Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
Question
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Chapter 19, Problem 19D.3P

(a)

Interpretation Introduction

Interpretation:

The zero-current potential of the Fe2+/Fe cathode and overpotential of each given potential have to be calculated.

(a)

Expert Solution
Check Mark

Explanation of Solution

The zero-current potential of the Fe2+/Fe cathode and overpotential are calculated as,

    Fe2++2e-Fe ν=2 E0=-0.447E0=E0-RTνFlnQ=E0-RTνF1Fe2+ γFe2+=1=-0.447V-25.693×10-3V2ln(moldm-31.70×10-6moldm3)=-0.618V(618mV)η=E'-E0 ηvaluesarereportedas,η=84,109,134,194.

(b)

Interpretation Introduction

Interpretation:

The cathodic current density from the rate of deposition of iron ion has to be calculated.

(b)

Expert Solution
Check Mark

Explanation of Solution

The cathodic current density from the rate of deposition of iron ion is calculated as,

    j=νFAdnFedt=2(96485C/mol)91cm2dnFedtj=j0{e(1-α)-e-αfη}=j0e-αfη{e{}-1} =-j0{e-1}j0=-je-1.

j0 values are reported as,

dnFedt(10-12mol/s)-E’/mV-η/mVj/(μAcm-12)jc/(μAcm-12)
1.47702840.03120.0324
2.187271090.04620.0469
3.117521340.06590.0663
7.268121940.1540.154

(c)

Interpretation Introduction

Interpretation:

The exchange-current density has to be determined by analysing the given data.

(c)

Expert Solution
Check Mark

Explanation of Solution

The exchange-current density using the given data is calculated as,

    j=j0e-αfηlnj0=lnj0Performinglinearregressionanalysisofthelnjcversusηdata,wefindlnj0=4.608 standarddeviation=0.015αf=0.0413standarddeviation=0.00011R=0.99994

The correlation coefficient and the standard deviation indicate that the plot provides an excellent description of the data.

    j0=e4.608 or j0=0.00997μAcm-2α=0.01413f=(0.01413mV-1)(26.693mV)α=0.363.

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