Problem 16 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

By how much would the oxidizing power of \(\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}\) couple change if the \(\mathrm{H}^{\circ}\) ions concentration is decreased 100 times at \(25^{\circ} \mathrm{C}\) ? (a) increases by \(189 \mathrm{mV}\) (b) decreases by \(189 \mathrm{mV}\) (c) will increase by \(19 \mathrm{mV}\) (d) will decrease by \(19 \mathrm{mV}\)

Step-by-Step Solution

Answer

The oxidizing power of the \(\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}\) couple decreases by \(189 \mathrm{mV}\).

Step 1: Understand the Concept

The problem involves a redox couple \(\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}\) and the effect of the hydrogen ion \(\mathrm{H}^{+}\) concentration on its oxidizing power. The Nernst equation is the key to relate the potential of the redox couple with the concentration of ions involved in the reaction.

Step 2: Write Down the Nernst Equation

The Nernst equation for a redox reaction at temperature \(25^\circ \mathrm{C}\) (298 K) is given by: \[E = E^\circ - \frac{0.0591}{n} \log Q\] Where \(E^\circ\) is the standard electrode potential, \(n\) is the number of moles of electrons transferred in the redox reaction, and \(Q\) is the reaction quotient.

Step 3: Analyze the Redox Reaction

The balanced redox reaction involving \(\mathrm{MnO}_{4}^{-}\) and \(\mathrm{Mn}^{2+}\) in acidic solution is: \[\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\] In this reaction, \(n = 5\), since 5 moles of electrons are transferred.

Step 4: Calculate the Change in Oxidizing Power

The change in oxidizing power is associated with the change in potential due to the change in \(\mathrm{H}^{+}\) ion concentration. By increasing \(\mathrm{H}^{+}\) concentration, the \(Q\) value decreases, which increases the potential \(E\). Conversely, by decreasing \(\mathrm{H}^{+}\) concentration 100 times, the reaction quotient \(Q\) increases. Using the Nernst equation, we can find the change in \(E\) as: \[\Delta E = -\frac{0.0591}{5}\log\left(\frac{Q_{\text{new}}}{Q_{\text{old}}}\right)\]. Since \(\mathrm{H}^{+}\) concentration decreases by 100 times, Q will increase by 100^8 = 10^{16} times. \[\Delta E = -\frac{0.0591}{5}\log(10^{16})\] \[\Delta E = -\frac{0.0591}{5} \cdot 16\] \[\Delta E = -0.1892 \mathrm{V}\] or \[\Delta E = -189.2 \mathrm{mV}\].

Step 5: Determine the Direction of the Change

Since the \(\Delta E\) is negative, this means that the oxidizing power of the \(\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}\) couple decreases due to the dilution of \(\mathrm{H}^{+}\) ions concentration.

Key Concepts

Redox Couple

A redox couple is composed of two species that can undergo a reduction-oxidation (redox) reaction among themselves. It consists of one reductant, which can donate electrons, and one oxidant, which can accept electrons. The redox couple in our exercise, \(\mathrm{MnO}_4^-/\mathrm{Mn}^{2+}\), involves the permanganate ion (oxidant) and the manganese ion (reductant).

In the context of electrochemistry, each redox couple has an associated 'half-reaction' that takes place at an electrode during electrolysis or galvanic cell operations. Changes in concentration of the reactants or products of these half-reactions can alter the electrode potential, which is a measure of the tendency of the redox couple to acquire electrons and thereby be reduced.

Oxidizing Power

The oxidizing power of a substance refers to its ability to accept electrons in a redox reaction. Strong oxidizing agents have a high affinity for electrons and hence a high oxidizing power. This characteristic is critical in determining the direction and spontaneity of chemical reactions. In practical terms, the greater the oxidizing power of a species, the more likely it is to oxidize other species.

In our exercise, the oxidizing power of \(\mathrm{MnO}_4^-\) is quantitatively expressed by the electrode potential. When the concentration of \(\mathrm{H}^+\) ions is decreased, the oxidizing power of the \(\mathrm{MnO}_4^-/\mathrm{Mn}^{2+}\) couple changes as a result of the altered electrode potential, as predicted by the Nernst equation.

Electrode Potential

Electrode potential is the voltage developed between an electrode and its surrounding electrolyte. This potential arises from the tendency of the electrode to lose or gain electrons compared to the potential of a standard hydrogen electrode (SHE), which is set at 0 volts. Electrode potential can be either positive or negative, indicating a tendency to act as an oxidizing agent or reducing agent, respectively.

Measurement of electrode potential requires a reference electrode and understanding this value is essential in predicting battery performance, corrosion rates, and the feasibility of redox reactions. In electrochemical cells, the difference in electrode potentials between two half-cells is responsible for the voltage of the cell. In our exercise, the Nernst equation allows us to calculate how changes in ion concentration affect the electrode potential of the \(\mathrm{MnO}_4^-/\mathrm{Mn}^{2+}\) redox couple.

The reaction quotient (Q) is a measure of the relative amounts of products and reactants present during a reaction at a given moment. It is calculated using the same formula as the equilibrium constant \(K\), but unlike \(K\), which is used at equilibrium, \(Q\) can be used at any point during a reaction. The magnitude of \(Q\) is instrumental in determining the direction in which a reaction will proceed to reach equilibrium.

The Nernst equation incorporates \(Q\) to relate concentrations of reactants and products to the electrode potential. This means that any change in the concentrations of species involved in the redox couple will impact the electrode potential by shifting \(Q\). As seen in the exercise, changing \(\mathrm{H}^+\) concentration affects \(Q\) and consequently the potential and oxidizing power of the redox couple.