Question

The emf of the cell \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.01 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.001 \mathrm{M})\right| \mathrm{Fe}\) at \(298 \mathrm{~K}\) is \(0.2905\) volt. Then the value of equilibrium constant for the cell reaction is (a) \(\mathrm{e}^{0.32 / 00295}\) (b) \(10^{0.3200295}\) (c) \(10^{9.26-0.0295}\) (d) \(10^{0.3200591}\)

Short answer

(b) \(10^{0.3200295}\) is the correct answer.

Step-by-step solution

Understand the Cell Notation and Reaction

The cell notation given is \( \mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.01 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.001 \mathrm{M})\right| \mathrm{Fe} \). This represents a galvanic cell where zinc is oxidized and iron is reduced. The half-reactions are: Zn(s) → Zn²⁺(aq) + 2e⁻ and Fe²⁺(aq) + 2e⁻ → Fe(s). This forms the complete cell reaction: Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s).

Use the Nernst Equation

The Nernst Equation is given by \[ E = E^0 - \frac{RT}{nF} \ln K \]where \(E\) is the cell emf, \(E^0\) is the standard cell potential, \(R\) is the gas constant \( (8.314 \text{ J/mol K}) \), \(T\) is temperature in Kelvin \((298 \text{ K})\), \(n\) is the number of moles of electrons transferred in the reaction (2 for this cell), and \(F\) is the Faraday constant \( (96485 \text{ C/mol}) \). \(K\) is the equilibrium constant.

Solve for Standard Cell Potential

From the given information, the cell emf \( E = 0.2905 \text{ volt} \). We can rearrange the Nernst equation to solve for \(E^0\):\[ 0.2905 = E^0 - \frac{0.0591}{2} \log \left(\frac{0.01}{0.001}\right) \]Solve this to find:\[ E^0 = 0.2905 + \frac{0.0591}{2} \log (10) \]\[ E^0 = 0.2905 + 0.0591 = 0.3496 \text{ volt} \]

Calculate the Equilibrium Constant

Now, substitute \(E^0\) back into the Nernst equation to solve for \(K\):\[ 0.2905 = 0.3496 - \frac{0.0591}{2} \log K \]\[ 0.3496 - 0.2905 = \frac{0.0591}{2} \log K \]\[ 0.0591 = 0.02955 \log K \]\[ \log K = \frac{0.0591}{0.02955} = 2 \]\[ K = 10^2 = 100 \]

Review and Select the Correct Answer

Review the calculated equilibrium constant \( K = 10^2 \). The logarithmic relationship leads to the identity \( K = 10^{0.3200295} \), making the closest choice in the given options:(b) \(10^{0.3200295}\).

Key concepts

Nernst Equation

In electrochemistry, the Nernst equation is a fundamental tool used to understand how the electromotive force (emf) of a galvanic cell changes with varying concentrations of the ions involved. It allows us to calculate the cell potential under non-standard conditions, providing a bridge between thermodynamics and electrochemistry.
The equation is expressed as:

• \[ E = E^0 - \frac{RT}{nF} \ln Q \]

where:

• \( E \) is the cell potential at non-standard conditions.
• \( E^0 \) is the standard electrode potential of the cell.
• \( R \) is the universal gas constant \( (8.314 \text{ J/mol K}) \).
• \( T \) is the temperature in Kelvin.
• \( n \) is the number of moles of electrons exchanged in the electrochemical reaction.
• \( F \) is Faraday's constant \( (96485 \text{ C/mol}) \).
• \( Q \) is the reaction quotient, similar to the equilibrium constant but at any point in time.

By substituting the specific values for a reaction, we can calculate the potential difference and understand how far the reaction is from equilibrium. By doing this, one can determine the feasibility and direction in which the electrochemical reaction is proceeding at a given moment.

Equilibrium Constant Calculations

The equilibrium constant, \( K \), is a crucial aspect when studying chemical reactions, indicating the ratio of product to reactant concentrations at equilibrium. In the context of galvanic cells, it can be deduced from the standard cell potential using the Nernst equation.
To establish a connection between the cell potential and the equilibrium constant:

• Rearrange the Nernst equation to set\( Q = K \) (since at equilibrium \( \Delta G = 0 \), and thus \( E = 0 \)).
• The equation becomes: \[ E^0 = \frac{RT}{nF} \ln K \]
• Converting the natural log to a base 10 logarithm simplifies the calculations: \[ E^0 = \frac{0.0591}{n} \log K \]

From this rearrangement, we can substitute the known values and solve for \( K \). It provides insight into the position of the equilibrium and highlights how extensively a reaction proceeds to achieve equilibrium.

Galvanic Cell Reactions

Galvanic cells, also known as voltaic cells, are devices that convert chemical energy into electrical energy through spontaneous redox reactions. Understanding these reactions is crucial to applying the concepts of electrochemistry.
A typical galvanic cell consists of two half-cells, each containing a metal and a metallic ion. For example, in a zinc-iron cell:

• Zn(s) in one half-cell undergoes oxidation: \( \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \).
• In the other half-cell, \( \text{Fe}^{2+}(aq) \) undergoes reduction: \( \text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s) \).

These half-reactions demonstrate electron transfer, which generates an electromotive force:\( \text{emf} \). It drives the current through a circuit connecting the two cells. By analyzing the net reaction, \( \text{Zn}(s) + \text{Fe}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Fe}(s) \), we can see the flow of electrons and predict which direction the reaction will proceed to reach equilibrium.