Use the tabulated electrode potentials to calculate K for the oxidation of zinc by H+ (at 25 ∘C): Zn(s)+2H+(aq)→Zn2+(aq)+H2(g) Express your answer using two significant figures.

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.101QE: At 298 K, the solubility product constant for PbC2O4 is 8.5 1010, and the standard reduction...
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Use the tabulated electrode potentials to calculate K for the oxidation of zinc by H+ (at 25 ∘C):

Zn(s)+2H+(aq)→Zn2+(aq)+H2(g)

Express your answer using two significant figures.
 
Standard reduction half-cell potentials at 25° C
E° (
V)
E°
Half-reaction
Half-reaction
|(V)
Au+ (aq) + 3e→Au(s)
1.50
Fe2+ (aq) + 2e-→Fe(s)
0.45
Ag+ (aq) + e→Ag(s)
0.80
Cr (aq) + e —уCr? (aq)
0.50
Fe+ (aq) + 3e →Fe2+ (aq) 0.77
Cr+ (aq) + 3e→Cr(s)
0.73
Cu" (аq) + е —Cu(s)
0.52
Zn2+ (aq) +2e-→Zn(s)
0.76
Cu?+ (aq) + 2e-→Cu(s)
Mn2+ (aq) + 2e →Mn(s)
0.34
1.18
2H*(aq) + 20→H2(g)
0.00
Al+ (aq) + 3e -→Al(s)
1.66
Fe+ (aq) + 3e-→Fe(s)
Mg2+ (aq) + 2e–→Mg(s)
2.37
0.036
Pb2+ (aq) + 2e →Pb(s)
Na+ (aq) + e¯→Na(s)
0.13
2.71
Sn2+ (aq) + 2e→Sn(s)
Ca2+ (aq) + 2e→Ca(s)
0.14
2.76
Ni?+ (aq) + 2e–→Ni(s)
Ва2 (аq) + 2е Ba(s)
0.23
2.90
Co2+ (aq) + 2e¯→Co(s)
K* (aq) + e→K(s)
0.28
2.92
Cd2+ (aq) + 2e→Cd(s)
Lit (aq) + E→L¡(s)
0.40
3.04
Transcribed Image Text:Standard reduction half-cell potentials at 25° C E° ( V) E° Half-reaction Half-reaction |(V) Au+ (aq) + 3e→Au(s) 1.50 Fe2+ (aq) + 2e-→Fe(s) 0.45 Ag+ (aq) + e→Ag(s) 0.80 Cr (aq) + e —уCr? (aq) 0.50 Fe+ (aq) + 3e →Fe2+ (aq) 0.77 Cr+ (aq) + 3e→Cr(s) 0.73 Cu" (аq) + е —Cu(s) 0.52 Zn2+ (aq) +2e-→Zn(s) 0.76 Cu?+ (aq) + 2e-→Cu(s) Mn2+ (aq) + 2e →Mn(s) 0.34 1.18 2H*(aq) + 20→H2(g) 0.00 Al+ (aq) + 3e -→Al(s) 1.66 Fe+ (aq) + 3e-→Fe(s) Mg2+ (aq) + 2e–→Mg(s) 2.37 0.036 Pb2+ (aq) + 2e →Pb(s) Na+ (aq) + e¯→Na(s) 0.13 2.71 Sn2+ (aq) + 2e→Sn(s) Ca2+ (aq) + 2e→Ca(s) 0.14 2.76 Ni?+ (aq) + 2e–→Ni(s) Ва2 (аq) + 2е Ba(s) 0.23 2.90 Co2+ (aq) + 2e¯→Co(s) K* (aq) + e→K(s) 0.28 2.92 Cd2+ (aq) + 2e→Cd(s) Lit (aq) + E→L¡(s) 0.40 3.04
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