/*! 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 105 (a) Write the reactions for the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 105

(a) Write the reactions for the discharge and charge of a nickelcadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+\begin{array}{c} 2 \mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{~V} \end{array} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

Short Answer

Expert verified
The discharge and charge reactions for a nickel-cadmium (nicad) rechargeable battery are given as: 1. Discharge: NiO(OH)(s) + H\(_{2}\)O(l) + Cd(OH)\(_{2}\)(s) 鈫� Ni(OH)\(_{2}\)(s) + Cd(s) + 2 OH\(^{-}\)(aq) 2. Charge: Ni(OH)\(_{2}\)(s) + Cd(s) + 2 OH\(^{-}\)(aq) 鈫� NiO(OH)(s) + H\(_{2}\)O(l) + Cd(OH)\(_{2}\)(s) Calculating the standard emf (E\(^{\circ}\)) gives 1.25 V. The difference between this value and the typical nicad cell emf (1.30 V) can be attributed to factors such as concentration, temperature, and overpotentials in the battery. Using the Nernst equation and the given values, the equilibrium constant (K) for the overall nicad reaction is found to be approximately 4.173.

Step by step solution

01

1. Writing discharge and charge reactions

First, we need to identify the discharge (forward) and charge (backward) reactions for the nickel-cadmium battery cell. The given reactions are: 1. Cd(OH)\(_{2}\)(s) + 2 e\(^{-}\) 鈫� Cd(s) + 2 OH\(^{-}\)(aq) with E\(^{\circ}_{\text{red}}\) = -0.76 V 2. NiO(OH)(s) + H\(_{2}\)O(l) + e\(^{-}\) 鈫� Ni(OH)\(_{2}\)(s) + OH\(^{-}\)(aq) with E\(^{\circ}_{\text{red}}\) = +0.49 V
02

2. Calculating the standard emf of the cell

Using the given reduction potentials, we can calculate the standard emf (E\(^{\circ}\)) of the cell by summing the reduction potentials of the two half-reactions. E\(^{\circ}_{\mathrm{cell}}\) = E\(^{\circ}_{\text{red (Ni)}}\) - E\(^{\circ}_{\text{red (Cd)}}\) = 0.49 V - (-0.76 V) = 1.25 V.
03

3. Analyzing the difference between the calculated emf and the typical value

The calculated standard emf (1.25 V) is slightly different from the given typical nicad cell emf (1.30 V). This difference can be attributed to various factors, such as the concentration of the electrolytes in the cell, temperature, and any overpotentials that may be present in a real-world battery.
04

4. Calculating the equilibrium constant for the overall nicad reaction

The overall reaction for the nickel-cadmium battery cell can be obtained by combining the half-reactions: NiO(OH)(s) + H\(_{2}\)O(l) + Cd(OH)\(_{2}\)(s) 鈫� Ni(OH)\(_{2}\)(s) + Cd(s) + 2 OH\(^{-}\)(aq) To find the equilibrium constant for the overall reaction, we can use the Nernst equation with the typical emf value: E\(_{\mathrm{cell}}\) = E\(^{\circ}_{\mathrm{cell}}\) - \(\frac{RT}{nF}\) lnK Here, E\(_{\mathrm{cell}}\) is the actual emf of the cell (+1.30 V), E\(^{\circ}_{\mathrm{cell}}\) is the standard emf of the cell (1.25 V), R is the gas constant (8.314 J mol\(^{-1}\) K\(^{-1}\)), T is the temperature (assumed to be 298 K), n is the number of electrons transferred (2 for the overall reaction), F is the Faraday constant (96485 C mol\(^{-1}\)), and K is the equilibrium constant. We first need to rearrange the Nernst equation to solve for K: lnK = \(\frac{nF(E_{\mathrm{cell}} - E^{\circ}_{\mathrm{cell}})}{RT}\) Now we can plug in the values and solve for K: lnK = \(\frac{2 \times 96485 \times (1.30 - 1.25)}{8.314 \times 298}\) = 1.428 To obtain the equilibrium constant, we take the exponential of the result: K = exp(1.428) 鈮� 4.173 So the equilibrium constant for the overall nicad reaction is approximately 4.173.

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 91影视!

