Problem 84
Question
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) \(1.0 \times 10^{10}\)
Step-by-Step Solution
Verified Answer
The equilibrium constant is approximately \(1.0 \times 10^{10}\). Choose option (d).
1Step 1: 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.
2Step 2: Convert Temperature to Kelvin
Convert the given temperature from Celsius to Kelvin:\[ T = 25^{\circ}C + 273.15 = 298.15\, K \]
3Step 3: Rearrange the Nernst Equation
Rearrange the Nernst equation to solve for the equilibrium constant (K):\[ \ln(K) = \frac{nFE_{cell}}{RT} \]
4Step 4: Substitute Known Values into Equation
Substitute the given values into the rearranged equation:\[ \ln(K) = \frac{(2)(96500)(0.295)}{(8.314)(298.15)} \]
5Step 5: Calculate \(\ln(K)\)
Perform the calculation for \(\ln(K)\):\[ \ln(K) = \frac{(2)(96500)(0.295)}{(8.314)(298.15)} \approx 23.026 \]
6Step 6: Solve for K with Exponentiation
Exponentiate both sides to find the equilibrium constant \(K\):\[ K = e^{23.026} \approx 10^{10} \]
7Step 7: Choose the Correct Answer
Among the given options, the value of K approximates to option (d): \(1.0 \times 10^{10}\).
Key Concepts
Nernst EquationStandard Electromotive Force (EMF)Faraday's Constant
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:
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:
The general form of the Nernst Equation is:
- \( E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln(Q) \)
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:
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.
- 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.
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:
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.
Other exercises in this chapter
Problem 81
The emf of a Daniell cell at \(298 \mathrm{~K}\) is \(E_{i}\) \(\mathrm{Zn}\left|\mathrm{ZnSO}_{4} \| \mathrm{CuSO}_{4}\right| \mathrm{Cu}\) \((0.01 \mathrm{M})
View solution Problem 82
In the ionic equation \(\mathrm{xBrO}_{3}^{-}+\mathrm{yCr}^{3+}+\mathrm{zH}_{2} \mathrm{O} \longrightarrow \mathrm{Br}_{2}+\mathrm{HCrO}_{4}^{-}+\mathrm{H}^{+}\
View solution Problem 86
If \(E^{\circ}\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)=-0.441 \mathrm{~V}\) and \(E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.771 \mathrm{~V}
View solution Problem 89
The values of standard oxidation potentials of following reactions are given below: \(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} ; E^{\circ}=
View solution