Exploring Chemical Analysis
Exploring Chemical Analysis
5th Edition
ISBN: 9781429275033
Author: Daniel C. Harris
Publisher: Macmillan Higher Education
Question
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Chapter 15, Problem 15.6P

(a)

Interpretation Introduction

Interpretation:

Titration reaction has to be written and equivalence volume has to be calculated.

Concept Introduction:

Cell potential is evaluated by expression as follows:

  Ecell°=Eoxidation°+Ereduction°

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.

Expression to compute E for silver ion electrode is given as follows:

  Ecell=0.558+0.05916log[Ag+]

(a)

Expert Solution
Check Mark

Explanation of Solution

Titration reaction is written as follows:

  Hg22++2ClHg2Cl2(s)        (1)

Dilution formula is given as follows:

  M1V1=M2V2        (2)

Substitute  0.100 M for M1, 0.100 M for M2 , 50.0 mL for V2, in equation (2).

  (0.100 M)V1=(0.100 M)(50.0 mL)        (3)

Rearrange equation (3) to calculate value of V1.

  V1=(0.100 M)(50.0 mL)(0.100 M)=50 mL

Since volume of Hg22+ required to neutralize Cl is half of its volume as per equation (1), so equivalence volume is calculated as follows:

  Ve=50 mL2=25 mL

Thus equivalence volume is 25 mL.

(b)

Interpretation Introduction

Interpretation:

Equation to compute cell voltage has to be derived.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Electrochemical cell for mercury electrode is written as follows:

  Hg22++2e2Hg(l)

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

  Ecell=Ecell°0.0592nlog[P][R]        (4)

Since Hg(l) is formed as product, value of [Hg(l)] is taken as unity.

Substitute 2 for n, [Hg22+] for [R] and 1 for [P] in equation (4).

  Ecell=Ecell°0.05922log1[Hg22+]        (5)

Rearrange equation (5).

  Ecell=Ecell°0.05922log[Hg22+]1=Ecell°0.05922(1)log[Hg22+]=Ecell°+0.05922log[Hg22+]

Reduction potential correspoding to mercury electrode is 0.796 V.

Cell potential is evaluated by expression as follows:

  Ecell°=Eright°Eleft°        (6)

Substitute 0.796 V+(0.0592/2)log[Hg22+] for Eright° and 0.241 V for Eleft° in equation (6).

  Ecell°=0.796 V+0.05922log[Hg22+]0.241 VEcell°=0.555+0.05922log[Hg22+]

Hence cell voltage can be obtained from the equation as follows:

  Ecell°=0.555+0.05922log[Hg22+]

(c)

Interpretation Introduction

Interpretation:

Value of [Ag+] and E at indicated volumes has to be calculated and titration curve has to be sketched.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Titration reaction is written as follows:

  Hg22++2ClHg2Cl2(s)

When 0.1 mL solution of Hg22+ is added, concentration of unreacted Cl is calculated as follows:

  [Cl]=(50.01) mmol(50+0.1) mL=0.0996 M

Expression to compute metal ion concentration before equivalence volume is given as follows:

  [Hg22+]=Ksp[Cl]2        (3)

Substitute 0.0996 M for [Cl] and 1.2×1018 for Ksp in equation (3).

  [Hg22+]=1.2×1018(0.0996 M)2=1.209×1016 M

Expression to compute E for silver ion electrode is given as follows:

  Ecell=0.555+0.059162log[Hg22+]        (4)

Substitute 1.209×1016 M for [Hg22+] in equation (4).

  Ecell=0.555+0.059162log(1.209×1016 M)=0.084 V

When 10 mL solution is added, concentration of unreacted Cl is calculated as follows:

  [Cl]=(51) mmol(50+10) mL=0.0667 M

Substitute 0.0667 M for [Cl] and 1.2×1018 for Ksp in equation (3).

  [Hg22+]=1.2×1018(0.0667 M)2=2.697×1016 M

Substitute 2.697×1016 M for [Hg22+] in equation (4).

  Ecell=0.555+0.059162log(2.697×1016 M)=0.102 V

Expression to compute [Hg22+] at equivalence volume is given as follows:

  [Hg22+]=(Ksp4)1/3        (5)

Substitute 1.2×1018 for Ksp in equation (5).

  [Hg22+]=(1.2×10184)1/3=6.7×107 M

Substitute 6.7×107 M for [Hg22+] in equation (4).

  Ecell=0.555+0.059162log(6.7×107 M)=0.372 V

Beyond equivalence volume when 30 mL solution is added excess [Hg22+] is calculated as follows:

  [Hg22+]=(50.1) mmol(50+30) mL=0.0063 M

Substitute 0.0063 M for [Hg22+] in equation (4).

  Ecell=0.555+0.059162log(0.0063)=0.490 V

Corresponding titration curve is drawn as follows:

Exploring Chemical Analysis, Chapter 15, Problem 15.6P

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