Problem 30 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the electrochemical cell, [Adv. 2018] \(\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\) the standard emf of the cell is \(2.70 \mathrm{~V}\) at \(300 \mathrm{~K}\). When the concentration of \(\mathrm{Mg}^{2+}\) is changed to \(x \mathrm{M}\), the cell potential changes to \(2.67 \mathrm{~V}\) at \(300 \mathrm{~K}\). The value of \(x\) is (given, \(\frac{F}{R}=11500 \mathrm{~K} \mathrm{~V}^{-1}\), where \(F\) is the Faraday constant and \(R\) is the gas constant, \(\ln 10=2.30\) )

Step-by-Step Solution

Answer

The concentration \( x \) is approximately 0.742 M.

Step 1: Identify the cell reaction

For the given electrochemical cell: \(\text{Mg(s)}|\text{Mg}^{2+}\text{(aq, 1M)}||\text{Cu}^{2+}\text{(aq, 1M)}|\text{Cu(s)}\), the cell reaction is \(\text{Mg(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Cu(s)}\). This is a redox reaction where magnesium oxidizes and copper ions reduce.

Step 2: Use the Nernst Equation

The Nernst equation is given by: \[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\]where \(E_{cell}\) is the cell potential, \(E^0_{cell}\) 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, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient.

Step 3: Calculate the reaction quotient Q

The reaction quotient \(Q\) is given by: \[Q = \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]}\]Since both concentrations are 1 M initially, when concentration of \([\text{Mg}^{2+}]\) changes to \(x\) M, \(Q\) becomes \(Q = \frac{x}{1} = x\).

Step 4: Plug values into Nernst Equation

Substitute the known values into the Nernst equation:\[2.67 = 2.70 - \frac{8.314 \times 300}{2 \times 96500} \ln x\]where \(E_{cell} = 2.67\) V, \(E^0_{cell} = 2.70\) V, \(R = 8.314\) J/(mol K), \(T = 300\) K, \(n = 2\), and \(F = 96500\) C/mol.

Step 5: Solve for x

Rearrange the Nernst equation to solve for \(\ln x\):\[0.03 = - \frac{8.314 \times 300}{2 \times 96500} \ln x\]Simplifying gives:\[ \ln x = \frac{-0.03 \times 2 \times 96500}{8.314 \times 300} \approx -0.2984\]Thus,\[x = e^{-0.2984}\approx 0.742\].

Step 6: Conclusion

The concentration of \( [\text{Mg}^{2+}] \) when the cell potential is 2.67 V is approximately \( 0.742 \) M.

Key Concepts

Nernst equation

The Nernst equation is a fundamental tool in electrochemistry. It allows you to calculate the cell potential of an electrochemical cell under non-standard conditions. Essentially, it adjusts the standard cell potential to account for the concentrations of the reactants and products involved. The general form of the equation is:\[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\]Where:

• \(E_{cell}\) is the cell potential under non-standard conditions.
• \(E^0_{cell}\) is the standard cell potential.
• \(R\) is the gas constant (8.314 J/(mol K)).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons exchanged in the electrochemical reaction.
• \(F\) is the Faraday constant (96500 C/mol).
• \(Q\) is the reaction quotient, which reflects the ratio of the concentrations of the products to the reactants.

The Nernst equation shows us how the potential difference between electrodes changes when concentrations vary. It is highly useful for determining how changes in chemical conditions affect the performance of the electrochemical cell. The closer the system gets to equilibrium, the smaller the difference from the standard cell potential becomes.

reaction quotient

The reaction quotient, often denoted by \(Q\), is used in the Nernst equation and helps determine the direction in which a chemical reaction will proceed. It is defined as the ratio of the product of the concentrations (or activities) of the products to the product of the concentrations (or activities) of the reactants, each raised to the power of their respective coefficients in the balanced chemical equation.For example, in a reaction of the form:\[aA + bB \rightarrow cC + dD\]The reaction quotient \(Q\) is given by:\[Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]In the context of the electrochemical cell described in the exercise:- The balanced overall reaction is \( \text{Mg(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Cu(s)} \).- Therefore, \(Q = \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]}\).The value of \(Q\) provides insight into whether the reaction will move forward to form more products or reverse to form more reactants until equilibrium is reached.

standard cell potential

The standard cell potential, denoted as \(E^0_{cell}\), is the voltage or electrical potential difference of an electrochemical cell when all reactants and products are at their standard states. A standard state typically refers to a concentration of 1 M for solutions, a pressure of 1 atm for gases, and pure solids or liquids for other phases, all measured at 25°C (298 K).For an electrochemical cell, the standard cell potential is a measure of the driving force behind the electron flow from the anode to the cathode. It is calculated from the standard reduction potentials of the two half-reactions involved in the cell reaction:\[E^0_{cell} = E^0_{cathode} - E^0_{anode}\]In the provided exercise:

• The standard emf of the cell is 2.70 V, indicating a significant driving force for the redox reaction \( \text{Mg(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Cu(s)} \).
• This value helps define the maximum voltage output under standard conditions before any concentrations of ions are varied.

Changes in these conditions, such as altering the concentrations of reactants or products, will cause the actual cell potential to deviate from this standard value, as determined by the Nernst equation.