EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 8220106637203
Author: Chang
Publisher: YUZU
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Chapter 19, Problem 19.129SP

a)

Interpretation Introduction

Interpretation:

Half reactions; the anode and cathode have to be labelled.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: Ecell0 is composed of a contribution from the anode and a contribution from the cathode is given by,

Ecello=EcathodeoEanodeo

Where both Ecathodeo and Eanodeo are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: ΔG0=-nFEcell0

Relation between Ecell0 and equilibrium constant (K) of a redox reaction:

Ecell0=RTnFlnKwhereRisgasconstant(8.314J/K.mol)TisTemperatureinKelvinnisno.ofelectronstransferredinredoxreactionFisFaradayconstant(96500J/V.mol)Kisequilibriumconstant

Relation between ΔG0 and K:ΔG0=-RTlnK

Effect of concentration on cell Emf:

The mathematical relationship between the emf of galvanic cell and the concentration of reactants and products in a redox reaction under nonstandard-state conditions is,

ΔG=ΔG0+RTlnQwhere, ΔG0isstandardGibb'sfreeenergy            Qisreactionquotient.

As known ΔG0=-nFEcell0 and ΔG=-nFEcell, above expression can be written as,

ΔG=ΔG0+RTlnQ-nFEcell=-nFEcell0+RTlnQ

Dividing by –nF, the above equation becomes,

-nFEcellnF=-nFEcell0nF+RTlnQnFEcell=Ecell0RTnFlnQNernst equation

Nernst equation: The Nernst equation is used to calculate the cell voltage under nonstandard-state conditions.

a)

Expert Solution
Check Mark

Explanation of Solution

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO, Chapter 19, Problem 19.129SP

        Figure.1

For the given redox reactions,

In the galvanic cell, Oxidation occurs at anode and reduction occurs at cathode.

Therefore,

Anode (Oxidation): Co(s)Co2+(aq)+2e

Cathode (Reduction): Mg2+(aq)+2eMg(s)

b)

Interpretation Introduction

Interpretation:

The minimum voltage needed to drive the reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: Ecell0 is composed of a contribution from the anode and a contribution from the cathode is given by,

Ecello=EcathodeoEanodeo

Where both Ecathodeo and Eanodeo are the standard reduction potentials of the electrodes.

b)

Expert Solution
Check Mark

Explanation of Solution

The emf values for the two given half-reactions are,

Anode (Oxidation): Co(s)Co2+(aq)+2eEanode=+0.28V

Cathode (Reduction): Mg2+(aq)+2eMg(s)Ecathode=2.37V

Calculated standard emf for galvanic cell as follows,

Ecello=Ecathodeo+Eanodeo=(-2.37)V+(0.28)V=-2.09V

The minimum voltage needed to drive the reaction is +2.09V

c)

Interpretation Introduction

Interpretation:

The emf of a given galvanic cell has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: Eocell is composed of a contribution from the anode and a contribution from the cathode is given by,

Ecello=EcathodeoEanodeo

Where both Ecathodeo and Eanodeo are the standard reduction potentials of the electrodes.

Effect of concentration on cell Emf:

The mathematical relationship between the emf of galvanic cell and the concentration of reactants and products in a redox reaction under nonstandard-state conditions is,

ΔG=ΔG0+RTlnQwhere, ΔG0isstandardGibb'sfreeenergy            Qisreactionquotient.

