/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 106 Consider a Daniell cell operatin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a Daniell cell operating under nonstandardstate conditions. Suppose that the cell's reaction is multiplied by 2 . What effect does this have on each of the following quantities in the Nernst equation? (a) \(E,\left(\right.\) b) \(E^{\circ},\) (c) \(Q,\) (d) \(\ln Q\), and (e) \(n\) ?

Short Answer

Expert verified
Multiplying the cell reaction by 2 does not affect \(E\) and \(E^{\circ}\), but it squares \(Q\), doubles \(\ln Q\), and doubles \(n\).

Step by step solution

01

Effects on \(E\) and \(E^{\circ}\)

Doubling the cell's reaction will not have an effect on the \(E\) and \(E^{\circ}\). The cell potential \(E\) is independent of the stoichiometry of the reaction. The standard cell potential \(E^{\circ}\) also remains unchanged as it is only dependent on the nature of the reactants.
02

Effects on \(Q\)

The reaction quotient \(Q\) will change as it depends on the concentration of the reactants and the products. As a result, when the reaction is multiplied by 2, \(Q\) will be squared.
03

Effects on \(\ln Q\)

The natural logarithm of the reaction quotient, \(\ln Q\), will be affected since \(\ln Q\) depends on \(Q\). When \(Q\) is squared (because the reaction was multiplied by 2), \(\ln Q\) will be doubled.
04

Effects on \(n\)

The number of moles of electrons transferred in the reaction \(n\) will be doubled when the reaction is multiplied by 2 because \(n\) is directly proportional to the reaction stoichiometry.

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 91Ó°ÊÓ!

Key Concepts

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

Cell potential
Understanding cell potential is crucial in electrochemistry, as it represents the driving force behind the flow of electrons in an electrochemical cell. To put it simply, cell potential, denoted as E, is the measure of the energy per unit charge available from the redox reaction occurring in the cell.

It's composed from the difference in potential between the two electrodes. The greater the difference, the higher the cell potential, which translates to more electrical energy that can be harnessed from the reaction. When conditions deviate from standard states, such as changes in concentration, pressure, or temperature, the cell potential can be calculated using the Nernst equation, accounting for these non-standard conditions. However, it is important to note that changes in the stoichiometry of the reaction, like doubling the reaction, will not influence the cell potential directly.
Reaction quotient (Q)
The reaction quotient, known as Q, is a snapshot of the reaction's progress, comparing the concentrations of products and reactants at a given moment. It's similar to the equilibrium constant but is used for any point during the reaction, not just at equilibrium.

Determining Q

For a general reaction aA + bB ↔ cC + dD, the reaction quotient is expressed as:\[\begin{equation}Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\end{equation}\]Where the concentrations of the products C and D are raised to their respective stoichiometric coefficients and divided by those of the reactants A and B. When the reaction is modified, such as being multiplied by a factor, this affects the reaction quotient accordingly.
Standard cell potential (·¡Â°)
The standard cell potential, ·¡Â°, is a key term in electrochemistry, as it indicates the potential of a cell under standard conditions (1 M concentration of all reactants and products, 1 atm pressure, and 25°C temperature).

This reference value allows chemists to predict the direction and spontaneity of a redox reaction. Like the cell potential E, the standard cell potential ·¡Â° is unaffected by changes in the stoichiometry of the reaction, such as doubling. As such, the ·¡Â° remains a constant for a given redox reaction unless the actual reactants or their nature are changed.
Natural logarithm in electrochemistry
The natural logarithm—often seen in equations as \[\begin{equation} ln \end{equation}\]—is a central part of the Nernst equation in electrochemistry. It helps relate the cell potential to the reaction quotient (\[\begin{equation} Q\end{equation}\]) and the number of moles of electrons transferred (\[\begin{equation} n\end{equation}\]).