Key Concepts

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

Electrochemical Cells
An electrochemical cell is the fundamental unit of devices like batteries and fuel cells that convert chemical energy into electrical energy through electrochemical reactions. Each cell consists of two electrodes, an anode and a cathode, separated by an electrolyte that permits the flow of ions but not electrons. In the case of a nickel-cadmium (NiCd) battery, the anode reaction involves cadmium transforming into cadmium hydroxide, while at the cathode, nickel oxyhydroxide gets reduced to nickel hydroxide.

During discharge, electrons flow from the anode to the cathode through an external circuit, powering any connected devices. Upon charging, an external power source drives electrons in the opposite direction, regenerating the initial chemical states of the electrodes. The ability of such cells to be recharged makes NiCd batteries a type of secondary cell. Understanding the reactions within these cells is essential for grasping how batteries store and release energy.
Reduction Potentials
Reduction potential, often represented as Ered掳, is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. In the context of a nickel-cadmium battery, we are presented with two half-reactions, each with its own reduction potential. Cadmium hydroxide has a reduction potential of -0.76 V, indicating its reduced form, cadmium, is relatively stable. Conversely, the reduction potential of nickel oxyhydroxide is +0.49 V, suggesting that it is more inclined to be reduced to nickel hydroxide.

When determining the voltage that a battery can provide, it is the difference in reduction potentials that matters. In the NiCd battery, the overall reaction is the sum of the reduction of nickel and the oxidation of cadmium. The sum of their potentials gives the standard electromotive force (emf) of the battery cell, an ideal value calculated when the reaction conditions are at standard state.
Nernst Equation
The Nernst equation provides a bridge between the standard cell potential, reaction quotient, temperature, and number of electrons transferred to calculate the actual emf of an electrochemical cell under non-standard conditions. It can be expressed as:
Ecell = E掳cell - (RT/nF) lnQ

Where Ecell is the emf under current conditions, E掳cell is the standard emf, R is the universal gas constant, T is the temperature in kelvin, n is the number of moles of electrons exchanged in the electrochemical reaction, F is the Faraday constant, and Q is the reaction quotient. For the nickel-cadmium battery, the Nernst equation allows us to account for real-world conditions like temperature and concentration that affect battery performance and can cause the actual emf to deviate from the standard emf calculated from reduction potentials.
Equilibrium Constant
The equilibrium constant, K, is a dimensionless number that provides insight into the position of equilibrium for a chemical reaction. In the context of our nickel-cadmium battery example, the equilibrium constant relates to the extent that the reaction proceeds under standard conditions. According to Le Chatelier's principle, a large K value implies a strong tendency toward products at equilibrium, whereas a small K value indicates a reaction that favors reactants.

Using the Nernst equation, we can calculate the equilibrium constant K for the overall redox reaction taking place in the NiCd battery. By rearranging the equation and inputting the actual cell voltage and the number of electrons transferred (two in this case), we solve for K to understand how favorable the reaction is. A K value significantly larger than 1, as we find here, suggests that at equilibrium, the products are favored, which is consistent with a battery that does a good job at storing and discharging electrical energy.

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

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Ni}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s) \quad\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s)\) (acidic solution) (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{S}(s)+\mathrm{HSO}_{4}^{-}(a q) \quad\) (acidic solution) (d) \(\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{ClO}^{-}(a q)\) (basic solution)

In the Br酶nsted-Lowry concept of acids and bases, acid-base reactions are viewed as proton-transfer reactions. The stronger the acid, the weaker is its conjugate base. In what ways are redox reactions analogous? [Sections 20.1 and 20.2\(]\)

(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cu}^{2+}, \mathrm{Cl}_{2}, \mathrm{O}_{2} .\) (b) Arrange the fol- lowing in order of increasing strength as reducing agents in acidic solution: \(\mathrm{Zn}, \mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Al}\).

By using the data in Appendix \(\mathrm{E}\), determine whether each of the following substances is likely to serve as an oxidant or a reductant: (a) \(\mathrm{Cl}_{2}(g),\) (b) \(\mathrm{MnO}_{4}^{-}(a q,\) acidic solution) (c) \(\mathrm{Ba}(s),\) (d) \(\mathrm{Zn}(s)\)

The capacity of batteries such as the typical AA alkaline battery is expressed in units of milliamp-hours (mAh). An AA alkaline battery yields a nominal capacity of \(2850 \mathrm{mAh}\). (a) What quantity of interest to the consumer is being expressed by the units of \(\mathrm{mAh} ?\) (b) The starting voltage of a fresh alkaline battery is \(1.55 \mathrm{~V}\). The voltage decreases during discharge and is \(0.80 \mathrm{~V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation
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.