At 298 K, what is E (in V) for the electrochemical cell below when [Cd2]= 3.600 M and [Cu?+] = 0.008 M ? Cd(s) | Cd2*(aq) || Cu²*(aq) | Cu(s) Half-Reaction E° (V) Cl2 (g) + 2e 2C1 (aq) 1.36 Br2 (1) + 2e - 2Br (aq) 1.07 Ag (aq) + e 0.80 Ag(s) Cu2+ (aq) + 2e → Cu(s) 0.34 2H*(aq) + 2e 0.00 H2(g) Pb2+ (aq) + 2e → Cd2+ (aq) + 2e™ -0.13 Pb(s) Cd(s) Al(s) Bе- (aд) + 2е — 0.40 A13+ (ад) + Зе — -1.66 Be(s) -1.85
At 298 K, what is E (in V) for the electrochemical cell below when [Cd2]= 3.600 M and [Cu?+] = 0.008 M ? Cd(s) | Cd2*(aq) || Cu²*(aq) | Cu(s) Half-Reaction E° (V) Cl2 (g) + 2e 2C1 (aq) 1.36 Br2 (1) + 2e - 2Br (aq) 1.07 Ag (aq) + e 0.80 Ag(s) Cu2+ (aq) + 2e → Cu(s) 0.34 2H*(aq) + 2e 0.00 H2(g) Pb2+ (aq) + 2e → Cd2+ (aq) + 2e™ -0.13 Pb(s) Cd(s) Al(s) Bе- (aд) + 2е — 0.40 A13+ (ад) + Зе — -1.66 Be(s) -1.85
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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![At 298 K, what is E (in V) for the electrochemical cell below when [Cd2]= 3.600 M and
[Cu?+] = 0.008 M ?
Cd(s) | Cd2*(aq) || Cu²*(aq) | Cu(s)
Half-Reaction
E° (V)
Cl2 (g) + 2e→
2C1 (aq)
1.36
Br2 (1) + 2e -
2Br (aq)
1.07
Ag" (aq) + e
0.80
Ag(s)
Cu2+ (ag) + 2e →
2H"(aq) + 2e
Cu(s)
0.34
0.00
H2(g)
Pb2+ (aq) + 2e →
Cd2+ (aq) + 2e¯
-0.13
Pb(s)
Cd(s)
Al(s)
Be(s)
0.40
A13+
(ад) + Зе —
-1.66
Bе- (aд) + 2е
-1.85](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe61668f1-2443-4f86-abac-631d13c8408e%2F14ef65e9-5b33-4c7a-86ee-c3ca3ff14371%2Fqbrauxk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:At 298 K, what is E (in V) for the electrochemical cell below when [Cd2]= 3.600 M and
[Cu?+] = 0.008 M ?
Cd(s) | Cd2*(aq) || Cu²*(aq) | Cu(s)
Half-Reaction
E° (V)
Cl2 (g) + 2e→
2C1 (aq)
1.36
Br2 (1) + 2e -
2Br (aq)
1.07
Ag" (aq) + e
0.80
Ag(s)
Cu2+ (ag) + 2e →
2H"(aq) + 2e
Cu(s)
0.34
0.00
H2(g)
Pb2+ (aq) + 2e →
Cd2+ (aq) + 2e¯
-0.13
Pb(s)
Cd(s)
Al(s)
Be(s)
0.40
A13+
(ад) + Зе —
-1.66
Bе- (aд) + 2е
-1.85
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