Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 17, Problem 86E

 (a)

Interpretation Introduction

Interpretation:

The information regarding the disproportionation reaction is given. Various questions based on the spontaneity are to be answered and the calculation of ΔG° and K is to be stated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specified as,

E=E°(0.0591n)log(Q)

To determine: The spontaneity of the given reaction (a) and further calculation of ΔG° and K .

 (a)

Expert Solution
Check Mark

Answer to Problem 86E

  1. a. The given reaction (a) is spontaneous and The value of ΔG° for the given reaction is calculated as -34.73kJ_ and The value of K is calculated as 1.2×106_ .

Explanation of Solution

Given

The disproportionation reaction is,

2Cu+(aq)Cu2+(aq)+Cu(s)

The reaction taking place at cathode is,

Cu+(aq)+eCu(s)E°red=0.52V

The reaction taking place at anode is,

Cu+(aq)Cu2+(aq)+eE°ox=0.16V

Add both the oxidation and reduction half-reaction,

Cu+(aq)+eCu(s)Cu+(aq)Cu2+(aq)+e

The final equation is,

2Cu+(aq)Cu2+(aq)+Cu(s)

The overall cell potential is calculated as,

E°cell=E°ox+E°red=0.16V+0.52V=0.36V

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • n is the number of electrons involved in the reaction.
  • F is the Faraday’s constant.
  • E°cell is the cell potential at the standard condition.

As the value of E°cell comes out to be positive then ΔG° becomes negative. Therefore it leads to a spontaneous reaction.

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

The reaction involves the transfer of 1mole

Substitute the values of n , E°cell and F in the above equation as,

ΔG°=1×96,485×0.36=34,734.6J=-34.73kJ_

The value of ΔG° for the given reaction is calculated as -34.73kJ_ .

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

ΔG°=RTlnK=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • R is the universal gas constant.
  • T is the temperature in Kelvin.
  • K is the equilibrium constant.

Rearranged equation at 25°C is obtained as,

lnK=((ΔG°)R×T)

Substitute the obtained values in above equation,

2.303logK=((34.73kJ).00831kJKmol×298K)logK=6.08K=106.08=1.2×106_

The value of K is calculated as 1.2×106_ .

(b)

Interpretation Introduction

Interpretation:

The information regarding the disproportionation reaction is given. Various questions based on the spontaneity are to be answered and the calculation of ΔG° and K is to be stated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specified as,

E=E°(0.0591n)log(Q)

To determine: The spontaneity of the given reaction (b) and the value of ΔG° and K if the reaction is spontaneous.

(b)

Expert Solution
Check Mark

Answer to Problem 86E

  1. b. The given reaction (b) is non-spontaneous in nature.

Explanation of Solution

The disproportionation reaction is,

3Fe2+(aq)2Fe3+(aq)+Fe(s)

The reaction taking place at cathode is,

Fe2+(aq)+2eFe(s)E°red=0.44V

The reaction taking place at anode is,

2Fe2+(aq)2Fe3+(aq)+2eE°ox=0.77V

Add both the oxidation and reduction half-reaction,

Fe2+(aq)+2eFe(s)2Fe2+(aq)2Fe3+(aq)+2e

The final equation is,

3Fe2+(aq)2Fe3+(aq)+Fe(s)

The overall cell potential is calculated as,

E°cell=E°ox+E°red=0.44V+(0.77V)=-1.21V_

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • n is the number of electrons involved in the reaction.
  • F is the Faraday’s constant.
  • E°cell is the cell potential at the standard condition.

As the value of E°cell comes out to be negative then ΔG° becomes positive. Therefore it leads to a non-spontaneous reaction.

(c)

Interpretation Introduction

Interpretation:

The information regarding the disproportionation reaction is given. Various questions based on the spontaneity are to be answered and the calculation of ΔG° and K is to be stated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specified as,

E=E°(0.0591n)log(Q)

To determine: The spontaneity of the given reaction (c) and further calculation of ΔG° and K if the reaction is spontaneous.

(c)

Expert Solution
Check Mark

Answer to Problem 86E

  1. c. The given reaction (c) is spontaneous and The value of ΔG° for the given reaction is calculated as -84.90kJ_ and The value of K is calculated as 7.5×1014_ .

