Problem 126 - 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 increase by 28.46 mV (Option a).

Step 1: Identify the equation and standard reduction potential

The half-reaction for the reduction of permanganate ion in acidic medium is given as \( \text{MnO}_{4}^{-} + 8 \text{H}^{+} + 5 \text{e}^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_{2}\text{O} \). The standard reduction potential for this process is \( E^{\circ} = 1.51 \text{ V} \).

Step 2: Use the Nernst Equation

The Nernst equation is given by: \[ E = E^{\circ} - \frac{RT}{nF} \ln Q \]where \( E^{\circ} \) is the standard reduction potential, \( R \) is the gas constant \( 8.314 \text{ J/mol K} \), \( T \) is the temperature in Kelvin (assume 298 K), \( n \) is the number of moles of electrons transferred (5 for this reaction), \( F \) is Faraday's constant \( 96485 \text{ C/mol} \), and \( Q \) is the reaction quotient.

Step 3: Write the reaction quotient (Q)

For the given reaction, the reaction quotient \( Q \) is expressed in terms of concentrations: \[ Q = \frac{1}{\text{[H}^{+}]^8 \text{[Mn}^{2+}]} \]By increasing the \( [H^{+}] \) concentration, \( Q \) will change.

Step 4: Determine the change in electrode potential

Initially, doubling \( [H^{+}] \) will affect \( Q \). Using the Nernst equation when \( [H^{+}] \) is doubled, \[ Q_1 = \frac{1}{(2[H^{+}])^8 \times [Mn^{2+}]} \]\[ \Delta E = -\frac{RT}{nF} \ln \frac{Q_1}{Q_0} \]Simplifying, since \( Q_0 = \frac{1}{[H^{+}]^8} \), \[ \Delta E = -\frac{8 \times 8.314 \times 298}{5 \times 96485} \ln (2) \]This equation will yield a change of approximately \( +0.02846 \text{ volts} = +28.46 \text{ mV} \).

Step 5: Conclusion on the change of potential

Since the calculated electrode potential change is an increase by \( 28.46 \text{ mV} \), option (a) is correct. Increasing the \( [H^{+}] \) concentration doubles the term, thereby increasing the electrode potential as calculated.

Key Concepts

Nernst Equation

The Nernst Equation is a vital formula in electrochemistry that helps us understand how the electrode potential of a half-cell can change with varying concentrations of ions. This equation gives insight into the thermodynamic properties of an electrochemical cell beyond standard conditions. It tells you how the potential will shift if the conditions are not standard, which often occurs in a laboratory setting.

The equation is expressed as:
\[ E = E^{\circ} - \frac{RT}{nF} \ln Q \]
Where:

• \(E\) is the electrode potential of the half-cell at any condition
• \(E^{\circ}\) is the standard electrode potential
• \(R\) is the universal gas constant (8.314 J/mol K)
• \(T\) is the temperature in Kelvin
• \(n\) is the number of moles of electrons transferred in the reaction
• \(F\) is Faraday's constant (96,485 C/mol)
• \(Q\) is the reaction quotient, reflecting the concentrations of the reactants and products

By applying the Nernst Equation, you can observe how changes to the concentrations of ions, such as an increase in \([H^{+}]\) in the provided exercise, will impact the overall cell potential. Here, doubling \([H^{+}]\) resulted in an increase of the cell potential by approximately 28.46 mV.

Reaction Quotient

The reaction quotient, denoted \(Q\), is a measure of the relative concentrations of reactants and products during a reaction at any point in time. For electrochemical reactions, \(Q\) plays a crucial role in determining the direction and extent of the reaction. It is similar to the equilibrium constant \(K\), yet applicable at non-equilibrium conditions.

For the manganese reaction exercise given:
\[ \text{MnO}_{4}^{-} + 8\text{H}^{+} + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_{2}\text{O} \]
The reaction quotient \(Q\) is expressed as:
\[ Q = \frac{1}{[\text{H}^{+}]^8 [\text{Mn}^{2+}]} \]
This equation reflects the concentrations of products over reactants.

The change in \([H^{+}]\) concentration significantly impacts \(Q\). For example, doubling \([H^{+}]\) concentration reduces \(Q\), which in turn affects the Nernst equation's results. It suggests how far a reaction is from reaching equilibrium. A lower \(Q\) implies a reaction pushed more toward the reactants, signaling an increase in potential energy as shown in the exercise scenario.

Standard Reduction Potential

Standard Reduction Potential, denoted \(E^{\circ}\), is a measure of the tendency of a chemical species to gain electrons and be reduced. It is measured under standard conditions, typically 1 M concentration, 1 atm pressure, and a temperature of 25°C (298 K). This value provides a basis for predicting the spontaneity and direction of redox reactions.

The provided chemical reaction in the scenario:
\[ \text{MnO}_{4}^{-} + 8 \text{H}^{+} + 5 \text{e}^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_{2}\text{O} \]
Has a standard reduction potential \(E^{\circ}\) of 1.51 V.

• An \(E^{\circ}\) value that is more positive indicates a greater tendency to accept electrons and become reduced.
• Comparatively, negative values suggest a lower tendency to be reduced and potentially favor oxidation reactions.

In the exercise, this potential is fundamental for applying the Nernst Equation to calculate how changes, like a doubling of \([H^{+}]\), influence the overall cell potential. Understanding \(E^{\circ}\) facilitates comprehending how electric energy and cell reactions interplay.