Calculate the potential developed by a voltaic cell using the following reaction dissolved species are 0.012 M at 25 °C. 2 Fe²+ (aq) + H₂O₂(aq) + 2 H+ (aq) → 2 Fe³+ (aq) + 2 H₂O(l) Acidic Solution Standard Electrode Potential, +1.77 H₂O₂(aq) + 2 H(aq) + 2 e- 2 H₂O(l) +1.68 Au*(aq) + e —»Au(s) Au*'(aq)+3 e — Au(s) +1.50 Br₂(l) +2 e2 Br(aq) +1.08 NO3(aq) + 4 H*(aq) + 3 e NO(g) + 2 H₂O +0.96 Ag (aq) + e Ag(s) - +0.80 Hg₂² (aq) + 2 e→→→→→2 Hg(l) +0.789 Fe³(aq) + e- Fe²(aq) +0.77 Cu²+ (aq) + 2 e Cu(s) +0.337 HgzClz(s) + 2 e 2 Hg(l) + 2 Cl(aq) +0.27 Sn¹¹(aq) + 2e →→→Sn²+ (aq) +0.15 0.00 2 H(aq) + 2 e - > H₂(g) Pb²+ (aq) + 2 e→→→→→→ Pb(s) -0.126 Sn²+ (aq) + 2 e Sn(s) -0.14 Ni²+ (aq) + 2 e- - Ni(s) -0.25 Cd(s) -0.40 Cd²+ (aq) + 2 e Cr³+ (aq) + e Cr²+ (aq) -0.408 Fe²+ (aq) + 2 e - Fe(s) -0.44 Zn²+ (aq) + 2 e. Zn(s) -0.763
Calculate the potential developed by a voltaic cell using the following reaction dissolved species are 0.012 M at 25 °C. 2 Fe²+ (aq) + H₂O₂(aq) + 2 H+ (aq) → 2 Fe³+ (aq) + 2 H₂O(l) Acidic Solution Standard Electrode Potential, +1.77 H₂O₂(aq) + 2 H(aq) + 2 e- 2 H₂O(l) +1.68 Au*(aq) + e —»Au(s) Au*'(aq)+3 e — Au(s) +1.50 Br₂(l) +2 e2 Br(aq) +1.08 NO3(aq) + 4 H*(aq) + 3 e NO(g) + 2 H₂O +0.96 Ag (aq) + e Ag(s) - +0.80 Hg₂² (aq) + 2 e→→→→→2 Hg(l) +0.789 Fe³(aq) + e- Fe²(aq) +0.77 Cu²+ (aq) + 2 e Cu(s) +0.337 HgzClz(s) + 2 e 2 Hg(l) + 2 Cl(aq) +0.27 Sn¹¹(aq) + 2e →→→Sn²+ (aq) +0.15 0.00 2 H(aq) + 2 e - > H₂(g) Pb²+ (aq) + 2 e→→→→→→ Pb(s) -0.126 Sn²+ (aq) + 2 e Sn(s) -0.14 Ni²+ (aq) + 2 e- - Ni(s) -0.25 Cd(s) -0.40 Cd²+ (aq) + 2 e Cr³+ (aq) + e Cr²+ (aq) -0.408 Fe²+ (aq) + 2 e - Fe(s) -0.44 Zn²+ (aq) + 2 e. Zn(s) -0.763
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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Question
![Calculate the potential developed by a voltaic cell using the following reaction if all
dissolved species are 0.012 M at 25 °C.
2 Fe²+ (aq) + H₂O₂(aq) + 2 H+(aq) → 2 Fe³+ (aq) + 2 H₂0 (1)
Acidic Solution
Standard Electrode Potential, E (volts)
+1.77
2 H₂O(l)
H₂O₂(aq) + 2 H*(aq) + 2 e
Au*(aq) + e —»Au(s)
Au*'(aq)+3e — Au(s)
+1.68
+1.50
+1.08
NO(g) + 2 H₂O
+0.96
Br₂(l) + 2 e2 Br(aq)
NO3(aq) + 4 H*(aq) + 3 e
Ag (aq) + e→→→→→→ Ag(s)
Hg₂²(aq) + 2 e→→→→→2 Hg(l)
+0.80
+0.789
Fe³(aq) + e-→→→→→→ Fe²¹(aq)
+0.77
Cu²+ (aq) + 2 e → Cu(s)
+0.337
HgzClz(s) + 2 e2 Hg() + 2 Cl(aq)
+0.27
Sn¹+ (aq) + 2e → Sn²+ (aq)
+0.15
2 H*(aq) + 2 e→→→→→→ H₂(g)
0.00
Pb²+ (aq) + 2 e
Pb(s)
-0.126
Sn²+ (aq) + 2 e → Sn(s)
-0.14
Ni²+ (aq) + 2 e → Ni(s)
-0.25
Cd²+ (aq) + 2 e → Cd(s)
-0.40
Cr³+ (aq) + e →Cr²+ (aq)
-0.408
Fe²+ (aq) + 2 e
→ > Fe(s)
-0.44
Zn²+ (aq) + 2 e→→→→→→ Zn(s)
-0.763
Cr²+ (aq) + 2 e →→ Cr(s)
-0.91
Al³+ (aq) + 3 e
→Al(s)
-1.66
Mg²+ (aq) + 2 e
-2.37
Mg(s)
V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92e192f1-f2df-4b15-8b23-eecf927ed87a%2Fc754c061-40b8-497d-8a7e-93ed0989a869%2Feoj0w3s_processed.png&w=3840&q=75)
Transcribed Image Text:Calculate the potential developed by a voltaic cell using the following reaction if all
dissolved species are 0.012 M at 25 °C.
2 Fe²+ (aq) + H₂O₂(aq) + 2 H+(aq) → 2 Fe³+ (aq) + 2 H₂0 (1)
Acidic Solution
Standard Electrode Potential, E (volts)
+1.77
2 H₂O(l)
H₂O₂(aq) + 2 H*(aq) + 2 e
Au*(aq) + e —»Au(s)
Au*'(aq)+3e — Au(s)
+1.68
+1.50
+1.08
NO(g) + 2 H₂O
+0.96
Br₂(l) + 2 e2 Br(aq)
NO3(aq) + 4 H*(aq) + 3 e
Ag (aq) + e→→→→→→ Ag(s)
Hg₂²(aq) + 2 e→→→→→2 Hg(l)
+0.80
+0.789
Fe³(aq) + e-→→→→→→ Fe²¹(aq)
+0.77
Cu²+ (aq) + 2 e → Cu(s)
+0.337
HgzClz(s) + 2 e2 Hg() + 2 Cl(aq)
+0.27
Sn¹+ (aq) + 2e → Sn²+ (aq)
+0.15
2 H*(aq) + 2 e→→→→→→ H₂(g)
0.00
Pb²+ (aq) + 2 e
Pb(s)
-0.126
Sn²+ (aq) + 2 e → Sn(s)
-0.14
Ni²+ (aq) + 2 e → Ni(s)
-0.25
Cd²+ (aq) + 2 e → Cd(s)
-0.40
Cr³+ (aq) + e →Cr²+ (aq)
-0.408
Fe²+ (aq) + 2 e
→ > Fe(s)
-0.44
Zn²+ (aq) + 2 e→→→→→→ Zn(s)
-0.763
Cr²+ (aq) + 2 e →→ Cr(s)
-0.91
Al³+ (aq) + 3 e
→Al(s)
-1.66
Mg²+ (aq) + 2 e
-2.37
Mg(s)
V
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