Explanation of Solution

Given

The disproportionation reaction is,

HClO2(aq)ClO3(aq)+HClO(aq)

The reaction taking place at cathode is,

HClO2(aq)+2H+(aq)+2eHClO(aq)+H2O(l)E°red=1.65V

The reaction taking place at anode is,

HClO2(aq)+H2O(l)ClO3(aq)+3H+(aq)+2eE°ox=1.21V

Add both the oxidation and reduction half-reaction,

HClO2(aq)+2H+(aq)+2eHClO(aq)+H2O(l)HClO2(aq)+H2O(l)ClO3(aq)+3H+(aq)+2e

The final equation is,

2HClO2(aq)HClO(aq)+ClO3(aq)+H+(aq)

The overall cell potential is calculated as,

E°cell=E°ox+E°red=1.21V+1.65V=0.44V_

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • n is the number of electrons involved in the reaction.
  • F is the Faraday’s constant.
  • E°cell is the cell potential at the standard condition.

As the value of E°cell comes out to be positive then ΔG° becomes negative. Therefore it leads to a spontaneous reaction.

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

The reaction involves the transfer of 2 moles of electrons.

Substitute the values of n , E°cell and F in the above equation as,

ΔG°=2×96,485×0.44=84,906.8J=-84.90kJ_

The value of ΔG° for the given reaction is calculated as -84.90kJ_ .

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

ΔG°=RTlnK=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • n is the number of electrons involved in the reaction.
  • F is the Faraday’s constant.
  • E°cell is the cell potential at the standard condition.

Rearranged equation at 25°C is obtained as,

lnK=((ΔG°)R×T)

Substitute the obtained values in above equation,

2.303logK=((84.90kJ)0.00831kJKmol×298K)logK=14.88K=1014.88=7.5×1014_

The value of K is calculated as 7.5×1014_ .

