CHEMISTRY:CENTRAL..(LL)W/S.G.+MASTERING
CHEMISTRY:CENTRAL..(LL)W/S.G.+MASTERING
14th Edition
ISBN: 9780134773223
Author: Brown
Publisher: PEARSON
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Question
Chapter 20, Problem 1DE

(a)

Interpretation Introduction

To determine: Sketch of a labeled voltaic cell and the concentration of the solution.

(a)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution:

The labeled sketch of the voltaic cell is,

    CHEMISTRY:CENTRAL..(LL)W/S.G.+MASTERING, Chapter 20, Problem 1DE , additional homework tip  1

The concentration of the copper and the manganese ion is 1M and the mass of manganese and copper present in the 50mL of their ionic solution is 2.747g and 3.177g , respectively.

Explanation of Solution

Given

The electrical potential output of the cell is 1.50V .

The external device draws the current of 0.50A in 2h=2×3600s .

The electric potential of the cell is calculated as,

    Electricpotentialofthecell=Oxidationpotential+Reductionpotential

According to the above equation, the difference in the oxidation and the reduction potential should be 1.50V .

The standard oxidation potential of manganese is +1.18V and the standard reduction potential of copper is +0.337V . The sum of these two values of standard oxidation and standard reduction potential gives,

    +1.18V++0.337V=1.51V

Therefore, the pair of manganese and copper is used to construct a voltaic cell with manganese being anode at which oxidation takes place and copper being cathode at which reduction takes place.

The standard reduction potential is the potential of the electrode dipped in the aqueous solution of its ion of concentration 1M .

Therefore, the concentration of the copper and the manganese ion is 1M .

We are given the beaker of the capacity of 100mL .

The volume of the solution of manganese and copper ion is assumed to be 50mL=0.05L .

The labeled sketch of the voltaic cell is,

    CHEMISTRY:CENTRAL..(LL)W/S.G.+MASTERING, Chapter 20, Problem 1DE , additional homework tip  2
    Figure. 1
The atomic mass of manganese and copper is 54.938g/mol and 63.546g/mol , respectively.

The mass of the manganese in the 50mL solution is calculated by the formula,
    Mass=Molarconcentration×Molarmass×Volumeofsolutioninliter

Substitute the value of molar concentration, molar mass and the volume of the solution of manganese in the above equation.

    Mass=1M×54.938g/mol×0.05L=2.747g

The mass of the copper in the 50mL solution is calculated by the formula,

    Mass=Molarconcentration×Molarmass×Volumeofsolutioninliter

Substitute the value of molar concentration, molar mass and the volume of the solution of copper in the above equation.

    Mass=1M×63.546g/mol×0.05L=3.177g

Therefore, the mass of manganese and copper present in the 50mL of their ionic solution is 2.747g and 3.177g , respectively.

Conclusion



The concentration of the copper and the manganese ion is 1M . The mass of manganese is 2.747g and copper is 3.177g present in the 50mL of their ionic solution.

(b)

Interpretation Introduction

To determine: The concentration of manganese and copper ion in the solution after 2 hour.

(b)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The concentration of manganese and copper ion in the solution after 2 hour is 1.37M and 0.63M , respectively.

Explanation of Solution

Given

The electrical potential output of the cell is 1.50V .

The external device draws the current of 0.50A in 2h=2×3600s .

The charge transport is calculated by the formula,

    Chargetransfer=Current×time

Substitute the value of current and time in the above formula,

    Chargetransfer=0.50A×2×3600s=3600C

Charge on each electron is 1.602×1019C and the value of Avogadro’s number is 6.023×1023 .

The moles of electron present in 3600C is calculated as,

    Numberofelectrons=ChargetransferChargeonelectron×Avogadro'snumber

Substitute the value of charge transfer, charge on each electron and Avogadro’s number in the above formula.

    Numberofelectrons=3600C1.602× 10 19C×6.023× 10 23=0.037

Two electron transfer takes place by oxidation and reduction of manganese and copper, respectively.

Therefore, the change in moles of manganese and copper is:

    0.0372=0.0185mol .

The initial concentration of the ions is 1M .

Thus, one mole of ion is present in one liter of the solution and 0.05 mole ion is present in 0.05 liter of the solution.

The number of moles of manganese ion increase due to oxidation of metal. Therefore, the molar concentration of manganese ion in 50mL=0.05L of the solution is:
    0.05+0.01850.05=1.37M .

