Problem 71 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Zinc granules are added in excess to a \(500 \mathrm{~mL}\). of \(1.0 \mathrm{M}\) nickel nitrate solution at \(25^{\circ} \mathrm{C}\) until the equilibrium is reached. If the standard reduction potential of \(\mathrm{Zn}^{2+} \mid \mathrm{Zn}\) and \(\mathrm{Ni}^{2+} \mid \mathrm{Ni}\) are \(-0.75 \mathrm{~V}\) and \(-0.24 \mathrm{~V}\) respectively, find out the concentration of \(\mathrm{Ni}^{2+}\) in solution at equilibrium.

Step-by-Step Solution

Answer

The concentration of \(\mathrm{Ni}^{2+}\) at equilibrium is essentially 0 M.

Step 1: Identify the Redox Reaction

We need to identify the redox reaction that takes place when zinc granules are added to nickel nitrate solution. The relevant reactions are as follows:\[\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\quad \text{(oxidation)}\] \[\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}\quad \text{(reduction)}\] The overall redox reaction is: \[\text{Zn} + \text{Ni}^{2+} \rightarrow \text{Zn}^{2+} + \text{Ni}\] This reaction indicates that zinc will reduce nickel ions.

Step 2: Calculate the Cell Potential

The standard reduction potential for zinc is \(E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.75\) V and for nickel is \(E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} = -0.24\) V. The cell potential, \(E_{\text{cell}}^{\circ}\), is given by:\[E_{\text{cell}}^{\circ} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\] Substituting the given potentials, we find:\[E_{\text{cell}}^{\circ} = (-0.24) - (-0.75) = 0.51\, \text{V}\] This positive cell potential indicates that the reaction is spontaneous.

Step 3: Use the Nernst Equation

The Nernst equation relates the concentration of ions at non-standard conditions to the standard electrode potential:\[E = E_{\text{cell}}^{\circ} - \frac{RT}{nF}\ln{Q}\]Where \(R = 8.314\, \text{J/mol K}\), \(T = 298\, \text{K}\), \(n = 2\), \(F = 96500\, \text{C/mol}\), and \(Q\) is the reaction quotient \(\frac{[\text{Zn}^{2+}]}{[\text{Ni}^{2+}]}\). At equilibrium, \(E = 0\).

Step 4: Solve the Nernst Equation for Concentration of Ni²⁺

At equilibrium, \(E = 0\), hence the Nernst equation becomes:\[0 = 0.51 - \frac{(8.314)(298)}{2(96500)}\ln{\frac{[\text{Zn}^{2+}]}{[\text{Ni}^{2+}]}}\] Rearranging gives:\[\ln{\frac{[\text{Zn}^{2+}]}{[\text{Ni}^{2+}]}} = \frac{0.51 \times 2 \times 96500}{8.314 \times 298}\] Calculating the right-hand side yields approximately 39.96. Exponentiating both sides, \(\frac{[\text{Zn}^{2+}]}{[\text{Ni}^{2+}]} = e^{39.96}\). Given \([\text{Zn}^{2+}]\) is in excess initially, approximate the concentration of \([\text{Ni}^{2+}]= e^{-39.96}\approx 0\). At equilibrium, the nickel ions are nearly completely replaced by zinc.

Key Concepts

Redox Reaction

In electrochemistry, understanding redox reactions is crucial since they form the basis of how galvanic cells work. A redox reaction involves two simultaneous processes: oxidation and reduction. Oxidation refers to the loss of electrons, whereas reduction means gaining electrons. These processes occur together in a redox pair. For instance, when zinc is added to a nickel nitrate solution:
- Zinc undergoes oxidation: \[\text{Zn} \to \text{Zn}^{2+} + 2e^-\]- Nickel ions are reduced: \[\text{Ni}^{2+} + 2e^- \to \text{Ni}\]
The overall redox reaction can be expressed as:\[\text{Zn} + \text{Ni}^{2+} \to \text{Zn}^{2+} + \text{Ni}\]
Here, zinc reduces nickel ions (\(\text{Ni}^{2+}\)) to nickel metal without requiring an external electric current. Such reactions are spontaneous if the overall cell potential is positive, indicating a natural tendency of the reaction to occur under given conditions.

Nernst Equation

The Nernst equation is a powerful tool in electrochemistry that allows us to calculate cell potentials under non-standard conditions. It essentially modifies the standard electrode potential to accommodate for varying concentrations of reactants or products. The equation is given by:
\[E = E_{\text{cell}}^{\circ} - \frac{RT}{nF} \ln Q\]
where:

• \(E\) is the cell potential under non-standard conditions,
• \(E_{\text{cell}}^{\circ}\) 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 exchanged in the reaction,
• \(F\) is Faraday's constant (96500 C/mol), and
• \(Q\) is the reaction quotient.

In the given problem, zinc and nickel form a redox system, and at equilibrium, the potential \(E = 0\). This relationship allows us to solve for unknown concentrations, such as when calculating the concentration of nickel ions \(\text{Ni}^{2+}\) at equilibrium.

Standard Reduction Potential

Standard reduction potential, represented as \(E^{\circ}\), is a measure of the tendency of a chemical species to be reduced, measured under standard conditions (1M concentration, 1 atm pressure, and usually 25°C). It's a key component in determining the direction and feasibility of electrochemical reactions.
For example, the standard reduction potential of zinc is \(-0.75 \text{ V}\) and for nickel, it is \(-0.24 \text{ V}\). These values tell us that nickel has a greater tendency to gain electrons and be reduced compared to zinc because it is less negative.
When comparing standard reduction potentials:

• A more positive \(E^{\circ}\) means a stronger oxidizing agent (more likely to gain electrons).
• A less positive \(E^{\circ}\) (or more negative \(E^{\circ}\)) means a stronger reducing agent (more likely to donate electrons).

Thus, the reduction of nickel ions and oxidation of zinc in the given exercise occurs spontaneously. The differences in their standard reduction potentials ultimately determine the cell's potential, guiding us on how the reaction will proceed.