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

Question

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

Step-by-Step Solution

Answer

The cell emf is approximately 1.0321 V, but none of the options exactly match.

Step 1: Determine the two half-cell reactions

The cell is composed of two half-reactions. The oxidation half-reaction occurs at the anode (Zn): \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \) which has a reduction potential of \( E^0 = -0.76 \text{ V} \). The reduction half-reaction occurs at the cathode (Cu): \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with a reduction potential of \( E^0 = +0.34 \text{ V} \).

Step 2: Calculate the standard cell potential

The standard cell potential \( E^0_{\text{cell}} \) is calculated by subtracting the anode potential from the cathode potential: \( E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = 0.34 - (-0.76) = 1.10 \text{ V} \).

Step 3: Apply the Nernst equation

The Nernst equation is used to calculate the actual cell potential under non-standard conditions: \[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{RT}{nF} \ln Q \] for which \( RT/F \approx 0.059 \) at standard temperature and \( n = 2 \) (number of electrons transferred).

Step 4: Calculate the reaction quotient \( Q \)

The reaction quotient \( Q \) is given by \( Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.05}{0.005} = 10 \).

Step 5: Calculate the cell potential

Substitute the values into the Nernst equation: \[ E_{\text{cell}} = 1.10 - \frac{0.059}{2} \ln(10) \]. Calculate \( \ln(10) \approx 2.302 \). Then, \[ E_{\text{cell}} = 1.10 - 0.0295 \times 2.302 = 1.10 - 0.0679 = 1.0321 \].

Key Concepts

Standard Reduction Potential

The standard reduction potential is a crucial concept in electrochemistry that indicates the tendency of a chemical species to gain electrons and thereby be reduced. It's measured in volts (V) relative to the standard hydrogen electrode which is defined at 0 V.

A positive standard reduction potential, like the +0.34 V for the copper ion (\( ext{Cu}^{2+}\)) indicates a strong affinity for gaining electrons, making it a good oxidizing agent. On the other hand, a negative value, such as the -0.76 V for zinc (\( ext{Zn}^{2+}\)), illustrates a lesser tendency to be reduced, indicating that it more readily loses electrons and acts as a reducing agent.

In electrochemical cells, these potentials help determine which species will undergo oxidation or reduction. The component with the higher reduction potential will usually act as the cathode, undergoing reduction, while the one with the lower potential will act as the anode, undergoing oxidation.

Cell EMF

The cell electromotive force (emf) is the driving force behind the flow of electrons in an electrochemical cell. It's the potential difference between two electrodes when no current is flowing, essentially the cell's "battery" power.

To determine the cell emf (\(E^0_{\text{cell}}\)), we subtract the standard reduction potential of the anode from that of the cathode as shown in the exercise. Mathematically, it's expressed as:

• \(E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}}\)

In our case, this calculated to 1.10 V, indicating the cell's natural potential to push electrons through the external circuit when both reactants and products are at unit activities.

Remember, a higher cell emf signifies a more spontaneous reaction, useful in applications requiring significant energy transfer, such as in batteries.

Nernst Equation

The Nernst Equation is a fundamental tool used to determine the actual electromotive force of a cell under non-standard conditions, which typically involve concentrations not equal to 1 M.

The Nernst Equation is often written as:

• \[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{RT}{nF} \ln Q \]

where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged in the reaction, \( F \) is the Faraday's constant, and \( Q \) is the reaction quotient.

In many problems (like this one), at standard room temperature,\( RT/F \approx 0.059 \) V. The equation helps predict the behavior of the cell's emf under various environmental or experimental conditions.

Reaction Quotient

The reaction quotient (\(Q\)) plays a pivotal role in determining the position of equilibrium and the current state of a reaction in an electrochemical cell.

It’s a dimensionless number that compares product concentrations to reactant concentrations, each raised to the power of their coefficients in the balanced chemical equation. Specifically, for the studied electrochemical cell:

• \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.05}{0.005} = 10 \]

Calculating \( Q \) is crucial when applying the Nernst Equation as it allows us to determine how far from equilibrium a cell is operating.

A higher value of \( Q \) usually signals more reactants have converted to products, influencing the cell potential and driving the reaction back toward equilibrium. As indicated in the exercise, understanding \( Q \) is key to adjusting the Nernst Equation appropriately and accurately predicting the cell's real-time voltage.