Problem 125 - A Complete Resource Book in Chemistry for JEE Main

Question

In acidic medium \(\mathrm{MnO}_{4}^{-}\)is an oxidizing agent \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} .\) If \(\mathrm{H}^{+}\)ion concentration is doubled, electrode potential of the half cell \(\mathrm{MnO}_{4}^{-}, \mathrm{Mn}^{2+} / \mathrm{Pt}\) will (a) increase by \(28.46 \mathrm{mV}\) (b) decrease by \(28.46 \mathrm{mV}\) (c) increase by \(14.23 \mathrm{mV}\) (d) decrease by \(142.30 \mathrm{mV}\)

Step-by-Step Solution

Answer

The electrode potential will decrease by 28.46 mV, hence the answer is (b).

Step 1: Understanding Nernst Equation

The Nernst equation relates the reduction potential of a half-cell at any point to its standard electrode potential, temperature, and activities or concentrations of the species undergoing reduction and oxidation. For the half-cell reaction, the Nernst equation is: \[ E = E^0 - \frac{RT}{nF} \ln\left( \frac{[\text{products}]}{[\text{reactants}]} \right) \] where \( E \) is the electrode potential, \( E^0 \) is the standard electrode potential, \( R \) is the gas constant \((8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1})\), \( T \) is temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, and \( F \) is Faraday's constant \((96485 \, \text{C} \, \text{mol}^{-1})\).

Step 2: Applying Nernst Equation to MnO_4^- + 8H^+ + 5e^- 0 Mn^{2+} + 4H_2O

The reaction given involves ions, so the concentration terms apply directly. Here, \([\text{MnO}_4^-]\) and \([\text{Mn}^{2+}]\) are involved, as well as \([\text{H}^+]\). The Nernst equation for this half-cell at 298 K becomes: \[ E = E^0 - \frac{0.059}{5} \log\left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][\text{H}^+]^8} \right) \]

Step 3: Assessing Impact of Doubling [H^+]

When \([\text{H}^+]\) doubles, its impact on the potential \( E \) must be calculated. Since \([\text{H}^+]^8\) is in the denominator, doubling \([\text{H}^+]\) causes \([\text{H}^+]^8\) to increase 256 times \((2^8)\), changing the logarithmic term. This adjustment in the logarithmic term will affect the electrode potential.

Step 4: Calculate Change in Electrode Potential

The change in electrode potential \(\Delta E\) due to the change in \([\text{H}^+]\) is given by: \[ \Delta E = -\frac{0.059}{5} \log(2^8) \] \[ = -\frac{0.059}{5} \times 8 \log(2) \] \[ = -\frac{0.472}{5} \times 0.3010 \approx -0.02846 \text{ V} \] This result corresponds to a decrease of 28.46 mV.

Key Concepts

Electrode Potential

Electrode potential is a crucial concept in electrochemistry. It measures the ability of a chemical species to be reduced, which means gaining electrons within a reaction. Electrode potential values are expressed in volts and can indicate how likely a substance is to undergo a reduction.
In the exercise, electrode potential is influenced by the concentration of ions like [H extsuperscript{+}]. According to the Nernst equation, changes in ion concentration can affect the potential in a half-cell. The equation helps us quantify how these changes alter the potential. If concentrations of reactants or products shift, the electrode potential adjusts too. It's fascinating because this shows the dynamic nature of chemical systems under different conditions.
Understanding electrode potential allows scientists to predict and control chemical reactions, such as in batteries, where knowing which species will be reduced or oxidized is crucial for generating electricity.

Oxidizing Agent

An oxidizing agent is a substance that promotes oxidation by accepting electrons from another substance during a chemical reaction. In simpler terms, it's the electron-stealing party in a chemical reaction.
In the exercise, \\(\text{MnO}_4^-\) acts as an oxidizing agent in the acidic medium. This means it gains electrons to form \(\text{Mn}^{2+}\), while simultaneously facilitating the reduction of the other species in the reaction.

• Oxidizing agents are critical in many processes, ranging from industrial applications to biological systems.
• Common properties include having a high affinity for electrons and a positive electrode potential.

Knowing the role of an oxidizing agent provides insights into how chemical processes are powered, such as those in breathing, metabolism, and as seen in the exercise, in electrochemical cells.

Half-Cell Reaction

A half-cell reaction represents one part of an electrochemical cell, involving either the oxidation or reduction process separately. These reactions are at the heart of battery technology and electrochemical methods. In every electrochemical cell, there are two half-cells: one for oxidation and another for reduction. The exercise focuses on the reduction of \(\text{MnO}_4^-\) in acidic conditions, where it turns into \(\text{Mn}^{2+}\).
Half-cell reactions are essential to understanding cell potential and how a battery produces voltage. They allow us to calculate the potential using the Nernst equation, which considers concentrations as shown in the example.

• Each half-cell is written with the reactants and products that it includes.
• Combining two half-cell reactions provides a complete picture of the electrochemical process.

Understanding half-cell reactions provides the basis for studying energy conversion and fuel cells, emphasizing their importance in sustainable technologies.