3. Estimate Eº for the half-reaction2H₂O + 2e → H₂ + 2OH-given the following values of AG :H₂O(l) = -237 kJ/molH₂(g) = 0.0OH(aq) = -157 kJ/mole = 0.0Compare this value of E° with the value of E° given in Table 18.1 3' Ž Standard Reduction Potentials at 25°C (298 K) for Many Common Half-ReactionsHalf-ReactionF₂+2e-2FAg+ e → Agcoi te coH₂O₂ + 2H+ 2e → 2H₂OCe +e→Ce"PbO₂ + 4H+SO+2e →PbSO. + 2H₂OMnO, + 4H+ + 3e →MnO₂ + 2H₂O2e + 2H+10 →10 + H₂OMnO, +8H + 5e →Mn² + 4H₂OAu + 3e → AuPbO₂ + 4H+ 2e →Pb² + 2H₂OCl₂ +2e2C1-Cr₂0,² + 14H+ + 6e →2Cr³+ + 7H₂0O₂ + 4H+ 4e → 2H₂OMnO₂ + 4H+ + 2e →Mn²+ + 2H₂O10₂ + 6H + 5e →→₂ + 3H₂OBr₂ +2e2BrVO₂ + 2H+ e→VO² + H₂OAuCl₂ + 3e-Au + 4CNO₂ + 4H+ 3e→NO + 2H₂OCIO₂ + e→CIO,2Hg2+ 2e → Hg₂²⁰Ag+ e → AgHg₂+ 2e →→ 2HgFe +eFe²+O₂ + 2H+ 2e → H₂O₂MnO, + e→→MnO²-1₂ + 2e → 21-Cut + e-Cu8° (V)2.871.991.821.781.701.691.681.601.511.501.461.361.331.231.211.201.091.000.990.960.9540.910.800.800.770.680.560.540.52Half-ReactionO₂ + 2H₂O + 4e →40H™Cu²+ 2e → CuHg,Cl₂ + 2e → 2Hg + 2C1-AgCl+eAg + CrSO2 + 4H+ 2e → H₂SO₂ + H₂OCu²+eCut2H + 2e → H₂Fe + 3e-FePb²+ + 2e →→ PbSn² +2e → SnNi²+ + 2e-NiPbSO4 + 2e →→ Pb + 50,²Cd²+2e →→ CdFe + 2e → FeCr³+ + e→Cr³Cr³+ + 3e-CrZn²+2e → Zn2H₂O + 2e →→H₂ + 20H™Mn +2e →→MnAP+3e AlH₂ + 2e → 2H™Mg2+2e → MgLa" + 3e-LaNa + e-NaCa²+2e →→ CaBa²+2e-BaK' + e-KLi' + e-Ligº (V)0.400.340.340.220.200.160.00-0.036<-0.13<-0.14-0.23<<-0.35-0.40-0.44-0.50-0.73-0.76-0.83-1.18-1.66-2.23-2.37-2.37<-2.71-2.76-2.90-2.92-3.05

3. Estimate Eº for the half-reaction 2H₂O + 2e → H₂ + 2OH* given the following values of AG : H₂O(l) = -237 kJ/mol H₂(g) = 0.0 OH(aq) = -157 kJ/mol e = 0.0 Compare this value of E° with the value of E° given in Table 18.1

3. Estimate Eº for the half-reaction 2H₂O + 2e → H₂ + 2OH* given the following values of AG : H₂O(l) = -237 kJ/mol H₂(g) = 0.0 OH(aq) = -157 kJ/mol e = 0.0 Compare this value of E° with the value of E° given in Table 18.1

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electrolyte, electrodes, and some form of external power source are the main components required for conducting electrolysis. A partition, such as an ion-exchange membrane or a salt bridge, may also be used, but this is optional. These are primarily used to prevent the products from diffusing near the opposing electrode.

Question

3. Estimate Eº for the half-reaction
2H₂O + 2e → H₂ + 2OH-
given the following values of AG :
H₂O(l) = -237 kJ/mol
H₂(g) = 0.0
OH(aq) = -157 kJ/mol
e = 0.0
Compare this value of E° with the value of E° given in Table 18.1

3' Ž Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions
Half-Reaction
F₂+2e-2F
Ag+ e → Ag
coi te co
H₂O₂ + 2H+ 2e → 2H₂O
Ce +e→Ce"
PbO₂ + 4H+SO+2e →PbSO. + 2H₂O
MnO, + 4H+ + 3e →MnO₂ + 2H₂O
2e + 2H+10 →10 + H₂O
MnO, +8H + 5e →Mn² + 4H₂O
Au + 3e → Au
PbO₂ + 4H+ 2e →Pb² + 2H₂O
Cl₂ +2e2C1-
Cr₂0,² + 14H+ + 6e →2Cr³+ + 7H₂0
O₂ + 4H+ 4e → 2H₂O
MnO₂ + 4H+ + 2e →Mn²+ + 2H₂O
10₂ + 6H + 5e →→₂ + 3H₂O
Br₂ +2e2Br
VO₂ + 2H+ e→VO² + H₂O
AuCl₂ + 3e-Au + 4C
NO₂ + 4H+ 3e→NO + 2H₂O
CIO₂ + e→CIO,
2Hg2+ 2e → Hg₂²⁰
Ag+ e → Ag
Hg₂+ 2e →→ 2Hg
Fe +eFe²+
O₂ + 2H+ 2e → H₂O₂
MnO, + e→→MnO²-
1₂ + 2e → 21-
Cut + e-Cu
8° (V)
2.87
1.99
1.82
1.78
1.70
1.69
1.68
1.60
1.51
1.50
1.46
1.36
1.33
1.23
1.21
1.20
1.09
1.00
0.99
0.96
0.954
0.91
0.80
0.80
0.77
0.68
0.56
0.54
0.52
Half-Reaction
O₂ + 2H₂O + 4e →40H™
Cu²+ 2e → Cu
Hg,Cl₂ + 2e → 2Hg + 2C1-
AgCl+eAg + Cr
SO2 + 4H+ 2e → H₂SO₂ + H₂O
Cu²+eCut
2H + 2e → H₂
Fe + 3e-Fe
Pb²+ + 2e →→ Pb
Sn² +2e → Sn
Ni²+ + 2e-Ni
PbSO4 + 2e →→ Pb + 50,²
Cd²+2e →→ Cd
Fe + 2e → Fe
Cr³+ + e→Cr³
Cr³+ + 3e-Cr
Zn²+2e → Zn
2H₂O + 2e →→H₂ + 20H™
Mn +2e →→Mn
AP+3e Al
H₂ + 2e → 2H™
Mg2+2e → Mg
La" + 3e-La
Na + e-Na
Ca²+2e →→ Ca
Ba²+2e-Ba
K' + e-K
Li' + e-Li
gº (V)
0.40
0.34
0.34
0.22
0.20
0.16
0.00
-0.036
<-0.13
<-0.14
-0.23
<<-0.35
-0.40
-0.44
-0.50
-0.73
-0.76
-0.83
-1.18
-1.66
-2.23
-2.37
-2.37
<-2.71
-2.76
-2.90
-2.92
-3.05

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