Problem 75 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The EMF of a cell corresponding to the reaction: \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}+(0.1 \mathrm{M})+\mathrm{H}_{2}(\mathrm{~g})(1 \mathrm{~atm})\) is \(0.28\) volt at \(25^{\circ} \mathrm{C}\). Write the half-cell reactions and calculate the \(\mathrm{pH}\) of the solution at the hydrogen electrode. \(E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{volt} ; E_{\mathrm{H}^{+} / \mathrm{H}_{2}}^{0}=0\)

Step-by-Step Solution

Answer

The pH of the solution is 8.62.

Step 1: Identifying Half Reactions

For the given reaction, the first half-cell reaction involves zinc: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-} \). The second half-cell reaction involves hydrogen: \( 2\text{H}^{+} + 2e^{-} \rightarrow \text{H}_{2} \).

Step 2: Calculating Standard Cell Potential

The standard cell potential, \( E^{\circ}_{\text{cell}} \), is calculated using the standard electrode potentials of the half reactions: \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0 - (-0.76) = 0.76 \text{ V} \).

Step 3: Using the Nernst Equation

The Nernst equation for the cell is: \[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF}\ln(Q) \]. At \( 25^{\circ} \text{C} \), this becomes: \[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n}\log(Q) \], where \( n = 2 \) and \( Q = \frac{[\text{Zn}^{2+}]}{[\text{H}^{+}]^2} \).

Step 4: Substituting Known Values in Nernst Equation

Given \( E_{\text{cell}} = 0.28 \text{ V} \), substitute the values: \[ 0.28 = 0.76 - \frac{0.0591}{2}\log\left(\frac{0.1}{[\text{H}^{+}]^2}\right) \].

Step 5: Solving for \([\text{H}^+]\)

Rearrange the equation to solve for \( [\text{H}^+] \): \[ 0.76 - 0.28 = \frac{0.0591}{2}\log\left(\frac{0.1}{[\text{H}^{+}]^2}\right) \]. This simplifies to \[ 0.48 = \frac{0.0591}{2}\log\left(\frac{0.1}{[\text{H}^{+}]^2}\right) \].

Step 6: Calculating Logarithmic Expression

Solve for the logarithmic term \( \log\left(\frac{0.1}{[\text{H}^{+}]^2}\right) \): \[ \log\left(\frac{0.1}{[\text{H}^{+}]^2}\right) = \frac{2 \times 0.48}{0.0591} = 16.24 \].

Step 7: Determining \([\text{H}^+]\) and \(\text{pH}\)

Since \( \log (\frac{0.1}{[\text{H}^{+}]^2}) = 16.24 \), we solve for \( \log([\text{H}^+]^2) = 16.24 - \log(0.1) = 16.24 + 1 = 17.24 \). Thus, \( [\text{H}^+]^2 = 10^{-17.24} \) and \( [\text{H}^+] = 10^{-8.62} \). Therefore, \( \text{pH} = 8.62 \).

Key Concepts

Nernst Equation

To understand how the Nernst Equation is applied in calculating cell potentials, let's begin with its general form: The Nernst Equation is expressed as:\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF}\ln(Q) \]where:

• \( E_{\text{cell}} \) is the electromotive force (EMF) of the cell;
• \( E^{\circ}_{\text{cell}} \) is the standard cell 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 cell reaction;
• \( F \) is the Faraday constant (approximately 96485 C/mol);
• \( Q \) is the reaction quotient.

At room temperature (25°C), the equation is simplified to:\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n}\log(Q) \]This allows us to compute the cell potential at non-standard conditions.

The Nernst Equation is valuable in electrochemistry because it connects the chemical reaction with its electrical output, predicting how changes in concentration affect the EMF.

Half-cell Reactions

In an electrochemical cell, we break down the overall cell reaction into half-cell reactions. Each half-reaction takes place in different compartments called half-cells.

For the given reaction:- **Oxidation Half-Reaction:** \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-} \) - Here, zinc metal (\( \text{Zn} \)) loses electrons to form zinc ions (\( \text{Zn}^{2+} \)). This is the anode reaction where oxidation occurs.
- **Reduction Half-Reaction:** \( 2\text{H}^{+} + 2e^{-} \rightarrow \text{H}_{2} \) - Protons (\( \text{H}^+ \)) in solution gain electrons to form hydrogen gas (\( \text{H}_2 \)). This is the cathode reaction where reduction takes place.

Understanding half-cell reactions helps us determine which substances are oxidized and which are reduced in the cell, and it is crucial for correctly applying the Nernst Equation later on.

pH Calculation

The hydrogen ion concentration is key to determining the pH of a solution using the relation:
\[ \text{pH} = -\log_{10}([\text{H}^+]) \]Calculating the pH involves finding the concentration of \( [\text{H}^+] \) ions in the solution.In the given exercise, after applying the Nernst Equation to find \( [\text{H}^+] \), we determined:- \( [\text{H}^+] = 10^{-8.62} \)
As such, the calculated pH is 8.62.

• Important to remember: a pH below 7 indicates acidity, while above 7 indicates basicity.
• Since the pH is 8.62, the solution is slightly basic.
• This calculation shows how altering conditions (like gas pressure and ion concentration) in galvanic cells can affect pH balance.

Standard Cell Potential

The standard cell potential, \( E^{\circ}_{\text{cell}} \), represents the voltage difference under standard conditions (1 M concentrations, 1 atm pressure, 25°C) between two electrodes in an electrochemical cell.

Given in the exercise:- **Zinc Reaction Potential:** \( E^{\circ}_{\text{Zn}^{2+} / \text{Zn}} = -0.76 \text{ V} \)- **Hydrogen Reaction Potential:** \( E^{\circ}_{\text{H}^{+} / \text{H}_{2}} = 0 \text{ V} \)
The standard cell potential is calculated by subtracting the anode potential from the cathode potential:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0 \text{ V} - (-0.76 \text{ V}) = 0.76 \text{ V} \]This value tells us the maximum potential difference achievable under ideal conditions.

• Larger standard potential implies a stronger tendency for reduction at the cathode.
• A positive \( E^{\circ}_{\text{cell}} \) generally indicates a spontaneous reaction under standard conditions.