Using natural logarithms allows us to find the relationship between the electrochemical properties and the concentrations of reactants and products in a reaction. Importantly, when \[\begin{equation} Q\end{equation}\] changes, such as being squared due to the doubling of the reaction, \[\begin{equation} ln(Q)\end{equation}\] will change. The properties of logarithms dictate that the natural logarithm of a squared number is double the logarithm of the original number. Thus, when analyzing changes in reaction stoichiometry, one must consider the impact on the natural logarithm's value.
Moles of electrons (n)
The mole of electrons, n, is an essential value in the Nernst equation, signifying the amount of electrons transferred during the redox reaction. This number is integral when calculating cell potential, as it impacts the potential's magnitude directly.

As n is determined by the stoichiometry of the reaction, any changes to the reactant coefficients, such as doubling the reaction, will subsequently alter the value of n. For instance, if the original reaction transfers 2 moles of electrons, doubling the reaction implies that now 4 moles of electrons would be involved. Understanding the role of moles of electrons within electrochemical calculations is vital for predicting and manipulating the electrical output of chemical cells.

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

Write the Nernst equation and explain all the terms.

A galvanic cell is constructed by immersing a piece of copper wire in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}\) solution and a zinc strip in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{ZnSO}_{4}\) solution. (a) Calculate the emf of the cell at \(25^{\circ} \mathrm{C}\) and predict what would happen if a small amount of concentrated \(\mathrm{NH}_{3}\) solution were added to (i) the \(\mathrm{CuSO}_{4}\) solution and (ii) the \(\mathrm{ZnSO}_{4}\) solution. Assume that the volume in each compartment remains constant at \(25.0 \mathrm{~mL}\). (b) In a separate experiment, \(25.0 \mathrm{~mL}\) of \(3.00 M \mathrm{NH}_{3}\) are added to the \(\mathrm{CuSO}_{4}\) solution. If the emf of the cell is \(0.68 \mathrm{~V},\) calculate the formation constant \(\left(K_{\mathrm{f}}\right)\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\).

The hydrogen-oxygen fuel cell is described in Section 19.6 . (a) What volume of \(\mathrm{H}_{2}(g)\), stored at \(25^{\circ} \mathrm{C}\) at a pressure of 155 atm, would be needed to run an electric motor drawing a current of \(8.5 \mathrm{~A}\) for \(3.0 \mathrm{~h} ?\) (b) What volume (liters) of air at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) will have to pass into the cell per minute to run the motor? Assume that air is 20 percent \(\mathrm{O}_{2}\) by volume and that all the \(\mathrm{O}_{2}\) is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behavior.

Which species in each pair is a better reducing agent under standard-state conditions? (a) \(\mathrm{Na}\) or \(\mathrm{Li},\) (b) \(\mathrm{H}_{2}\) or \(\mathrm{I}_{2},\) (c) \(\mathrm{Fe}^{2+}\) or \(\mathrm{Ag}\), (d) \(\mathrm{Br}^{-}\) or \(\mathrm{Co}^{2+}\).

To remove the tarnish \(\left(\mathrm{Ag}_{2} \mathrm{~S}\right)\) on a silver spoon, a student carried out the following steps. First, she placed the spoon in a large pan filled with water so the spoon was totally immersed. Next, she added a few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolved. Finally, she placed some aluminum foil at the bottom of the pan in contact with the spoon and then heated the solution to about \(80^{\circ} \mathrm{C}\). After a few minutes, the spoon was removed and rinsed with cold water. The tarnish was gone and the spoon regained its original shiny appearance. (a) Describe with equations the electrochemical basis for the procedure. (b) Adding \(\mathrm{NaCl}\) instead of \(\mathrm{NaHCO}_{3}\) would also work because both compounds are strong electrolytes. What is the added advantage of using \(\mathrm{NaHCO}_{3}\) ? (Hint: Consider the \(\mathrm{pH}\) of the solution.) (c) What is the purpose of heating the solution? (d) Some commercial tarnish removers containing a fluid (or paste) that is a dilute \(\mathrm{HCl}\) solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two disadvantages of using this procedure compared to the one described here.

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.