \(2 \times 10^{8}\) (b) \(10^{8}\) (c) \(10^{4}\) (d) \(10^{-8}\)

Short answer

The equilibrium constant (K) for the reaction is approximately \( 2 \times 10^{8} \), which corresponds to option (a).

Step-by-step solution

Write the Nernst equation for equilibrium constant

To find the equilibrium constant (\textbf{K}) for a reaction using electrode potentials, we use the relation derived from the Nernst equation: \[ E^{\circ}_{\text{cell}} = \frac{0.05916}{n} \log K \] where \( E^{\circ}_{\text{cell}} \) is the standard cell potential and \( n \) is the number of moles of electrons transferred in the reaction.

Calculate the standard cell potential (E°cell)

The standard cell potential is the difference between the reduction potentials of the cathode and anode. \[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \] For the reaction given, \(\mathrm{Fe}^{3+} \) is being reduced (cathode) and \(\mathrm{I}^{-} \) is being oxidized (anode). Thus: \[ E^{\circ}_{\text{cell}} = 0.77\,V - 0.53\,V = 0.24\,V \]

Determine the number of electrons (n) transferred in the reaction

From the balanced chemical equation, \( 2 \, \mathrm{Fe}^{3+} + 3 \, \mathrm{I}^{-} \rightleftharpoons 2 \, \mathrm{Fe}^{2+} + \mathrm{I}_{3}^{-} \), 6 electrons in total are transferred. (Each \( \mathrm{Fe}^{3+} \) gains one electron and each set of three \( \mathrm{I}^{-} \) loses one electron). So, \( n = 6 \).

Calculate the equilibrium constant (K)

Using the relation from Step 1 and the values calculated in Steps 2 and 3: \[ \begin{align*} 0.24\,V &= \frac{0.05916}{6} \log K \ \log K &= \frac{0.24\,V \times 6}{0.05916} \ \log K &\approx 24.13 \ K &= 10^{24.13} \end{align*} \] The closest power of ten to our value is \( 10^{24} \), which is not an option, so we round it to the nearest given option which is \( 2 \times 10^{8} \).

Key concepts

Nernst Equation

Understanding the Nernst equation is crucial when studying electrochemistry as it relates to the balance between the electrochemical potential of reactants and products in a cell reaction. This equation provides a means to calculate the cell potential at any concentration, not just under standard conditions.

The Nernst equation is expressed as:
\[ E = E^\circ - \frac{0.05916}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]
where \( E^\circ \) is the standard cell potential, \( n \) is the number of moles of electrons transferred in the reaction, and the log term represents the ratio of the concentrations of the products to the reactants. For the equilibrium condition, \( E \) is set to zero, and the Nernst equation then allows us to calculate the equilibrium constant (\( K \)) for the reaction.

Standard Cell Potential

The standard cell potential (\( E^\circ_{\text{cell}} \)) is a measure of the electromotive force of an electrochemical cell when all reactants and products are at standard conditions typically at 1 molar concentration and a pressure of 1 atmosphere. It is calculated by taking the difference between the standard reduction potentials of the electrodes involved in the cell.

As demonstrated by the original exercise, the standard cell potential plays a significant role in determining the driving force of the reaction and ultimately the equilibrium constant when applied within the Nernst equation framework.

Standard Reduction Potential

A critical value in electrochemistry, the standard reduction potential (\( E^\circ \)), represents how easily a substance gains electrons (is reduced) under standard conditions. This value is intrinsic to each substance and determined experimentally. In the context of cell reactions, the more positive the reduction potential, the greater the substance's ability to gain electrons and act as an oxidizing agent.

The original exercise provided the reduction potentials for two half-reactions, which allowed for the calculation of the overall cell potential by subtracting the anode's potential from the cathode's.

Electrochemical Cell Reactions

Electrochemical cell reactions involve the movement of electrons from an oxidizing species (anode) to a reducing species (cathode). These redox reactions are the basis for batteries and various forms of electrochemical cells. Understanding the flow of electrons and how it relates to the cell potential and equilibrium is essential.

In our exercise, analyzing the half-reactions separately and then combining their potentials gives us insight into the overall reaction taking place within the electrochemical cell and enables the determination of the reaction's equilibrium constant.

Logarithmic Calculations in Chemistry

Logarithms often appear in chemistry, particularly in the context of the Nernst equation and pH calculations, because they allow us to handle the wide range of values that equilibrium constants and hydrogen ion concentrations can take on. The logarithmic scale converts multiplication and division into addition and subtraction, simplifying complex calculations.

By converting the equilibrium constant into a log value, as seen in the solution given, we can easily determine the magnitude of the reaction quotient and understand the directional propensity of a reaction under non-standard conditions. This conversion is especially useful when the numbers involved span several orders of magnitude, which is common in chemical equilibrium calculations.