Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
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Chapter 17, Problem 17.BE

(a)

Interpretation Introduction

Interpretation:

The cathode potential at which Sb deposition will happen from SbO+ has to be calculated.

Concept Introduction:

When the electric current is too small, the voltage of cell is given as

E = E(cathode)-E(anode)

E(cathode) is electrode’s potential which is attached to negative terminal of current source.

E(anode) is electrode’s potential which is attached to positive  terminal of current source.

Concentration Polarization:  It is the change in concentration of products and reactants at electrode’s surface unlike they are same in solution.

(a)

Expert Solution
Check Mark

Answer to Problem 17.BE

The cathode potential at which Sb deposition will happen from SbO+ is 0.028 V

Explanation of Solution

To determine: The cathode potential at which Sb deposition will happen from SbO+ .

E(cathode)  =0.208 -0.059 162log1[SbO+][H+]2=0.208 -0.059 162log1(0.010) (1.0)2=0.169VE(cathode versus Ag |AgCl) =E(versus S.H.E.) - E(Ag |AgCl)=0.169 -0.197=0.028 V

(b)

Interpretation Introduction

Interpretation:

The percentage of Cu2+ ion gets converted into copper has to be calculated.

Concept Introduction:

When the electric current is too small, the voltage of cell is given as

E = E(cathode)-E(anode)

E(cathode) is electrode’s potential which is attached to negative terminal of current source.

E(anode) is electrode’s potential which is attached to positive  terminal of current source.

Concentration Polarization:  It is the change in concentration of products and reactants at electrode’s surface unlike they are same in solution.

(b)

Expert Solution
Check Mark

Answer to Problem 17.BE

The percentage of Cu2+ ion gets converted into copper is 99.998% .

Explanation of Solution

To determine: The percentage of Cu2+ ion gets converted into copper.

The concentration of copper ion equilibrium with copper at 0.169V is calculated as follows

Cu2++ 2e- Cu(s) Eo=0.339E(cathode)  =0.339 -0.059 162log1[Cu2+]0.169=0.339 -0.059 162log1[Cu2+][Cu2+]=1.8×10-6M

The percentage of copper ion not reduced is determined as

=1.8×10-60.10×100=1.8×10-3%%ofreducedcopperionis99.998%

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