TABLE 20.1 Standard Reduction Potentials in Water at 25°C E°red(V) Reduction Half-Reaction → 2 F (aq) Mno, (aq) + 8 H*(aq) + 5 e- → 2 CI (aq) +2.87 F(g) + 2e +1.51 Mn2"(aq) + 4 H,O(1) +1.36 Cl2(8) + 2 e +1.33 Ct20,7 (ag) + 14 H*(aq) + 6 e 2 Cr"(aq) + 7 H20(I) O2(8) + 4 H*(aq) + 4e → 2 Br (aq) +1.23 2 H20(I) +1.06 Br,(1) + 2 e +0.96 NO3 (ag) + 4 H*(aq) + 3 e NO(g) + 2 H20(1) +0.80 Ag*(aq) + e - Ag(s) +0.77 Fe*(aq) + e Fe2* (aq) O2(8) + 2 H*(ag) + 2e ► H,O2(aq) +0.68 +0.59 MnO, (aq) + 2 H20(I) + 3 e MnO2(s) + 4 OH(aq) +0.54 2(s) + 2e 21 (aq) -4Он (ад) ► Cu(s) +0.40 O2(8) + 2 H20(1) + 4 e +0.34 Cu2+(ag) + 2 e → 0 [defined] 2H*(ag) + 2e H2(8) -0.28 Ni2*(aq) + 2e Ni(s) -0.44 Fe2+ (aq) + 2 e Fe(s) -0.76 Zn2+(aq) + 2 e · Zn(s) -0.83 2 H,0(1) + 2 e H2(8) + 2 OH (aq) -1.66 Al3* (aq) + 3 e Al(s) -2.71 Na* (aq) + e Na(s) -3.05 Li* (aq) + e - Li(s)
TABLE 20.1 Standard Reduction Potentials in Water at 25°C E°red(V) Reduction Half-Reaction → 2 F (aq) Mno, (aq) + 8 H*(aq) + 5 e- → 2 CI (aq) +2.87 F(g) + 2e +1.51 Mn2"(aq) + 4 H,O(1) +1.36 Cl2(8) + 2 e +1.33 Ct20,7 (ag) + 14 H*(aq) + 6 e 2 Cr"(aq) + 7 H20(I) O2(8) + 4 H*(aq) + 4e → 2 Br (aq) +1.23 2 H20(I) +1.06 Br,(1) + 2 e +0.96 NO3 (ag) + 4 H*(aq) + 3 e NO(g) + 2 H20(1) +0.80 Ag*(aq) + e - Ag(s) +0.77 Fe*(aq) + e Fe2* (aq) O2(8) + 2 H*(ag) + 2e ► H,O2(aq) +0.68 +0.59 MnO, (aq) + 2 H20(I) + 3 e MnO2(s) + 4 OH(aq) +0.54 2(s) + 2e 21 (aq) -4Он (ад) ► Cu(s) +0.40 O2(8) + 2 H20(1) + 4 e +0.34 Cu2+(ag) + 2 e → 0 [defined] 2H*(ag) + 2e H2(8) -0.28 Ni2*(aq) + 2e Ni(s) -0.44 Fe2+ (aq) + 2 e Fe(s) -0.76 Zn2+(aq) + 2 e · Zn(s) -0.83 2 H,0(1) + 2 e H2(8) + 2 OH (aq) -1.66 Al3* (aq) + 3 e Al(s) -2.71 Na* (aq) + e Na(s) -3.05 Li* (aq) + e - Li(s)
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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For the reaction
3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH-(aq)---->3 Ni(s) + 2 CrO42-(aq) + 8H2O(l)
ΔG° = +87 kJ/mol. Given the standard reduction potential
of Ni2+(aq)in Table , what value do you calculate for the
standard reduction potential of the half-reaction
CrO42-(aq) + 4 H2O(l) + 3 e----> Cr(OH)3(s) + 5 OH-(aq)?
