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 decreases by \( 28.46 \mathrm{mV} \).

Step 1: Understanding the Nernst Equation

The Nernst equation is used to calculate the electrode potential of a half-cell under non-standard conditions. The equation is \( E = E^0 - \left( \frac{RT}{nF} \right) \ln Q \), where \( E \) is the electrode potential, \( E^0 \) is the standard electrode potential, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons transferred in the reaction, \( F \) is Faraday's constant, and \( Q \) is the reaction quotient. In this equation, \( Q = \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][H^+]^8} \).

Step 2: Simplifying with Given Conditions

For this specific half-reaction under acidic conditions, we know that the number of electrons transferred \( n = 5 \), and if concentration changes only affect \([H^+]\), then the Nernst equation simplifies to \( E = E^0 - \left( \frac{RT}{5F} \right) \ln \left( \frac{1}{{[H^+]}^8} \right) \). Doubling \([H^+]\) means \( Q \) will change as \( \ln \left( 2^8 \right) \), which is \( 8 \ln 2 \).

Step 3: Calculating Potential Change Due to Concentration Change

The change in electrode potential \( \Delta E \) can be expressed as \( \Delta E = - \frac{RT}{nF} \times 8 \ln 2 \). Substituting \( R = 8.314 \: \text{J/mol K} \), \( T = 298 \:K \), \( F = 96500 \: \text{C/mol} \), and using \( \ln 2 \approx 0.693 \), we calculate: \( \Delta E = - \frac{8.314 \times 298}{5 \times 96500} \times 8 \times 0.693 = -0.02846 \: V \). Multiply by 1000 to convert to mV: \( \Delta E = -28.46 \: mV \).

Step 4: Interpreting Results

Since \( \Delta E \) is negative, this indicates a decrease in electrode potential. Thus, doubling the concentration of \( H^+ \) results in a decrease of the electrode potential by 28.46 mV.

Key Concepts

Electrode Potential

Electrode potential is a fundamental concept in electrochemistry that describes the ability of a half-cell in an electrochemical cell to gain or lose electrons. Think of it almost like the cell's desire to undergo chemical change. When a redox reaction occurs, electrons are exchanged, altering the chemical state of the components involved. The electrode potential quantifies these changes in terms of electric potential difference.

Key points about electrode potential:

• Standard Electrode Potential (E⁰): This is the potential measured under standard conditions (1 M concentration, 1 atm pressure, 25°C). It acts like a baseline to compare different reactions.
• Actual Electrode Potential (E): This is what we calculate when conditions aren't standard, using the Nernst Equation. It accounts for real-world variables like concentration changes.

The Nernst Equation is essential for determining the actual electrode potential. By adjusting for concentration changes (like the doubling of \( H^+ \) ions), the Nernst Equation helps us understand how such variations influence the overall reaction potential.

Oxidizing Agents

Oxidizing agents are substances that gain electrons during a chemical reaction, causing another substance to be oxidized. In simple terms, they take electrons away. In the context of the exercise, \( \text{MnO}_4^- \) acts as an oxidizing agent in the acidic medium, accepting electrons to transform into \( \text{Mn}^{2+} \).

Characteristics of oxidizing agents:

• Electron Acceptance: They readily accept electrons from other substances. This is key to driving the oxidation process.
• Change in Oxidation State: Typically, they undergo a reduction themselves, meaning their oxidation state decreases. For example, \( \text{MnO}_4^- \) with manganese in the +7 state, reduces to +2 in \( \text{Mn}^{2+} \).
• Influence on Electrode Potential: The likelihood of gaining electrons and oxidizing another compound directly influences the electrode potential established by the reaction.

Understanding the role of oxidizing agents is crucial in predicting and explaining the direction of redox reactions.

Acidic Medium

An acidic medium contributes significantly to the behavior of electrochemical cells, influencing reactions involving ions and redox processes. It acts as a facilitator, often providing protons (\( H^+ \) ions) necessary for specific reactions.

Importance of acidic medium in electrochemistry:

• Provision of Protons: An acidic solution supplies necessary \( H^+ \) ions which are often involved in balancing redox reactions. In our given exercise, \( 8 H^+ \) ions participate, facilitating the conversion of \( \text{MnO}_4^- \) to water and \( \text{Mn}^{2+} \).
• Effect on Electrode Potential: The concentration of \( H^+ \) ions directly impacts the electrode potential, as seen with the Nernst Equation adjustment. For instance, increasing \( H^+ \) concentration in the exercise results in a decrease in potential.
• pH Influence: The presence of an acidic medium (lower pH) can also affect the solubility and stability of reaction products, further impacting the reaction's electrode potential.

Understanding how an acidic medium affects chemical reactions helps elucidate the changes in electrode potential therein.