As known ΔG0=-nFEcell0 and ΔG=-nFEcell, above expression can be written as,

ΔG=ΔG0+RTlnQ-nFEcell=-nFEcell0+RTlnQ

Dividing by –nF, the above equation becomes,

-nFEcellnF=-nFEcell0nF+RTlnQnFEcell=Ecell0RTnFlnQNernst equation

Nernst equation: The Nernst equation is used to calculate the cell voltage under nonstandard-state conditions.

c)

Expert Solution
Check Mark

Explanation of Solution

Given: Current= 10.0 A; Time, t=2hr=2×60×60=7200sec

Convert Current into coulomb:

(Current)×(time)=no.ofCoulomb

As known, 1Ampere=1CoulombSec

(Current)=no.ofCoulombtimeno.ofCoulomb=(Current)(time)=(10.0C.Sec1)(7200sec)=72000C

Convert number of Coulombs into mole of electrons:

moleofe-=no.ofCoulombFaradayconstant=72000C96500Cmolee-=0.746molee-1

Convert mole of electrons into number of moles:

Co(s)Co2+(aq)+2eEanode=+0.28V

1 mole of Cobalt 2 mole of e-

‘X’of Cobalt = 0.746molee-1 moles of e-

Number of moles of Co=(1moleofCo)(0.746molee-1)2molofe=0.373molofCo

Therefore, no.of moles of Cobalt is 0.373molofCo

Assuming solution volumes of 1.00L, the concentration of Co2+ in solution after  2hours is 2.373 M, and the concentration of Mg2+ in solution after 2 hours is 1.627 M. we use the Nernst equation to solve for Ecell

Mg(s)+Co2+(aq)Mg2+(aq)+Co(s)

Calculation of non-standard emf value using Nernst equation:

The reaction quotient for the given reaction is, Q=[Mg2+][Co][Mg][Co2+]

The concentration of pure solids and pure liquids do not appear in the expression for Q.

Hence, the reaction quotient becomes, Q=[Mg2+][Co2+]

Substitute known constant values of R, T and F into Nernst equation becomes as follows,

Ecell=2.09V-(0.0257V)nln[Mg2+][Co2+]

The number of electrons transferred in the given redox reaction is TWO (n=2) and Ecell0=+2.09V

Ecell(+2.09V)-(0.0257V)2ln1.6272.373=(+2.09V)-(0.01285)ln(0.686)=(+2.09V)-(0.01285)(0.377)=(+2.09V)+0.00485=+2.095V

The emf of the given cell reaction is +2.095V

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Chapter 19 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

