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What is the equation that relates the equilibrium constant to the cell potential?

Short Answer

Expert verified
The equation relating the equilibrium constant (K) to the cell potential (\( E^0 \)) is \( K = e^{\left(\frac{nF}{RT} E^0\right)} \).

Step by step solution

01

Understanding the Nernst Equation

At equilibrium, the cell potential, E, is zero. The Nernst equation, which relates the cell potential to the reaction quotient (Q), is given by the formula: \( E = E^0 - \frac{RT}{nF} ln(Q) \), where \( E^0 \) is the standard cell potential, R is the universal gas constant, T is the temperature in kelvin, n is the number of moles of electrons transferred, and F is the Faraday constant.
02

Defining Equilibrium Constant from Reaction Quotient

At equilibrium, the reaction quotient Q becomes the equilibrium constant K. Therefore, substituting Q with K in the Nernst equation, and since the cell potential E is zero at equilibrium, we get: \( 0 = E^0 - \frac{RT}{nF} ln(K) \).
03

Deriving the Relationship

To find the explicit relation between the equilibrium constant (K) and the standard cell potential (\( E^0 \)), rearrange the equation to solve for K: \( ln(K) = \frac{nF}{RT} E^0 \), and then exponentiate both sides to get the equilibrium constant in terms of \( E^0 \): \( K = e^{\left(\frac{nF}{RT} E^0\right)} \).

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Key Concepts

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

Equilibrium Constant and its Relation to Cell Potential
The equilibrium constant, denoted as K, plays a pivotal role in the study of chemical reactions, particularly those taking part in electrochemical cells. It quantifies the extent to which a reaction proceeds until it reaches equilibrium, at which point the rate of the forward reaction equals the rate of the reverse reaction.

In the context of electrochemistry, the equilibrium constant is intrinsically related to the electrical energy potential of a cell, often referred to as cell potential. This relationship is articulated through the Nernst equation, a fundamental equation that connects the thermodynamic quantities of cell potential with the positions of equilibrium of reactions.

The Nernst equation for cell potential (E) at equilibrium reflects that no net electromotive force is driving the reaction in either direction since E is zero. Consequently, the equation simplifies, allowing us to isolate K and express it directly in terms of the standard cell potential (\( E^0 \)), which characterizes the driving force of the reaction when all components are at standard conditions. Thus, the equilibrium constant provides us with insight into the cell's behavior under equilibrium and the propensity of a cell's reaction to proceed.
The Cell Potential
Cell potential, symbolized by E, is a measure of the chemical driving force behind a reaction occurring in an electrochemical cell. It signifies the potential difference between the electrodes of the cell when no current is flowing. A specific case is the standard cell potential (\( E^0 \) ), which is determined under standard conditions—that is, each reactant and product at a concentration of 1 M, pressure of 1 atm, and a temperature of 25℃ (298 K).

The Nernst equation allows us to determine the cell potential for any set of conditions by using the reaction quotient (Q) and taking into account the number of electrons transferred (n) in the reaction, temperature (T), as well as the universal constants R (ideal gas constant) and F (Faraday constant).

Understanding the variation of cell potential with reaction conditions is crucial. It enables us to predict not only the direction in which a reaction will proceed but also to calculate the theoretical voltage that an electrochemical cell can produce, which is of utmost importance in the design and utilization of batteries and fuel cells.
Reaction Quotient and its Transition to Equilibrium Constant
The reaction quotient, Q, measures the relative quantities of reactants and products at a given instant during a reaction, irrespective of whether the reaction has reached equilibrium. It is computed using the same expression as the equilibrium constant, but with the current concentrations instead of the equilibrium concentrations.

As the reaction proceeds, the value of Q changes until the system reaches equilibrium, at this point it becomes equal to the equilibrium constant, K. Understanding the progression from Q to K is essential for predicting the behavior of a reaction over time and for calculating the current cell potential using the Nernst equation.

Although K and Q have similar mathematical forms, their implications are different. While K indicates the position of equilibrium, Q gives a snapshot of the reaction's position at any point in time before reaching equilibrium. By comparing Q with K, chemists can predict the direction in which a reaction will spontaneously proceed to achieve equilibrium.

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Most popular questions from this chapter

Erom the half-reactions below, determine the cell reaction and standard cell potential. $$ \begin{aligned} \mathrm{BrO}_{3}^{-}+6 \mathrm{H}^{+}+6 e^{-} \rightleftharpoons \mathrm{Br}^{-}+3 \mathrm{H}_{2} \mathrm{O} & \\ & E_{\mathrm{BrO}^{-}}^{\circ}=1.44 \mathrm{~V} \\ \mathrm{I}_{2}+2 e^{-} \rightleftharpoons 2 \mathrm{I}^{-} & E_{\mathrm{l}_{2}}^{\circ}=0.54 \mathrm{~V} \end{aligned} $$

The value of \(K_{\mathrm{g}}\) for \(\mathrm{AgBr}\) is \(5.4 \times 10^{-13}\). What will be the potential of a cell constructed of a standard hydrogen electrode as one half-cell and a silver wire coated with AgBr dipping into \(0.10 M \mathrm{HBr}\) as the other halfcell. For the \(\mathrm{Ag} / \mathrm{AgBr}\) electrode, $$ \begin{aligned} \mathrm{AgBr}(s)+e^{-} \rightleftharpoons \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q) & \\ E_{\mathrm{AgBr}}^{\circ} &=+0.070 \mathrm{~V} \end{aligned} $$

Suppose a galvanic cell was constructed at \(25^{\circ} \mathrm{C}\) using a \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell (in which the molar concentration of \(\mathrm{Cu}^{2+}\) was \(1.00 \mathrm{M}\) ) and a hydrogen electrode having a partial pressure of \(\mathrm{H}_{2}\) equal to 1 atm. The hydrogen electrode dips into a solution of unknown hydrogen ionconcentration, and the two half- cells are connected by a salt bridge. The precise value of \(E_{\mathrm{cell}}^{\circ}\) is \(+0.3419 \mathrm{~V}\). (a) Derive an equation for the \(\mathrm{pH}\) of the solution with the unknown hydrogen ion concentration, expressed in terms of \(E_{\text {cell }}\) and \(E_{\text {cell }}^{\circ}\) (b) If the \(\mathrm{pH}\) of the solution were \(5.15,\) what would be the observed potential of the cell? (c) If the potential of the cell were \(0.645 \mathrm{~V}\), what would be the \(\mathrm{pH}\) of the solution?

If two electrolytic cells are placed in series, the same number of electrons must pass through both cells. One student argues that you can get twice as much product if two cells are placed in series compared to a single cell and therefore the cost of production (i.e., the cost of electricity) will decrease greatly and profits will increase. Is the student correct? Explain your reasoning based on the principles of electrochemistry.

In a galvanic cell, do electrons travel from anode to cathode, or from cathode to anode? Explain.

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