The number of moles of copper ion increase due to oxidation of metal. Therefore, the molar concentration of copper ion in 50mL=0.05L of the solution is:
    0.050.01850.05=0.63M .

Thus, the concentration of manganese and copper ion after two hours is 1.37M and 0.63M , respectively.
Conclusion



The concentration of manganese and copper ion after two hours is 1.37M and 0.63M , respectively.

(c)

Interpretation Introduction

To determine: The voltage that a cell register at the end of the discharge.

(c)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The voltage that a cell register at the end of the discharge is 1.405V .

Explanation of Solution

Given

The electrical potential output of the cell is 1.50V .

The external device draws the current of 0.50A in 2h=2×3600s .

The concentration of manganese and copper ion after two hours is 1.37M and 0.63M , respectively at discharge.

The standard oxidation potential of manganese is +1.18V and the standard reduction potential of copper is +0.337V .

The standard reduction potential is the potential of the electrode dipped in the aqueous solution of its ion of concentration 1M .

Thus, one mole of manganese ion gives the oxidation potential of +1.18V .

Therefore, 1.37M of manganese ion gives the oxidation potential of +1.18V×1.37=+1.617V .

One mole of copper ion gives the reduction potential of +0.337V .

Therefore, 0.63M of manganese ion gives the oxidation potential of +0.337V×0.63=+0.212V .

The electric potential of the cell at discharge is calculated as,

    Electricpotentialofthecell=OxidationpotentialReductionpotential

Substitute the value of oxidation and reduction potential in the above equation.

    Electricpotentialofthecell=+1.617V0.212V=1.405V

Thus, the end cell potential is 1.405V .

Conclusion



The end cell potential is 1.405V .

(d)

Interpretation Introduction

To determine: The time taken for the reactant of one cell to get completely consumed.

(d)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The time taken for the reactant of one cell to get completely consumed is 108.11 .

Explanation of Solution

Given

The electrical potential output of the cell is 1.50V .

The external device draws the current of 0.50A in 2h=2×3600s .

The charge transport is calculated by the formula,

    Chargetransfer=Current×time

Substitute the value of current and time in the above formula,

    Chargetransfer=0.50A×2×3600s=3600C

Charge on each electron is 1.602×1019C and the value of Avogadro’s number is 6.023×1023 .

The moles of electron present in 3600C is calculated as,

    Numberofelectrons=ChargetransferChargeonelectron×Avogadro'snumber

Substitute the value of charge transfer, charge on each electron and Avogadro’s number in the above formula.

    Numberofelectrons=3600C1.602× 10 19C×6.023× 10 23=0.037

Two electron transfer takes place by oxidation and reduction of manganese and copper, respectively.

Therefore, the change in moles of manganese and copper is 0.0372=0.0185mol .

Charge is directly proportional to time at constant current flow.

The charge transferred is 0.0185M in two hours.

The total initial concentration of the reactant of one half cell is one molar.

Therefore, the charge transfer of one molar takes place in 20.0185=108.11 hours.

Conclusion



The charge transfer of one molar takes place in 108.11 hours.