(a) -0.43 V (b) -0.28 V (c) 0.02 V (d) -0.13 V (e) -0.15 V
![TABLE 20.1 Standard Reduction Potentials in Water at 25°C
E°red(V)
Reduction Half-Reaction
→ 2 F (aq)
Mno, (aq) + 8 H*(aq) + 5 e-
→ 2 CI (aq)
+2.87
F(g) + 2e
+1.51
Mn2"(aq) + 4 H,O(1)
+1.36
Cl2(8) + 2 e
+1.33
Ct20,7 (ag) + 14 H*(aq) + 6 e
2 Cr"(aq) + 7 H20(I)
O2(8) + 4 H*(aq) + 4e →
2 Br (aq)
+1.23
2 H20(I)
+1.06
Br,(1) + 2 e
+0.96
NO3 (ag) + 4 H*(aq) + 3 e
NO(g) + 2 H20(1)
+0.80
Ag*(aq) + e
- Ag(s)
+0.77
Fe*(aq) + e
Fe2* (aq)
O2(8) + 2 H*(ag) + 2e
► H,O2(aq)
+0.68
+0.59
MnO, (aq) + 2 H20(I) + 3 e
MnO2(s) + 4 OH(aq)
+0.54
2(s) + 2e
21 (aq)
-4Он (ад)
► Cu(s)
+0.40
O2(8) + 2 H20(1) + 4 e
+0.34
Cu2+(ag) + 2 e →
0 [defined]
2H*(ag) + 2e
H2(8)
-0.28
Ni2*(aq) + 2e
Ni(s)
-0.44
Fe2+ (aq) + 2 e
Fe(s)
-0.76
Zn2+(aq) + 2 e
· Zn(s)
-0.83
2 H,0(1) + 2 e
H2(8) + 2 OH (aq)
-1.66
Al3* (aq) + 3 e
Al(s)
-2.71
Na* (aq) + e
Na(s)
-3.05
Li* (aq) + e
- Li(s)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c310cd9-bb8a-431d-8676-838170dfe8f3%2F24cb6f67-5410-4ca2-baa4-0abfe56c2750%2Faoc21vi.png&w=3840&q=75)
Transcribed Image Text:TABLE 20.1 Standard Reduction Potentials in Water at 25°C
E°red(V)
Reduction Half-Reaction
→ 2 F (aq)
Mno, (aq) + 8 H*(aq) + 5 e-
→ 2 CI (aq)
+2.87
F(g) + 2e
+1.51
Mn2"(aq) + 4 H,O(1)
+1.36
Cl2(8) + 2 e
+1.33
Ct20,7 (ag) + 14 H*(aq) + 6 e
2 Cr"(aq) + 7 H20(I)
O2(8) + 4 H*(aq) + 4e →
2 Br (aq)
+1.23
2 H20(I)
+1.06
Br,(1) + 2 e
+0.96
NO3 (ag) + 4 H*(aq) + 3 e
NO(g) + 2 H20(1)
+0.80
Ag*(aq) + e
- Ag(s)
+0.77
Fe*(aq) + e
Fe2* (aq)
O2(8) + 2 H*(ag) + 2e
► H,O2(aq)
+0.68
+0.59
MnO, (aq) + 2 H20(I) + 3 e
MnO2(s) + 4 OH(aq)
+0.54
2(s) + 2e
21 (aq)
-4Он (ад)
► Cu(s)
+0.40
O2(8) + 2 H20(1) + 4 e
+0.34
Cu2+(ag) + 2 e →
0 [defined]
2H*(ag) + 2e
H2(8)
-0.28
Ni2*(aq) + 2e
Ni(s)
-0.44
Fe2+ (aq) + 2 e
Fe(s)
-0.76
Zn2+(aq) + 2 e
· Zn(s)
-0.83
2 H,0(1) + 2 e
H2(8) + 2 OH (aq)
-1.66
Al3* (aq) + 3 e
Al(s)
-2.71
Na* (aq) + e
Na(s)
-3.05
Li* (aq) + e
- Li(s)
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