EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
Question
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Chapter 16, Problem 16.1P
Interpretation Introduction

Interpretation:

Values of Eo and K for below titration reaction have to be determined.

  Ce4++Fe2+Ce3++Fe3+

Concept Introduction:

Expression to calculate value of K is as follows:

  K=10nEo/0.05916        (1)

Here,

n is number of electrons.

Eo is standard cell potential.

Expert Solution & Answer
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Answer to Problem 16.1P

Values of Eo and K are 0.933 V and 6×1015.

Explanation of Solution

Given titration reaction is as follows:

  Ce4++Fe2+Ce3++Fe3+

Reduction half-cell reaction of electrochemical cell is as follows:

  Ce4++eCe3+

Reduction potential of right half-cell reaction is 1.70 V.

Oxidation half-cell reaction of electrochemical cell is as follows:

  Fe2+Fe3++e

Reduction potential of left half-cell reaction is 0.767 V.

Value of Eo can be calculated as follows:

  Eo=ERedoEOxo=1.70 V0.767 V=0.933 V

Expression to calculate value of K is as follows:

  K=10nEo/0.05916        (1)

Substitute 1 for n and 0.933 V for Eo in equation (1).

  K=10(1)(0.933 V)/0.05916=6×1015

Hence value of K is 6×1015.

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