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Chapter 17 Solutions

Chemistry: An Atoms First Approach

Ch. 17 - Prob. 2ALQCh. 17 - Prob. 3ALQCh. 17 - Prob. 4ALQCh. 17 - Sketch a cell that forms iron metal from iron(II)...Ch. 17 - Which of the following is the best reducing agent:...Ch. 17 - Prob. 7ALQCh. 17 - Prob. 8ALQCh. 17 - Explain why cell potentials are not multiplied by...Ch. 17 - What is the difference between and ? When is equal...Ch. 17 - Prob. 11ALQCh. 17 - Look up the reduction potential for Fe3+ to Fe2+....Ch. 17 - Prob. 13ALQCh. 17 - Is the following statement true or false?...Ch. 17 - Prob. 15RORRCh. 17 - Assign oxidation numbers to all the atoms in each...Ch. 17 - Specify which of the following equations represent...Ch. 17 - The Ostwald process for the commercial production...Ch. 17 - Prob. 19QCh. 17 - Prob. 20QCh. 17 - When magnesium metal is added to a beaker of...Ch. 17 - How can one construct a galvanic cell from two...Ch. 17 - The free energy change for a reaction, G, is an...Ch. 17 - What is wrong with the following statement: The...Ch. 17 - When jump-starting a car with a dead battery, the...Ch. 17 - Prob. 26QCh. 17 - Prob. 27QCh. 17 - Consider the following electrochemical cell: a. If...Ch. 17 - Balance the following oxidationreduction reactions...Ch. 17 - Prob. 30ECh. 17 - Prob. 31ECh. 17 - Prob. 32ECh. 17 - Chlorine gas was first prepared in 1774 by C. W....Ch. 17 - Gold metal will not dissolve in either...Ch. 17 - Prob. 35ECh. 17 - Consider the following galvanic cell: a. Label the...Ch. 17 - Prob. 37ECh. 17 - Sketch the galvanic cells based on the following...Ch. 17 - Prob. 39ECh. 17 - Prob. 40ECh. 17 - Prob. 41ECh. 17 - Prob. 42ECh. 17 - Prob. 43ECh. 17 - Give the standard line notation for each cell in...Ch. 17 - Prob. 45ECh. 17 - Prob. 46ECh. 17 - Prob. 47ECh. 17 - Prob. 48ECh. 17 - Prob. 49ECh. 17 - The amount of manganese in steel is determined by...Ch. 17 - Prob. 51ECh. 17 - Prob. 52ECh. 17 - Estimate for the half-reaction 2H2O+2eH2+2OH given...Ch. 17 - Prob. 54ECh. 17 - Glucose is the major fuel for most living cells....Ch. 17 - Direct methanol fuel cells (DMFCs) have shown some...Ch. 17 - Prob. 57ECh. 17 - Using data from Table 17-1, place the following in...Ch. 17 - Answer the following questions using data from...Ch. 17 - Prob. 60ECh. 17 - Consider only the species (at standard conditions)...Ch. 17 - Prob. 62ECh. 17 - Prob. 63ECh. 17 - Prob. 64ECh. 17 - Prob. 65ECh. 17 - Prob. 66ECh. 17 - A galvanic cell is based on the following...Ch. 17 - Prob. 68ECh. 17 - Consider the concentration cell shown below....Ch. 17 - Prob. 70ECh. 17 - The overall reaction in the lead storage battery...Ch. 17 - Prob. 72ECh. 17 - Consider the cell described below:...Ch. 17 - Consider the cell described below:...Ch. 17 - Prob. 75ECh. 17 - Prob. 76ECh. 17 - Prob. 77ECh. 17 - Prob. 78ECh. 17 - Prob. 79ECh. 17 - An electrochemical cell consists of a nickel metal...Ch. 17 - An electrochemical cell consists of a standard...Ch. 17 - Prob. 82ECh. 17 - Consider a concentration cell that has both...Ch. 17 - Prob. 84ECh. 17 - Prob. 85ECh. 17 - Prob. 86ECh. 17 - Consider the following galvanic cell at 25C:...Ch. 17 - Prob. 88ECh. 17 - Prob. 89ECh. 17 - Prob. 90ECh. 17 - Prob. 91ECh. 17 - The solubility product for CuI(s) is 1.1 102...Ch. 17 - How long will it take to plate out each of the...Ch. 17 - The electrolysis of BiO+ produces pure bismuth....Ch. 17 - What mass of each of the following substances can...Ch. 17 - Prob. 96ECh. 17 - An unknown metal M is electrolyzed. It took 74.1 s...Ch. 17 - Electrolysis of an alkaline earth metal chloride...Ch. 17 - What volume of F2 gas, at 25C and 1.00 atm, is...Ch. 17 - What volumes of H2(g) and O2(g) at STP are...Ch. 17 - Prob. 101ECh. 17 - A factory wants to produce 1.00 103 kg barium...Ch. 17 - It took 2.30 min using a current of 2.00 A to...Ch. 17 - A solution containing Pt4+ is electrolyzed with a...Ch. 17 - A solution at 25C contains 1.0 M Cd2+, 1.0 M Ag+,...Ch. 17 - Consider the following half-reactions: A...Ch. 17 - In the electrolysis of an aqueous solution of...Ch. 17 - Copper can be plated onto a spoon by placing the...Ch. 17 - Prob. 109ECh. 17 - Prob. 110ECh. 17 - Prob. 111ECh. 17 - What reaction will take place at the Cathode and...Ch. 17 - Gold is produced electrochemically from an aqueous...Ch. 17 - Prob. 114AECh. 17 - The saturated calomel electrode. abbreviated SCE....Ch. 17 - Consider the following half-reactions: Explain why...Ch. 17 - Consider the standard galvanic cell based on the...Ch. 17 - Prob. 118AECh. 17 - The black silver sulfide discoloration of...Ch. 17 - Prob. 120AECh. 17 - When aluminum foil is placed in hydrochloric acid,...Ch. 17 - Prob. 122AECh. 17 - Prob. 123AECh. 17 - The overall reaction and equilibrium constant...Ch. 17 - What is the maximum work that can be obtained from...Ch. 17 - The overall reaction and standard cell potential...Ch. 17 - Prob. 127AECh. 17 - Prob. 128AECh. 17 - Prob. 129AECh. 17 - Prob. 130AECh. 17 - Prob. 131AECh. 17 - Prob. 132AECh. 17 - Prob. 133AECh. 17 - Prob. 134CWPCh. 17 - Consider a galvanic cell based on the following...Ch. 17 - Consider a galvanic cell based on the following...Ch. 17 - Consider a galvanic cell based on the following...Ch. 17 - An electrochemical cell consists of a silver metal...Ch. 17 - An aqueous solution of PdCl2 is electrolyzed for...Ch. 17 - Prob. 140CPCh. 17 - Prob. 141CPCh. 17 - The overall reaction in the lead storage battery...Ch. 17 - Consider the following galvanic cell: Calculate...Ch. 17 - Prob. 144CPCh. 17 - A galvanic cell is based on the following...Ch. 17 - Prob. 146CPCh. 17 - The measurement of pH using a glass electrode...Ch. 17 - Prob. 148CPCh. 17 - A galvanic cell is based on the following...Ch. 17 - Prob. 150CPCh. 17 - Prob. 151CPCh. 17 - Prob. 152CPCh. 17 - Consider the following galvanic cell: A 15 0-mole...Ch. 17 - When copper reacts with nitric acid, a mixture of...Ch. 17 - The following standard reduction potentials have...Ch. 17 - An electrochemical cell is set up using the...Ch. 17 - Three electrochemical cells were connected in...Ch. 17 - A silver concentration cell is set up at 25C as...Ch. 17 - A galvanic cell is based on the following...Ch. 17 - Prob. 160MP
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