Problem 102 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard reduction potentials of \(\mathrm{Zn}^{2} \mid \mathrm{Zn}\) and \(\mathrm{Cu}^{2}\) \(\mid \mathrm{Cu}\) are \(-0.76 \mathrm{~V}\) and \(+0.34 \mathrm{~V}\) respectively. What is the cell emf (in V) of the following cell? \((\mathrm{RT} / F=0.059)\) \(\mathrm{Zn}\left|\mathrm{Zn}^{2}(0.05 \mathrm{M}) \| \mathrm{Cu}^{21}(0.005 \mathrm{M})\right| \mathrm{Cu}\) (a) \(1.1295\) (b) \(1.0705\) (c) \(1.1\) (d) \(1.041\)

Step-by-Step Solution

Answer

The cell emf is 1.0705 V, option (b).

Step 1: Identify the Half-Cell Reactions

Identify the oxidation and reduction reactions from the electrode potentials given. The two reactions are: 1. Oxidation: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \) 2. Reduction: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)

Step 2: Determine Standard Cell Potential (E°cell)

Calculate the standard cell potential using the formula:\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]Given, \( E^\circ_{\text{reduction}} = +0.34 \text{ V} \) (for copper) and \( E^\circ_{\text{oxidation}} = -(-0.76) \text{ V} = +0.76 \text{ V} \) (for zinc since oxidation is reverse of reduction reaction), we find:\[ E^\circ_{\text{cell}} = 0.34 V - (-0.76 V) = 1.10 V \]

Step 3: Write the Nernst Equation

The Nernst equation is used to calculate the cell potential under non-standard conditions:\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \left(\frac{RT}{nF}\right) \ln{Q} \]Given \(\frac{RT}{F} = 0.059 \text{ V}\) and \(n = 2\) for the transfer of 2 electrons.

Step 4: Calculate the Reaction Quotient (Q)

The reaction quotient \(Q\) is calculated as:\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]Substitute the concentration values given in the problem:\[ Q = \frac{0.05}{0.005} = 10 \]

Step 5: Plug Values into the Nernst Equation

Use the values to find the cell emf (electromotive force):\[ E_{\text{cell}} = 1.10 - \frac{0.059}{2} \cdot \log{10} \]\[ E_{\text{cell}} = 1.10 - 0.0295 \times 1 \]\[ E_{\text{cell}} = 1.10 - 0.0295 = 1.0705 \text{ V} \]

Step 6: Select the Correct Option

The calculated cell emf is 1.0705 V. Therefore, the correct answer is option (b).

Key Concepts

Standard Reduction Potential

In electrochemistry, the standard reduction potential is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is represented by the symbol \(E^\circ\) and measured in volts (V). These potentials are determined under standard conditions: solute concentration of 1 M, pressure of 1 atmosphere, and a temperature of 25°C. The standard reduction potential is crucial for predicting the direction of redox reactions and determining the electromotive force (EMF) of an electrochemical cell.

A higher positive \(E^\circ\) value indicates a greater tendency to gain electrons, while a more negative value suggests a stronger tendency to lose them. In the given exercise, the standard reduction potential for zinc \((\text{Zn}^{2+} | \text{Zn})\) is \(-0.76\,\mathrm{V}\), while for copper \((\text{Cu}^{2+} | \text{Cu})\) it is \(+0.34\,\mathrm{V}\).

Knowing these values allows us to identify which species will undergo oxidation and which will experience reduction. In any galvanic cell, the species with a lower reduction potential (zinc in this case) typically undergoes oxidation while the one with a higher potential (copper) undergoes reduction.

Nernst Equation

The Nernst equation provides a way to calculate the cell potential for an electrochemical cell under non-standard conditions. While standard conditions are rarely met in practice, the Nernst equation helps us understand how changes in concentration impact the cell's electromotive force (EMF).

Here's the formula: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \left(\frac{RT}{nF}\right) \ln Q \] Where:

• \(E_{\text{cell}}\): Cell potential at non-standard conditions
• \(E^\circ_{\text{cell}}\): Standard cell potential
• \(R\): Universal gas constant \(8.314 \,\text{J/mol⋅K}\)
• \(T\): Temperature in Kelvin
• \(n\): Number of moles of electrons transferred in the reaction
• \(F\): Faraday constant \(96485\,\text{C/mol}\)
• \(Q\): Reaction quotient

For the given exercise, with \(RT/F = 0.059\,\text{V}\) and \(n = 2\) electrons, the equation adapts simplified form suitable for room temperature. It is critical to account for the reaction quotient when solving for \(E_{\text{cell}}\), as it reflects the current state of the reaction relative to equilibrium.

Reaction Quotient (Q)

The reaction quotient (\(Q\)) is a measure of the relative amounts of products and reactants present during a reaction at a particular point in time. It is used in the Nernst equation to gauge how far a reaction has proceeded from its standard state.

For a simple reaction \( a \text{A} + b \text{B} \rightleftharpoons c \text{C} + d \text{D} \), \(Q\) is expressed as:\[ Q = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \] Concentrations are expressed in molarity. In the context of the electrochemical cell \((\mathrm{Zn} | \mathrm{Zn}^{2+})\) || \((\mathrm{Cu}^{2+} | \mathrm{Cu})\), \(Q\) focuses on the ions in solution:

• Products: \([\text{Zn}^{2+}]\)
• Reactants: \([\text{Cu}^{2+}]\)

For the given exercise: \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.05}{0.005} = 10 \]
This result shows the concentration gradient present across the electrochemical cell, signifying non-standard conditions that affect the cell potential.

Electromotive Force (EMF)

Electromotive force, commonly referred to as EMF, is the voltage or potential difference that an electrochemical cell can provide under ideal conditions. It's essentially the driving force responsible for the flow of electrons from the anode to the cathode.

The standard EMF \(E^\circ_{cell}\) is calculated using the formula:\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]In this scenario, the standard potentials of the cathode \((\text{Cu}^{2+} | \text{Cu})\) and anode \((\text{Zn}|\text{Zn}^{2+})\) are \(+0.34\,\text{V}\) and \(-0.76\,\text{V}\) respectively, yielding:\[ E^\circ_{\text{cell}} = 0.34 - (-0.76) = 1.10 \,\text{V} \]This calculation tells us the maximum voltage the cell can generate when no current flows through it. However, under actual conditions, EMF differs due to factors like concentration variations, captured by the Nernst equation, which further refines the EMF to 1.0705 V in this exercise. Understanding EMF is crucial for predicting the cell's performance in real-world applications. An accurate EMF allows for insights into the efficiency and usability of potential electrochemical systems.