Problem 68 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The standard reduction potential for the half-cell \(\mathrm{NO}_{3}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e} \rightarrow \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}\) is \(0.78 \mathrm{~V}\) (i) Calculate the reduction potential in \(8 \mathrm{M} \mathrm{H}^{+}\) (ii) What will be the reduction potential of the half-cell in a neutral solution? Assume all the other species to be at unit concentration.

Step-by-Step Solution

Answer

(i) 0.8571 V (ii) 0.4202 V

Step 1: Understand the Nernst Equation

The Nernst equation is used to calculate the reduction potential under non-standard conditions. It is given by: \[E = E^{\circ} - \frac{RT}{nF} \ln Q\]where:- \(E\) is the reduction potential,- \(E^{\circ}\) is the standard reduction potential,- \(R\) is the ideal gas constant \(8.314 \text{ J mol}^{-1} \text{K}^{-1}\),- \(T\) is the temperature in Kelvin,- \(n\) is the number of moles of electrons transferred in the reaction,- \(F\) is the Faraday constant \(96485 \text{ C mol}^{-1}\),- \(Q\) is the reaction quotient.

Step 2: Identify the Reaction Components and Values

For the given half-cell reaction:\[\mathrm{NO}_{3}^{-} + 2 \mathrm{H}^{+} + \mathrm{e}^- \rightarrow \mathrm{NO}_{2} + \mathrm{H}_{2} \mathrm{O}\]we have the standard reduction potential \(E^{\circ} = 0.78 \text{ V}\), and \(n = 1\) since one electron is involved.

Step 3: Write the Reaction Quotient for the Nernst Equation

In the reaction quotient \(Q\), only the concentrations of aqueous species and gases should be included:\[Q = \frac{1}{\left[ \mathrm{H}^{+} \right]^2 \left[ \mathrm{NO}_{3}^{-} \right]}\]Given conditions: - For part (i): \(\left[ \mathrm{NO}_{3}^{-} \right] = 1 M\), \(\left[ \mathrm{H}^{+} \right] = 8 M\). - For part (ii) in neutral solution: \(\left[ \mathrm{H}^{+} \right] = 1 \times 10^{-7} M\).

Step 4: Calculate the Potential in 8 M H⁺

Substitute the values into the Nernst equation for 8 M \(\mathrm{H}^{+}\) and simplify:\[E = 0.78 - \frac{0.0257}{1} \log\left(\frac{1}{8^2 \times 1}\right)\]\[E = 0.78 - 0.0257 \log\left(\frac{1}{64}\right)\]\[E \approx 0.78 + 0.0771\]\[E \approx 0.8571 \text{ V}\]

Step 5: Calculate the Potential in Neutral Solution

Substitute the neutral condition into the Nernst equation:\[E = 0.78 - \frac{0.0257}{1} \log\left(\frac{1}{(1 \times 10^{-7})^2 \times 1}\right)\]\[E = 0.78 - 0.0257 \log\left(\frac{1}{10^{-14}}\right)\]\[E \approx 0.78 - 0.3598\]\[E \approx 0.4202 \text{ V}\]

Key Concepts

Nernst equation

The Nernst equation is an essential concept in electrochemistry. It helps us determine the reduction potential of an electrochemical cell under non-standard conditions. In simple terms, standard conditions assume all involved species are at 1 M concentration, but these conditions rarely reflect real-world scenarios. The equation for the Nernst equation is:\[E = E^{\circ} - \frac{RT}{nF} \ln Q\]Here's a breakdown of its components:

• \(E\): Actual reduction potential of the cell at non-standard state, usually reported in volts (V).
• \(E^{\circ}\): Standard reduction potential under standard conditions, also in volts.
• \(R\): Ideal gas constant, valued at \(8.314 \text{ J mol}^{-1} \text{K}^{-1}\).
• \(T\): Temperature in Kelvin, usually taken as 298 K for standard calculations unless otherwise specified.
• \(n\): Number of moles of electrons exchanged in the electrode reaction.
• \(F\): Faraday constant, \(96485 \text{ C mol}^{-1}\), representing the total charge of one mole of electrons.
• \(Q\): Reaction quotient which indicates the ratio of product activities to reactant activities at any point in time.

By calculating the value of \(E\) using the Nernst equation, we gain insights into the behavior and efficiency of electrochemical reactions as conditions change.

Standard reduction potential

The standard reduction potential, represented as \(E^{\circ}\), is a crucial aspect of predicting the feasibility and direction of redox reactions.
It essentially reflects a substance's ability to gain electrons, acting as a measure of its oxidizing ability.
The higher the \(E^{\circ}\), the greater the tendency for the species to be reduced. Standard reduction potentials are measured under standardized conditions:

• All reactants and products are at an activity level under standard state, often 1 M concentration.
• The temperature is typically 298 K (25°C).
• All gases, if involved, are at 1 atm pressure.

This value allows us to compare different electrodes and predict which species can potentially undergo reduction or oxidation in an electrochemical cell.For example, in our original exercise, the standard reduction potential for the reaction \(\mathrm{NO}_{3}^{-} + 2\mathrm{H}^{+} + \mathrm{e}^- \rightarrow \mathrm{NO}_{2} + \mathrm{H}_{2}\mathrm{O}\) is given as 0.78 V. This information forms the baseline from which we apply the Nernst equation to calculate the change under different concentrations of reactants or products.

Reaction quotient

The reaction quotient, \(Q\), is another central concept when using the Nernst equation to determine reduction potentials at non-standard conditions.
It provides an instantaneous measure of the concentration of products relative to reactants in a given reaction mixture.
The expression for \(Q\) is similar to that used for calculating the equilibrium constant but provides a snapshot at any moment:

• Consider only the concentrations or partial pressures of reactants and products that are aqueous or gaseous.
• For the given half-cell reaction, \(\mathrm{NO}_{3}^{-} + 2 \mathrm{H}^{+} + \mathrm{e}^- \rightarrow \mathrm{NO}_{2} + \mathrm{H}_{2}\mathrm{O}\), the expression for \(Q\) includes those species whose concentrations change: \[Q = \frac{1}{[\mathrm{H}^{+}]^2 [\mathrm{NO}_{3}^{-}]}\]

In the exercise, \(Q\) changes based on the concentration of \(\mathrm{H}^{+}\):

• For part (i), where \(\mathrm{H}^{+}\) is 8 M, \(Q\) is calculated using these values, thus impacting the potential calculated through the Nernst equation.
• For part (ii), in a neutral solution, \([\mathrm{H}^{+}] = 1 \times 10^{-7} M\), significantly impacting \(Q\) and subsequently, the reduction potential \(E\).

Understanding \(Q\) and how it fluctuates with changing concentrations is key to mastering electrochemical predictions and applications.