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The standard emf of a galvanic cell involving cell reaction with \(\mathrm{n}=2\) is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction would be (Given \(\left.F=96500 \mathrm{C} \mathrm{mol}^{-1} ; \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(2.0 \times 10^{11}\) (b) \(4.0 \times 10^{12}\) (c) \(1.0 \times 10^{2}\) (d) \(1.0 \times 10^{10}\)

Short Answer

Expert verified
The equilibrium constant is approximately \(1.0 \times 10^{10}\). Choose option (d).

Step by step solution

01

Identify the Nernst Equation

To find the equilibrium constant (K), we can use the Nernst Equation in the form:\[ E_{cell} = \frac{RT}{nF} \ln(K) \] where \(E_{cell}\) is the standard EMF, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons exchanged, and \(F\) is Faraday's constant.
02

Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin:\[ T = 25^{\circ}C + 273.15 = 298.15\, K \]
03

Rearrange the Nernst Equation

Rearrange the Nernst equation to solve for the equilibrium constant (K):\[ \ln(K) = \frac{nFE_{cell}}{RT} \]
04

Substitute Known Values into Equation

Substitute the given values into the rearranged equation:\[ \ln(K) = \frac{(2)(96500)(0.295)}{(8.314)(298.15)} \]
05

Calculate \(\ln(K)\)

Perform the calculation for \(\ln(K)\):\[ \ln(K) = \frac{(2)(96500)(0.295)}{(8.314)(298.15)} \approx 23.026 \]
06

Solve for K with Exponentiation

Exponentiate both sides to find the equilibrium constant \(K\):\[ K = e^{23.026} \approx 10^{10} \]
07

Choose the Correct Answer

Among the given options, the value of K approximates to option (d): \(1.0 \times 10^{10}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nernst Equation
The Nernst Equation serves as a vital tool in electrochemistry, offering insights into how the electromotive force (EMF) varies with changes in ion concentration. It relates the cell potential to the reaction quotient and provides a connection between equilibrium constants and standard electromotive force.

The general form of the Nernst Equation is:
  • \( E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln(Q) \)
Here, \(E_{cell}\) represents the actual cell potential, \(E^0_{cell}\) is the standard cell potential, \(R\) is the ideal gas constant, \(T\) refers to the temperature in Kelvin, \(n\) indicates the number of electrons transferred in the cell reaction, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.
When a system reaches equilibrium, \(Q\) equals the equilibrium constant \(K\), and \(E_{cell}\) drops to zero. This insight allows us to rearrange the Nernst Equation as:
  • \( 0 = E^0_{cell} - \frac{RT}{nF} \ln(K) \)
  • \( E^0_{cell} = \frac{RT}{nF} \ln(K) \)
Standard Electromotive Force (EMF)
The Standard Electromotive Force (EMF) denotes the potential difference between two half-cells in an electrochemical cell under standard conditions. It acts as a measure of the energy per electron transferred in a redox reaction. Standard conditions specify that:
  • the concentration of ions is maintained at 1 M,
  • the pressure of gases is at 1 atm, and
  • the temperature is 25°C or 298.15 K.
Understanding the Standard EMF allows us to assess the tendency of a redox reaction to occur. A higher value of EMF indicates a more spontaneous reaction. It also plays a critical role in calculating the equilibrium constant via the Nernst Equation. The calculated standard EMF in the original exercise is given as 0.295 V.

To compute equilibrium constants using standard EMF, we incorporate it into the Nernst Equation, a principal step in deriving the relationship between EMF and the equilibrium constant.
Faraday's Constant
Faraday's Constant is a crucial figure in electrochemistry. It designates the amount of electric charge carried by one mole of electrons, equating to approximately 96500 Coulombs per mole \((\text{C/mol})\).

It's pivotal for converting between moles of electrons and the total charge in coulombs. This constant appears in many electrochemical equations, including the Nernst Equation, to relate cell potentials to stoichiometric elements of chemical reactions.

The Faraday constant provides a means to connect macroscopic measurements, like current and charge, with microscopic entities, such as the mole of electrons in a reaction. For example:
  • When calculating \(\ln(K)\)in the Nernst Equation, Faraday's Constant allows us to factor in the total electric charge involved in the process, ensuring accurate descriptions of electrochemical behaviors.
Faraday's Constant remains a fundamental undercurrent in tackling real-world applications of electrochemistry, such as battery science and electrolysis.

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Most popular questions from this chapter

Time required to deposit 1 millimol of aluminium metal by the passage of \(9.65\) ampere of current through aqueous solution of aluminium ion, is (a) \(30 \mathrm{~s}\) (b) \(10 \mathrm{~s}\) (c) \(30,000 \mathrm{~s}\) (d) \(10,000 \mathrm{~s}\)

A dilute aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is electrolyzed using platinum electrodes. The product at the anode and cathode are (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{Na}\) (c) \(\mathrm{O}_{2}, \mathrm{Na}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{H}_{2}\)

When \(\mathrm{Cl}_{2}\) is passed through hot \(\mathrm{NaOH}\) solution, oxidation number of chlorine changes from (a) 0 to \(-1\) (b) 0 to \(+5\) (c) 0 to \(+7\) (d) \(-1\) to 0

Match the following $$ \begin{array}{ll} \text { Column-I } & \text { Column-II } \\ \hline \begin{array}{l} \text { (a) } 50 \text { \% solution of } \mathrm{H}_{2} \mathrm{SO}_{4} \\ \text { using Pt electrodes } \end{array} & \text { (p) } \mathrm{H}_{2} \text { is evolved at } \\ \text { cathode } \\ \text { (b) } \begin{array}{l} \text { Dilute solution } \mathrm{NaCl} \\ \text { using Pt electrodes } \end{array} & \text { (q) } \mathrm{O}_{2} \text { is evolved at } \\ \text { (c) } \begin{array}{l} \text { Dilute solution of } \mathrm{H}_{2} \mathrm{SO}_{4} \\ \text { using Cu electrodes } \end{array} & \text { (r) } \mathrm{Cl}_{2} \text { is evolved at } \\ \text { (d) } \begin{array}{c} \text { Concentrated solution of } \\ \text { LiCl using Pt electrodes. } \end{array} & \text { (s) } \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8} \text { is } \\ \text { formed at anode } \\ & \text { (t) non-spontaneous } \\ \text { process } \end{array} $$

In an electrolytic cell, electrolysis is carried out. Electrical energy is converted into chemical energy. In an electrochemical cell, chemical reaction, i.e., redox reaction occurs and electricity is generated. So chemical energy is converted into electrical energy. Electrolysis is governed by Faraday's laws. The potential difference between the electrodes which is called electromotive force is responsible for the generation of electric energy in the electrochemical cells. Two electrolytic cells, one containing acidified \(\mathrm{FeCl}_{2}\) and another acidified \(\mathrm{FeCl}_{3}\) are connected in series. The ratio of iron deposited at the cathodes in the tow cells will be (a) \(3: 1\) (b) \(2: 1\) (c) \(2: 3\) (d) \(3: 2\)

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