EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
Question
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Chapter 14, Problem 14.13P

(a)

Interpretation Introduction

Interpretation:

Reduction recations corresponding to each half cell along with potential and overall cell voltage have to be calculated.

Concept Introduction:

Expression to compute Ecell as per the Nernst equation is written as follows:

  Ecell=Ecell°0.0592nlog[P][R]

Here,

  • Ecell denotes overall cell potential.
  • Ecell° denotes standard cell potential.
  • n is the moles of electrons transferred in each reaction.
  • [P] denotes concentration of products.
  • [R] denotes concentration of reactant.

(a)

Expert Solution
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Explanation of Solution

Reduction half cell reaction in anodic half cell is as follows:

  Br2(l)+2e2Br

Reduction half cell in cathodic-half cell is as follows:

  Al3++3eAl(s)

Expression to compute Ecell  as per Nernst equation is as follows:

  Ecell=Ecell°0.0592nlog[P][R]        (1)

Substitute 1.078 V for Ecell°, 2 for n, [Br]2 for [P] and 1 for [R] in equation (1) to obtain Ecell for anodic half cell.

  Ecell=1.078 V0.05922log[Br]2        (2)

Substitute 0.10 M for [H+] in equation (2).

  Ecell=1.078 V0.05922log(0.10 M)2=1.1372 V

Substitute 1.677 V for Ecell°, 3 for n, [Al3+] for [R] and 1 for [P] in equation (1) to obtain Ecell for cathodic half cell.

  Ecell=1.677 V0.05923log1[Al3+]        (3)

Substitute 0.010 M for [Al3+] in equation (3).

  Ecell=1.677 V0.05923log(10.010 M)=1.716 V

Cell potential is evaluated by expression as follows:

  Ecell°=ErightEleft        (4)

Substitute 1.716 V for Eright and 1.1372 V for Eleft in equation (4).

  Ecell°=1.716 V(1.1372 V)=2.854 V

Hence net cell voltage is 2.854 V.

(b)

Interpretation Introduction

Interpretation:

Cell-diagram and net cell reaction in spontaneous direction has to be given.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
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Explanation of Solution

Cell-diagram is illustrated as follows:

EBK EXPLORING CHEMICAL ANALYSIS, Chapter 14, Problem 14.13P

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. Thus electrons move spontaneously from Al towards Pt electrode.

Net cell reaction in spontaneous direction is as follows:

  32Br2(l)+Al(s)3Br+Al3+

(c)

Interpretation Introduction

Interpretation:

Limiting reagent has to be identified.

Concept Introduction:

Reagent present in less quantity governs the overall reaction as it is consumed first. Such reagent is termed Limiting reagent.

(c)

Expert Solution
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Explanation of Solution

Moles of Br2 in 14.3 mL are calculated as follows:

  Moles of Br2=(3.12 g1 mL)(14.3 mL)(1 mol159.808 g)=0.2791 mol

Moles of Al are calculated as follows:

  Moles of Al=(12 g1 s)(1 mol26.98 g)=0.4447 mol

Since Br2 is present in lesser amount it acts as limiting reagent as it would be consumed rapidly.

(d)

Interpretation Introduction

Interpretation:

Electrical work due to 0.231 mL of Br2 has to be calculated.

Concept Introduction:

Work done is evaluated by expression as follows:

  W=VQ

Here,

Q denotes charge .

V denotes voltage.

W denotes electrical work done.

(d)

Expert Solution
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Explanation of Solution

Moles of Br2 in 0.231 mL are calculated as follows:

  Moles of Br2=(3.12 g1 mL)(0.231 mL)(1 mol159.808 g)=0.004507 mol

Work done in terms of Faradays is evaluated by expression as follows:

  W=VnZF        (5)

Substitute 0.004507 mol for n, 3 for Z, 9.649×104 C/mol for F and 1.50 V for V in equation (5).

  W=(1.50 V)(0.004507 mol)(3)(9.649×104 C/mol)=1.149×103 J

Hence, electrical work due to 0.231 mL of Br2 is 1.149×103 J.

(e)

Interpretation Introduction

Interpretation:

Rate of dissolution of Al(s) has to be calculated.

Concept Introduction:

On faraday is equivalent to flow of 9.649×104 C charge per mol.

(e)

Expert Solution
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Explanation of Solution

Since 2.89×104C/s flows and one faraday constitutes 9.649×104 C/mol thus amount of electron that flow per hour is calculated as follows:

  Flow rate for 3e=13(2.89×104C1 s)(1 mol9.649×104 C)=9.98×1010 mol/s

Since 1 mol Al has mass of 26.98 g/mol so rate in terms of g/s is calculated as follows:

  Flow rate(g/s)=(9.98×1010 mol1 s)(26.98 g1 mol)=2.69×108 g/s

Hence, rate of dissolution of Al(s) is 2.69×108 g/s.

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