Standard Reduction (Electrode) Potentials at 25 °C Half-Cell Reaction E° (volts) Fe(CN)6 (aq) + e. Fe(CN),*(aq) 0.48 Cu2 (aq) + 2 e Cu(s) 0.337 Cu2*(aq) + e¯ – Cu*(aq) 0.153 S(s) +2 H"(aq) + 2 e → H2S(aq) 0.14 2 H (aq) + 2 e –→H2(g) 0.0000 Pb2 (aq) + 2 e → Pb(s) -0.126 Sn2+(ag) + 2 e → Sn(s) -0.14
Standard Reduction (Electrode) Potentials at 25 °C Half-Cell Reaction E° (volts) Fe(CN)6 (aq) + e. Fe(CN),*(aq) 0.48 Cu2 (aq) + 2 e Cu(s) 0.337 Cu2*(aq) + e¯ – Cu*(aq) 0.153 S(s) +2 H"(aq) + 2 e → H2S(aq) 0.14 2 H (aq) + 2 e –→H2(g) 0.0000 Pb2 (aq) + 2 e → Pb(s) -0.126 Sn2+(ag) + 2 e → Sn(s) -0.14
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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Transcribed Image Text:Standard Reduction (Electrode) Potentials at 25 °C
Half-Cell Reaction
E° (volts)
Fe(CN)6 (aq) + e→
→ Fe(CN),*(aq)
0.48
2+
Cu"(aq) + 2 e →Cu(s)
0.337
Cu2"(aq) + e → Cu*(aq)
0.153
S(s) + 2 H"(aq)+2 e →
H2S(aq)
0.14
2 H (aq) + 2 e –→H2(g)
0.0000
Pb2*(aq) + 2 e
→ Pb(s)
-0.126
Sn2*(aq) + 2 e → Sn(s)
-0.14
Ni2 (aq) + 2 e - Ni(s)
-0.25
Co2*(aq) + 2 e –→ Co(s)
-0.28
Cd2*(aq) + 2 e
→ Cd(s)
-0.403
Cr3*(aq) + e Cr*(aq)
-0.41
Fe2 (aq) + 2 e –→ Fe(s)
-0.44
3+
(aq)+3 e → Cr(s)
-0.74
Zn2*(aq) + 2 e →
Zn(s)
-0.763
2 H20(1) + 2 e →
H2(g) + 2 OH (aq)
-0.83
Mn2"(aq) + 2 e → Mn(s)
-1.18
Al³* (aq) + 3 e–→ Al(s)
-1.66
2+
Mg"(aq) + 2 e → Mg(s)
-2.37
Transcribed Image Text:Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:
2Cu?*(aq) + Fe(s) 2Cu*(aq) + Fe²*(aq)
Answer: 114
kJ
K for this reaction would be greater
than one.
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