1. Construct the potential-pH (Pourbaix) diagram for zinc using the following reactions and assuming the activity of all dissolved substances (i.e. any metal cations or anions) is 106. Label the stable species in each region. Designate regions as Active (Corrosion), Passive, or Immune. Ranges for potential -2.5 VSHE to +2 VSHE (Y-axis) and for pH -2 to 14 (X-axis) while plotting the results. {1} Zn²++2e−=Zn(s) (E。= -0.764 V vs.SHE) {2} ZnO(s)+2H++2e¯¯=Zn(s)+H₂O (E。=-0.439 V vs.SHE) {3} ZnO2+4H+2e¯=Zn(s)+ 2H2O (E0= +0.441V) {4}Zn²++H20=Zn0+2H+ (equilibrium constant = 3.16 X 10-³) {5} ZnO+H20=ZnO₂²+2H+ (equilibrium constant = 1.58 X 10-30)
1. Construct the potential-pH (Pourbaix) diagram for zinc using the following reactions and assuming the activity of all dissolved substances (i.e. any metal cations or anions) is 106. Label the stable species in each region. Designate regions as Active (Corrosion), Passive, or Immune. Ranges for potential -2.5 VSHE to +2 VSHE (Y-axis) and for pH -2 to 14 (X-axis) while plotting the results. {1} Zn²++2e−=Zn(s) (E。= -0.764 V vs.SHE) {2} ZnO(s)+2H++2e¯¯=Zn(s)+H₂O (E。=-0.439 V vs.SHE) {3} ZnO2+4H+2e¯=Zn(s)+ 2H2O (E0= +0.441V) {4}Zn²++H20=Zn0+2H+ (equilibrium constant = 3.16 X 10-³) {5} ZnO+H20=ZnO₂²+2H+ (equilibrium constant = 1.58 X 10-30)
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter15: Acid–base Equilibria
Section: Chapter Questions
Problem 82AP
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Transcribed Image Text:1. Construct the potential-pH (Pourbaix) diagram for zinc using the following reactions and
assuming the activity of all dissolved substances (i.e. any metal cations or anions) is 106. Label
the stable species in each region. Designate regions as Active (Corrosion), Passive, or Immune.
Ranges for potential -2.5 VSHE to +2 VSHE (Y-axis) and for pH -2 to 14 (X-axis) while plotting
the results.
{1} Zn²++2e−=Zn(s) (E。= -0.764 V vs.SHE)
{2} ZnO(s)+2H++2e¯¯=Zn(s)+H₂O (E。=-0.439 V vs.SHE)
{3} ZnO2+4H+2e¯=Zn(s)+ 2H2O (E0= +0.441V)
{4}Zn²++H20=Zn0+2H+ (equilibrium constant = 3.16 X 10-³)
{5} ZnO+H20=ZnO₂²+2H+ (equilibrium constant = 1.58 X 10-30)
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