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Chapter 20 Solutions

CHEMISTRY:CENTRAL..(LL)W/S.G.+MASTERING

Ch. 20.4 - Prob. 20.6.1PECh. 20.4 - Prob. 20.6.2PECh. 20.4 - Prob. 20.7.1PECh. 20.4 - Prob. 20.7.2PECh. 20.4 - Prob. 20.8.1PECh. 20.4 - Practice Exercise 2 Using Table 20.1, rank...Ch. 20.5 - Practice Exercise 1 Which of the following...Ch. 20.5 - Prob. 20.9.2PECh. 20.5 - Prob. 20.10.1PECh. 20.5 - Prob. 20.10.2PECh. 20.6 - Prob. 20.11.1PECh. 20.6 - Prob. 20.11.2PECh. 20.6 - Prob. 20.12.1PECh. 20.6 - Prob. 20.12.2PECh. 20.6 - Prob. 20.13.1PECh. 20.6 - Prob. 20.13.2PECh. 20.9 - Prob. 20.14.1PECh. 20.9 - Prob. 20.14.2PECh. 20 - Prob. 1DECh. 20 - Prob. 1ECh. 20 - 20.2 You may have heard that “antioxidants” are...Ch. 20 - Prob. 3ECh. 20 - Prob. 4ECh. 20 - Prob. 5ECh. 20 - Prob. 6ECh. 20 - 20.7 Consider a redox reaction for which Eo is a...Ch. 20 - Prob. 8ECh. 20 - Prob. 9ECh. 20 - Prob. 10ECh. 20 - Prob. 11ECh. 20 - Prob. 12ECh. 20 - 20.13 What is meant by the term oxidation? On...Ch. 20 - Prob. 14ECh. 20 - Prob. 15ECh. 20 - Prob. 16ECh. 20 - Prob. 17ECh. 20 - Prob. 18ECh. 20 - Prob. 19ECh. 20 - Prob. 20ECh. 20 - Prob. 21ECh. 20 - Prob. 22ECh. 20 - Complete and balance the following half-reactions....Ch. 20 - Complete and balance the following half-reaction,...Ch. 20 - Complete and balance the following equations, and...Ch. 20 - Complete and balance the following equations, and...Ch. 20 - Prob. 27ECh. 20 - Prob. 28ECh. 20 - Prob. 29ECh. 20 - Prob. 30ECh. 20 - Prob. 31ECh. 20 - Prob. 32ECh. 20 - Prob. 33ECh. 20 - Prob. 34ECh. 20 - Prob. 35ECh. 20 - Prob. 36ECh. 20 - Prob. 37ECh. 20 - Prob. 38ECh. 20 - Prob. 39ECh. 20 - Prob. 40ECh. 20 - Prob. 41ECh. 20 - Prob. 42ECh. 20 - Prob. 43ECh. 20 - Prob. 44ECh. 20 - Prob. 45ECh. 20 - Prob. 46ECh. 20 - Assuming standard conditions, arrange the...Ch. 20 - Prob. 48ECh. 20 - Prob. 49ECh. 20 - Prob. 50ECh. 20 - Prob. 51ECh. 20 - For each of the following reactions, write a...Ch. 20 - Prob. 53ECh. 20 - Prob. 54ECh. 20 - Prob. 55ECh. 20 - Prob. 56ECh. 20 - A cell has a standard cell potential of +0.177 V...Ch. 20 - Prob. 58ECh. 20 - Prob. 59ECh. 20 - Prob. 60ECh. 20 - Prob. 61ECh. 20 - Prob. 62ECh. 20 - Prob. 63ECh. 20 - A voltaic cell utilizes the following reaction:...Ch. 20 - Prob. 65ECh. 20 - Prob. 66ECh. 20 - Prob. 67ECh. 20 - Prob. 68ECh. 20 - Prob. 69ECh. 20 - Prob. 70ECh. 20 - Prob. 71ECh. 20 - 20. 72 A voltaic cell is constructed that is based...Ch. 20 - Prob. 73ECh. 20 - Prob. 74ECh. 20 - Prob. 75ECh. 20 - Prob. 76ECh. 20 - Prob. 77ECh. 20 - In some applications nickel-cadmium batteries have...Ch. 20 - Prob. 79ECh. 20 - Prob. 80ECh. 20 - Prob. 81ECh. 20 - Prob. 82ECh. 20 - Prob. 83ECh. 20 - Prob. 84ECh. 20 - Prob. 85ECh. 20 - Prob. 86ECh. 20 - Prob. 87ECh. 20 - Prob. 88ECh. 20 - Prob. 89ECh. 20 - Prob. 90ECh. 20 - Prob. 91ECh. 20 - Metallic magnesium can be made by the electrolysis...Ch. 20 - 20.93 Calculate the mass of Li formed by...Ch. 20 - Prob. 94ECh. 20 - Prob. 95ECh. 20 - Prob. 96ECh. 20 - Prob. 97AECh. 20 - Prob. 98AECh. 20 - Prob. 99AECh. 20 - [20.100] Gold exists in two common positive...Ch. 20 - Prob. 101AECh. 20 - Prob. 102AECh. 20 - Prob. 103AECh. 20 - Prob. 104AECh. 20 - Prob. 105AECh. 20 - Prob. 106AECh. 20 - Prob. 107AECh. 20 - Prob. 108AECh. 20 - Prob. 109AECh. 20 - Prob. 110AECh. 20 - Prob. 111IECh. 20 - Prob. 112IECh. 20 - Prob. 113IECh. 20 - Prob. 114IECh. 20 - Prob. 115IECh. 20 - Prob. 116IECh. 20 - Prob. 117IECh. 20 - Prob. 118IECh. 20 - Prob. 119IE
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