Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

Given that \(E^{\circ} \quad\left(\mathrm{Zn}^{2+} / \mathrm{Zn}\right)=-0.763 \mathrm{~V}\) and \(\mathrm{E}^{\circ}\left(\mathrm{Cd}^{2+} / \mathrm{Cd}\right)=-0.403 \mathrm{~V}\), the emf of the following cell \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{a}=0.04) \| \mathrm{Cd}^{2+}(\mathrm{a}=0.2)\right| \mathrm{Cd}\) is given by (a) \(E=+0.36+[0.059 / 2][\log (0.2 / 0.004)]\) (b) \(E=-0.36+[0.059 / 2][\log (0.2 / 0.004)]\) (c) \(E=+0.36+[0.059 / 2][\log (0.004 / 0.2)]\) (d) \(E=-0.36+[0.059 / 2][\log (0.004 / 0.2)]\)

Step-by-Step Solution

Answer

(a) is correct because \( E = +0.36 - [0.059/2]\log(0.2) \).

Step 1: Identify Cell Reactions

First, determine the cell reaction from the half-reactions of the electrodes. The anode (oxidation) reaction is \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \) with \( E^{\circ}_{\text{anode}} = -0.763 \, \text{V} \). The cathode (reduction) reaction is \( \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \) with \( E^{\circ}_{\text{cathode}} = -0.403\, \text{V} \). Exchange of electrons indicates the overall cell reaction is: \( \text{Zn} + \text{Cd}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cd} \).

Step 2: Calculate Standard EMF (E°cell)

The standard EMF for the cell, \( E^{\circ}_{\text{cell}} \), can be calculated using the formula: \[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]Substitute in the given values:\[ E^{\circ}_{\text{cell}} = (-0.403) - (-0.763) = 0.36\, \text{V} \]

Step 3: Apply the Nernst Equation

The Nernst equation to calculate the cell's emf is: \[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log Q \]where \( n = 2 \) (number of electrons transferred) and \( Q \) is the reaction quotient: \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]} = \frac{0.04}{0.2} \]

Step 4: Calculate Reaction Quotient (Q)

The reaction quotient, \( Q \), is calculated as:\[ Q = \frac{0.04}{0.2} = 0.2 \]

Step 5: Substitute into Nernst Equation

Substitute \( E^{\circ}_{\text{cell}} = 0.36 \) V, \( n = 2 \), and \( Q = 0.2 \) into the Nernst equation:\[ E_{\text{cell}} = 0.36 - \frac{0.059}{2} \cdot \log(0.2) \]Since \( \log(0.2) = \log\left(\frac{0.2}{1}\right) = \log(0.2) \). Simplifying,\( \frac{0.059}{2}\log(0.2) \) gives a negative contribution. Use the given options to find the match.

Key Concepts

Nernst equation

The Nernst equation is an essential tool in electrochemistry. It helps in determining the electromotive force (EMF) of a galvanic cell under non-standard conditions. This equation allows us to see how the cell potential changes with the concentration of reactants and products.
The equation is given by:

• \[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log Q \]

Where:

• \( E^{\circ}_{\text{cell}} \) is the standard electrode potential.
• \( n \) is the number of electrons transferred in the reaction.
• \( Q \) is the reaction quotient.

The key takeaway from the Nernst equation is understanding its components. The equation adjusts the standard cell potential to reflect the actual concentrations or activities of the cell reactants and products.

Standard electrode potential

Standard electrode potential, denoted as \( E^{\circ} \), is the voltage that measures the propensity of a chemical species to be reduced. This is a benchmark measurement taken under standard conditions - where all solutions are at a concentration of 1 mol/L, gases at 1 atm pressure, and the temperature is 25°C.
For any half-reaction, the standard electrode potential can be expressed as:

• Reduction Reaction: \( ext{Cd}^{2+} + 2e^- \rightarrow ext{Cd}, \, E^{\circ} = -0.403 \, \text{V} \)
• Oxidation Reaction: \( ext{Zn} \rightarrow ext{Zn}^{2+} + 2e^-, \, E^{\circ} = -0.763 \, \text{V} \)

The standard cell potential \( E^{\circ}_{\text{cell}} \) is calculated by subtracting the anode potential from the cathode potential. This provides a standard value to compare different electrochemical reactions.

Galvanic cell

A galvanic cell, also known as a voltaic cell, converts chemical energy into electrical energy through a spontaneous redox reaction. These cells consist of two half-cells, each with an electrode and an electrolyte solution. In a typical setup, one half-cell undergoes oxidation (loss of electrons) while the other undergoes reduction (gain of electrons).
In our zinc-cadmium cell example:

• The \( \text{Zn} \) serves as the anode where oxidation occurs: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \).
• \( \text{Cd}^{2+} \) ions are reduced at the cathode: \( \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \).

Electrons flow from the zinc electrode (anode) to the cadmium electrode (cathode), creating an electrical current. The cell potential or EMF is directly measured by the difference in potential between the two electrodes.

Reaction quotient

The reaction quotient, represented by \( Q \), is a dimensionless number. It provides an ongoing measure of the relative activities or concentrations of the products and reactants for any reaction at a given moment in time, and helps in understanding how far or close the reaction is from reaching equilibrium.
It is calculated as follows:

• For the zinc-cadmium galvanic cell: \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]} = \frac{0.04}{0.2} = 0.2 \]

The smaller the value of \( Q \), the further the reaction is from equilibrium. In the context of the Nernst equation, \( Q \) helps determine the correction factor required for the cell potential, as it incorporates the effect of any deviation from standard conditions of reactant and product concentrations.