Problem 196 - A Complete Resource Book in Chemistry for JEE Main

Question

The cell \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu} \mathrm{E}_{\mathrm{cell}}^{\circ}=1.10\) V), was allowed to be completely discharged at 298 \(K\). The relative concentration of \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cu}^{2+}\left(\left[\mathrm{Zn}^{2+}\right] /\right.\) \(\left.\left[\mathrm{Cu}^{2+}\right]\right)\) is [2007] (a) \(37.3\) (b) \(10^{37.3}\) (c) \(9.65 \times 10^{4}\) (d) antilog \((24.08)\)

Step-by-Step Solution

Answer

The answer is (b) \(10^{37.3}\).

Step 1: Introduction to the Nernst Equation

The Nernst equation is used to determine the cell potential at any concentration or pressure. It is given by the equation: \[E = E^{\circ} - \frac{RT}{nF} \ln Q\]where \(E^{\circ}\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred in the balanced equation, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.

Step 2: Identification of Parameters

For the given cell reaction, the balanced equation is:\[\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}\]Here, \(n = 2\) because two electrons are transferred from zinc to copper.

Step 3: Substituting Constants and Variables into Nernst Equation

Given that \(E_{\text{cell}}^{\circ} = 1.10\) V, \(T = 298\) K, \(R = 8.314\) J/(mol·K), and \(F = 96500\) C/mol. Substituting into the Nernst equation gives:\[0 = 1.10 - \frac{8.314 \times 298}{2 \times 96500} \ln(\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]})\]

Step 4: Solving the Nernst Equation for the Reaction Quotient Q

Set \(E = 0\) for a fully discharged cell and solve for \(Q\):\[0 = 1.10 - \frac{8.314 \times 298}{2 \times 96500} \ln(\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]})\]Simplifying, \[\ln(\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}) = \frac{1.10 \times 2 \times 96500}{8.314 \times 298}\]

Step 5: Calculating the Reaction Quotient and Final Answer

Calculate the right side of the equation:\[\ln(\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}) = \frac{(1.10 \times 2 \times 96500)}{8.314 \times 298} \approx 73.94\]Thus, \(\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \approx e^{73.94} \approx 10^{37.3}\).

Step 6: Conclusion

Therefore, the relative concentration of \([\text{Zn}^{2+}]/[\text{Cu}^{2+}]\) is approximately \(10^{37.3}\). The correct answer is option (b): \(10^{37.3}\).

Key Concepts

Electrochemical cell

An electrochemical cell is a device that generates electrical energy from a chemical reaction or facilitates a chemical change through the introduction of electrical energy. It consists of two half-cells, each containing an electrode and an electrolyte solution.
An important aspect of the electrochemical cell is its ability to convert chemical energy into electrical energy via redox reactions.

• Each half-cell houses one half of the redox reaction.
• Typically, one half-cell performs oxidation while the other handles reduction.

The classic example is the Daniell cell, where zinc and copper electrodes are placed in their ion solutions. This setup harnesses the movement of electrons from the zinc electrode (anode) to the copper electrode (cathode), allowing us to capture the energy flow as electrical energy.

Reaction quotient

The reaction quotient, denoted as \(Q\), is a crucial concept in understanding the direction of a chemical reaction at any given point. It is analogous to the equilibrium constant \(K\), except \(Q\) is calculated using the current, not equilibrium, concentrations or pressures of the reactants and products.
For a general reaction \[aA + bB \rightarrow cC + dD\] the reaction quotient \(Q\) is calculated as:\[Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]
In an electrochemical context, \(Q\) is essential for applying the Nernst equation, which adjusts the cell potential based on the deviations from standard conditions.When \(Q < K\), the reaction tends to shift to the right, producing more products. Conversely, if \(Q > K\), the reaction shifts left, favoring reactants.

Standard cell potential

Standard cell potential, represented as \(E^{\circ}_{\text{cell}}\), is the measure of the potential difference (or electromotive force) between two electrodes when all components are in their standard states, usually with concentrations of 1 mol/L for solutions and pressures of 1 atm for gases.
It is determined under conditions where the only driving force is the natural propensity for electrons to flow from the anode to the cathode.

• It can be thought of as a benchmark or baseline from which changes in cell potential under non-standard conditions can be calculated.
• Standard reduction potentials from tables can be used to compute \(E^{\circ}_{\text{cell}}\) by subtracting the anode potential from the cathode potential.

In this context, understanding \(E^{\circ}_{\text{cell}}\) allows us to predict not only the spontaneity of the reaction but also to quantify maximal voltage the cell can achieve under ideal conditions.

Electrode reaction

Electrode reactions are the processes that take place at the surface of the electrodes where electron transfer occurs. This is crucial as these reactions drive the electrochemical cell's overall function, allowing chemical energy to be converted to electrical energy.
Each electrode hosts a half-reaction:

• Oxidation occurs at the anode, where a species loses electrons.
• Reduction takes place at the cathode, where a species gains electrons.

In our example of the zinc-copper cell, zinc undergoes oxidation, releasing electrons and producing zinc ions \(\text{Zn}^{2+}\). Meanwhile, copper ions \(\text{Cu}^{2+}\) are reduced and deposit onto the copper electrode.
The collective processes at both electrodes enable the flow of electrons through an external circuit, thus generating an electric current. Understanding these electrode reactions is vital for controlling and optimizing the efficiency of electrochemical cells.