Ch. 19.5 - Prob. 2PECh. 19.5 - Prob. 1RCCh. 19.7 - Prob. 1RCCh. 19.8 - Prob. 1RCCh. 19.8 - An aqueous solution of Mg(NO3)2 is electrolyzed....Ch. 19.8 - Prob. 2RCCh. 19.8 - Prob. 2PECh. 19.8 - Prob. 3RCCh. 19 - Prob. 19.1QPCh. 19 - Prob. 19.2QPCh. 19 - Prob. 19.3QPCh. 19 - Prob. 19.4QPCh. 19 - Prob. 19.5QPCh. 19 - Prob. 19.6QPCh. 19 - 19.7 What is the difference between the...Ch. 19 - Prob. 19.8QPCh. 19 - Prob. 19.9QPCh. 19 - Prob. 19.10QPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - 19.14 Which of the following reagents can oxidize...Ch. 19 - 19.15 Consider the following half-reactions: (aq)...Ch. 19 - 19.16 Predict whether the following reactions...Ch. 19 - Prob. 19.17QPCh. 19 - Prob. 19.18QPCh. 19 - Prob. 19.19QPCh. 19 - Prob. 19.20QPCh. 19 - Prob. 19.21QPCh. 19 - Prob. 19.22QPCh. 19 - Prob. 19.23QPCh. 19 - Prob. 19.24QPCh. 19 - Prob. 19.25QPCh. 19 - Prob. 19.26QPCh. 19 - Prob. 19.27QPCh. 19 - Prob. 19.28QPCh. 19 - Prob. 19.29QPCh. 19 - Prob. 19.30QPCh. 19 - Prob. 19.31QPCh. 19 - Prob. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - 19.35 Explain the differences between a primary...Ch. 19 - Prob. 19.36QPCh. 19 - Prob. 19.37QPCh. 19 - Prob. 19.38QPCh. 19 - Prob. 19.39QPCh. 19 - Prob. 19.40QPCh. 19 - Prob. 19.41QPCh. 19 - Prob. 19.42QPCh. 19 - 19.43 What is the difference between a galvanic...Ch. 19 - Prob. 19.44QPCh. 19 - Prob. 19.45QPCh. 19 - Prob. 19.46QPCh. 19 - Prob. 19.47QPCh. 19 - Prob. 19.48QPCh. 19 - Prob. 19.49QPCh. 19 - Prob. 19.50QPCh. 19 - 19.51 Calculate the amounts of Cu and Br2 produced...Ch. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - 19.55 What is the hourly production rate of...Ch. 19 - Prob. 19.56QPCh. 19 - Prob. 19.57QPCh. 19 - Prob. 19.58QPCh. 19 - Prob. 19.59QPCh. 19 - Prob. 19.60QPCh. 19 - Prob. 19.61QPCh. 19 - Prob. 19.62QPCh. 19 - Prob. 19.63QPCh. 19 - Prob. 19.64QPCh. 19 - Prob. 19.65QPCh. 19 - 19.66 A sample of iron ore weighing 0.2792 g was...Ch. 19 - Prob. 19.67QPCh. 19 - Prob. 19.68QPCh. 19 - Prob. 19.69QPCh. 19 - Prob. 19.70QPCh. 19 - Prob. 19.71QPCh. 19 - Prob. 19.72QPCh. 19 - Prob. 19.73QPCh. 19 - Prob. 19.74QPCh. 19 - Prob. 19.75QPCh. 19 - Prob. 19.76QPCh. 19 - Prob. 19.77QPCh. 19 - Prob. 19.78QPCh. 19 - Prob. 19.79QPCh. 19 - Prob. 19.80QPCh. 19 - Prob. 19.81QPCh. 19 - Prob. 19.82QPCh. 19 - Prob. 19.83QPCh. 19 - Prob. 19.84QPCh. 19 - Prob. 19.86QPCh. 19 - Prob. 19.87QPCh. 19 - Prob. 19.88QPCh. 19 - Prob. 19.89QPCh. 19 - Prob. 19.90QPCh. 19 - Prob. 19.91QPCh. 19 - Prob. 19.92QPCh. 19 - Prob. 19.93QPCh. 19 - Prob. 19.94QPCh. 19 - Prob. 19.95QPCh. 19 - Prob. 19.96QPCh. 19 - Prob. 19.97QPCh. 19 - Prob. 19.98QPCh. 19 - Prob. 19.99QPCh. 19 - Prob. 19.100QPCh. 19 - Prob. 19.101QPCh. 19 - 19.102 The magnitudes (but not the signs) of the...Ch. 19 - Prob. 19.103QPCh. 19 - Prob. 19.104QPCh. 19 - Prob. 19.105QPCh. 19 - Prob. 19.106QPCh. 19 - Prob. 19.107QPCh. 19 - Prob. 19.108QPCh. 19 - Prob. 19.109QPCh. 19 - Prob. 19.110QPCh. 19 - 19.111 A spoon was silver-plated electro lyrically...Ch. 19 - Prob. 19.112QPCh. 19 - Prob. 19.113QPCh. 19 - Prob. 19.114QPCh. 19 - Prob. 19.115QPCh. 19 - Prob. 19.116QPCh. 19 - Prob. 19.117QPCh. 19 - Prob. 19.118QPCh. 19 - Prob. 19.119QPCh. 19 - Prob. 19.120QPCh. 19 - Prob. 19.121SPCh. 19 - Prob. 19.122SPCh. 19 - Prob. 19.123SPCh. 19 - Prob. 19.124SPCh. 19 - Prob. 19.125SPCh. 19 - Prob. 19.126SPCh. 19 - Prob. 19.128SPCh. 19 - Prob. 19.129SPCh. 19 - Prob. 19.130SP
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