Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

At \(298 \mathrm{~K}\) a cell reaction has a standard emf of \(+0.17 \mathrm{~V}\). The equilibrium constant for the cell reaction is \(5.5 \times 10^{5} .\) What is the value of \(n\) for the cell reaction?

Short Answer

Expert verified
The value of \(n\) for the cell reaction is approximately 2.

Step by step solution

01

Recall the Nernst equation

The Nernst equation is given by \[E = E^0 - \frac{RT}{nF} \ln{Q}\] where E is the cell potential, \(E^0\) is the standard cell potential, R is the gas constant (8.314 J/mol K), T is the temperature in kelvins, n is the number of moles of electrons transferred, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient. Since the cell potential is at equilibrium, E is equal to zero, and the reaction quotient Q equals the equilibrium constant K.
02

Recall the equilibrium constant relationship

At equilibrium, we can relate the cell potential to the equilibrium constant (K) using the formula: \[0 = E^0 - \frac{RT}{nF} \ln{K}\] Step 2: Plug in the given values and solve for n
03

Insert given values into the equilibrium constant relationship

The given values are the temperature T = 298 K, the standard cell potential (\(E^0\)) = 0.17 V, and the equilibrium constant K = \(5.5 \times 10^5\). Now, substitute these values into the equilibrium constant relationship: \[0 = 0.17 - \frac{(8.314)(298)}{n(96485)} \ln{(5.5 \times 10^5)}\]
04

Isolate n in the equation

Let's isolate n and solve for its value: \[n = \frac{(8.314)(298)}{(96485)(0.17)} \ln{(5.5 \times 10^5)}\]
05

Calculate the value of n

By using a calculator, we can compute the value of n: \[n = 1.999 \approx 2\] So, the value of n for the cell reaction is approximately 2.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Electrochemistry
Electrochemistry explores the link between electricity and chemical changes. It is a fascinating branch of chemistry that involves the study of chemical reactions which produce electrical currents and the use of electricity to bring about chemical transformations. It's very much the science behind batteries, fuel cells, and electrolysis. The discipline has countless applications ranging from the development of renewable energy sources to the plating of metals.

Electrochemical reactions are characterized by the transfer of electrons. Losing electrons is called oxidation, while gaining them is referred to as reduction. These two processes occur simultaneously in a redox reaction. An essential component is the electrochemical cell, which consists of two half-cells. Each half-cell includes an electrode and an electrolyte. The potential difference between electrodes drives electron flow through an external circuit, and ions move through the electrolyte to balance the charge.
Cell Potential
Cell potential, denoted by E, represents the energy per charge available from a redox reaction and is measured in volts (V). The cell potential depends on the relative ease with which the chemical species involved undergo oxidation or reduction, and it is a measure of the driving force behind the reaction. It determines whether the electrochemical cell will work spontaneously as a battery or if it will need external energy to proceed, as in the case of electrolysis.

The cell potential at standard conditions (also called standard electrode potential) is symbolized by E°. This value offers a benchmark for predicting the direction of electron flow. The actual working cell potential can differ from E° due to different concentrations, pressures, and temperatures within the electrochemical cell. It's the Nernst equation that helps us account for these variations.
Equilibrium Constant
The equilibrium constant (K) provides a quantitative measure of the position of equilibrium in a chemical reaction. At equilibrium, the rates of the forward and reverse reactions are equal, meaning there's no net change in the concentrations of reactants and products over time. K is unitless and is derived from the ratio of the concentrations of products to the reactants, each raised to the power of their coefficients in the balanced equation.

The magnitude of K indicates the extent to which a reaction proceeds to form products. A large K, much greater than 1, suggests the reaction heavily favors the formation of products, while a small K, much less than 1, indicates the reverse. In electrochemistry, the standard cell potential and the equilibrium constant are related; the larger the value of K, the more positive E° is, indicating a greater tendency for the cell reaction to proceed spontaneously.
Standard EMF
Standard Electromotive Force (EMF), denoted by E°, refers to the cell potential under standard conditions. These conditions are typically 1 M concentrations for all aqueous solutions, a 1 atm pressure for gases, and pure solids or liquids for other phases, all at a temperature of 298 K (25 °C).

Standard EMF is a critical concept in electrochemistry as it gives us the maximum potential difference between the two electrodes of an electrochemical cell when no current is being drawn. In other words, it is the inherent 'voltage' of the cell when it's at full capacity to do electrical work. If the standard EMF is positive, the reaction is spontaneous in the direction written, while a negative standard EMF indicates that the reaction is non-spontaneous without external energy.
Reaction Quotient
The reaction quotient (Q) indicates the current state of a reaction, whether it's at equilibrium or not. It is expressed in the same form as the equilibrium constant K but with the actual concentrations of the reactants and products at any moment in time, not just at equilibrium. Hence, Q is calculated as the ratio of the product of the concentrations of products to the product of the concentrations of reactants, each raised to their respective stoichiometric coefficients.

Comparing Q to K can predict the direction in which the reaction will proceed to reach equilibrium. If Q < K, the reaction will move forward, producing more products. Conversely, if Q > K, the reaction will go backward, producing more reactants. At equilibrium, Q equals K. In the context of the Nernst equation, Q is replaced by K when the reaction reaches equilibrium, meaning that the cell potential E equals zero.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{-}(a q) \pm \mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(\mathrm{~s})+\mathrm{OCl}^{-}(a q)-{ }^{-\rightarrow} \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q)-\cdots\) \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)\) (acidic solution) (e) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q)-\mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{BrO}_{3}^{-}(a q)\) (basic solution) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}{ }^{2-}(a q)+\mathrm{ClO}^{-}(a q)-\cdots \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)\) (basic solution)

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{3}\), and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\). The overall cell reaction is $$ \mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two electrode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode, or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

Two important characteristics of voltaic cells are their cell potential and the total charge that they can deliver. Which of these characteristics depends on theamount of reactants in the cell, and which one depends on their concentration?

A voltaic cell is constructed with two \(\mathrm{Zn}^{2+}-\mathrm{Zn}\) electrodes. The two cell compartments have \(\left[\mathrm{Zn}^{2+}\right]=1.8 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=1.00 \times 10^{-2} M\), respectively. (a) Which electrode is the anode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Zn}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

(a) What happens to the emf of a battery as it is used? Why does this happen? (b) The AA-size and D-size alkaline batteries are both \(1.5\) - \(\mathrm{V}\) batteries that are based on the same electrode reactions. What is the major difference between the two batteries? What performance feature is most affected by this difference?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free