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 1.0705 V, which is option (b).

Step 1: Identify the half-reactions

In the given voltaic cell, the half-reactions are: \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) with a standard reduction potential of \( E^\circ = -0.76 \text{ V} \) and \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with \( E^\circ = +0.34 \text{ V} \). Zinc will act as the anode and copper as the cathode since zinc has a more negative reduction potential.

Step 2: Calculate the standard cell potential

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

Step 3: Apply the Nernst equation

The Nernst equation, which compensates for non-standard conditions, is: \[ E = E^\circ - \frac{RT}{nF} \ln Q \] where \( \frac{RT}{F} = 0.059 \text{ V} \), \( n = 2 \) (since two electrons are transferred), and \( Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.05}{0.005} = 10 \).

Step 4: Calculate the cell emf

Using the Nernst equation: \[ E = E^\circ - \frac{0.059}{2} \times \log_{10}(10) = 1.10 \text{ V} - 0.0295 \text{ V} = 1.0705 \text{ V}. \]

Step 5: Compare with given options

The calculated cell emf, \( 1.0705 \text{ V} \), matches option (b).

Key Concepts

Standard Reduction Potential

In electrochemistry, the standard reduction potential is a measure of the tendency of a chemical species to gain electrons and be reduced. It is denoted by \(E^\circ\) and measured in volts (V). Standard conditions are defined as solutions with concentrations of 1 M, at a pressure of 1 atmosphere, and a temperature of 25°C (298 K). Each species has a characteristic standard reduction potential indicating how easily it can be reduced.

For example, in the provided exercise, we consider the half-reactions for zinc and copper:

• \(\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) with \( E^\circ = -0.76 \text{ V} \).
• \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with \( E^\circ = +0.34 \text{ V} \).

The more positive the standard reduction potential, the greater the tendency of the species to gain electrons. Hence, copper, with a positive \(E^\circ\), is more likely to be reduced than zinc.

Voltaic Cell

Voltaic cells, also known as galvanic cells, are devices that convert chemical energy into electrical energy through spontaneous redox reactions. They consist of two electrodes (anode and cathode) immersed in electrolyte solutions and connected via a conductive path.

In our example, zinc acts as the anode, where oxidation occurs, and copper acts as the cathode, where reduction takes place. Electrons flow from the anode to the cathode, generating an electric current. The cell is often represented as:

• \(\text{Anode} \rightarrow \text{Salt Bridge} \rightarrow \text{Cathode}\)
• \(\text{Zn} | \text{Zn}^{2+} (\text{solution}) \parallel \text{Cu}^{2+} (\text{solution}) | \text{Cu}\)

The salt bridge helps maintain electrical neutrality by allowing ions to flow between the two half-cells.

Nernst Equation

The Nernst equation is critical in electrochemistry for calculating the cell potential under non-standard conditions. It enables us to estimate how the voltage of a voltaic cell changes with different concentrations of the involved species. The equation is given by:\[E = E^\circ - \frac{RT}{nF} \ln Q\]Here,

• \(E\) is the cell potential under non-standard conditions.
• \(R\) is the universal gas constant (8.314 J/mol K).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred in the reaction.
• \(F\) is Faraday's constant (96485 C/mol).
• \(Q\) is the reaction quotient, equivalent to the concentration of the products over the reactants.

In our case, with \(\frac{RT}{F} = 0.059 \text{ V}\), \(n = 2\), and \(Q = 10\), applying the Nernst equation helps us where standard conditions are not met.

Cell EMF Calculation

Electromotive force (EMF) is the voltage generated by a cell. In standard conditions, it is calculated using the standard reduction potentials of the cathode and anode:\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 \text{ V} - (-0.76 \text{ V}) = 1.10 \text{ V}\]For non-standard conditions, we use the Nernst equation to adjust this standard potential:\[E = 1.10 \text{ V} - \frac{0.059}{2} \log_{10}(10) = 1.10 \text{ V} - 0.0295 \text{ V} = 1.0705 \text{ V}\]This calculation shows the actual EMF, 1.0705 V, which corresponds to option (b) in the problem statement. By comparing the calculated EMF with the given options, we determine the correct answer.