Question

For a cell reaction involving two electrons, the standard emf of the cell is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction at \(25^{\circ} \mathrm{C}\) will be [2003] (a) \(1 \times 10^{-10}\) (b) \(29.5 \times 10^{-2}\) (c) 10 (d) \(1 \times 10^{10}\)

Short answer

Option (d) \(1 \times 10^{10}\) is correct.

Step-by-step solution

Recall the Nernst Equation

The Nernst equation relates the cell potential ( E ) to the standard cell potential ( E^0 ) at any moment in the reaction. The equation is:\[ E = E^0 - \frac{RT}{nF} \ln K \]where \( E^0 \) is the standard cell potential, \( n \) is the number of moles of electrons transferred, \( R \) is the universal gas constant \(8.314 \mathrm{~J \, mol^{-1} \, K^{-1}}\), \( T \) is the temperature in kelvins (298 K for \(25^{\circ} \mathrm{C}\)), \( F \) is Faraday's constant \(96485 \, \text{C/mol e}^- \), and \( K \) is the equilibrium constant.

Initialize given values

For this problem:- \( E^0 = 0.295 \, \text{V} \)- \( n = 2 \) electronsLet's substitute these values into the rearranged Nernst equation to solve for \( K \):\[ \ln K = \frac{nFE^0}{RT} \]

Plug in the known values

Substitute the known constants and given values in the equation:\[ \ln K = \frac{2 \times 96485 \, \text{C/mol} \times 0.295 \, \text{V}}{8.314 \, \text{J/molK} \times 298 \, \text{K}} \]Calculate the value.

Calculate \( \ln K \)

Perform the calculation:\[ \ln K = \frac{2 \times 96485 \times 0.295}{8.314 \times 298} \approx 23.902 \]

Solve for \( K \)

Now, solve for \( K \) by exponentiating both sides to remove the natural log:\[ K = e^{23.902} \]Calculate the exponential value.

Final calculation and comparison with options

Perform the final calculation:\[ K \approx e^{23.902} \approx 1 \times 10^{10} \]Compare with the given options.

Key concepts

Understanding Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. This field of study focuses on how electrons are transferred during chemical reactions, leading to the generation of electricity in electrochemical cells. These cells can be as simple as a traditional battery or as integral as the electrochemical processes occurring within living organisms. In electrochemistry, reactions at the electrode surfaces either produce an electric current, or the electric current is used to drive a chemical reaction in the cell. - **Oxidation-reduction reactions (redox reactions)**: These are foundational to electrochemistry. In redox reactions, oxidation refers to the loss of electrons and reduction refers to the gain of electrons. - **Electrochemical cells**: These cells can be galvanic (producing electricity spontaneously) or electrolytic (requiring electric energy for non-spontaneous reactions). Understanding the movement and flow of electrons in electrochemical systems allows chemists to harness these reactions for various applications, including power generation and metal plating.

The Equilibrium Constant and Reaction Dynamics

The equilibrium constant (K) is a crucial concept in chemistry, indicating the extent to which a reaction proceeds at a given temperature. It provides insights into the concentrations of reactants and products when a reaction is at equilibrium.- **Equilibrium state**: A chemical reaction reaches equilibrium when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant.- **Expression of the equilibrium constant**: For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is expressed as: \[ K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \] Here, the square brackets represent the concentration of each species.- **Relation to free energy**: The equilibrium constant can also be related to the Gibbs free energy change, \( \Delta G^0 \), reflecting the energy favorability of the reaction.In electrochemistry, \( K \) is directly related to the standard cell potential, often calculated using the Nernst equation to understand the feasibility and extent of redox reactions.

Exploring Cell Potential

The cell potential, also known as electromotive force (emf), is a measure of the potential difference between two electrodes in an electrochemical cell. It reflects how strongly electrons are driven from the anode to the cathode.- **Standard cell potential \( (E^0) \)**: This is the cell potential under standard conditions (: \( 1 \) Molar concentrations, \( 1 \, ext{atm} \) pressure, and \( 25^{\circ} \, \text{C} \)). It is a crucial metric for measuring the driving force behind an electrochemical reaction.- **Calculation of cell potential**: The cell potential can be calculated by subtracting the potential of the anode from the potential of the cathode: \[ E^0 = E^0_{\text{cathode}} - E^0_{\text{anode}} \] - **Nernst equation**: This equation relates the real-time cell potential to the standard cell potential and the concentrations of the reactants and products. It is denoted by: \[ E = E^0 - \frac{RT}{nF} \ln Q \] Where \( Q \) is the reaction quotient at any point in time.Understanding cell potential is essential for predicting how much voltage a cell can supply and determining the direction of electron flow in the cell, which is vital in both theoretical and practical applications of electrochemistry.