Give the balanced cell equation and determine & for the galvanic cells based on the following half-reactions. (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.) a. H₂O2 + 2H+ + 2e¯ → 2H₂O € = 1.78 V 2- Cr₂O7²- + 14H+ + 6e¯ →2 Cr³+ + 7H₂O €° = 1.33 V V + b. 2H+ + 2e →H₂ La³+ + 3e¯→La €° = V + + + = 0.00 V ɛ° = −2.37 V +
Give the balanced cell equation and determine & for the galvanic cells based on the following half-reactions. (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.) a. H₂O2 + 2H+ + 2e¯ → 2H₂O € = 1.78 V 2- Cr₂O7²- + 14H+ + 6e¯ →2 Cr³+ + 7H₂O €° = 1.33 V V + b. 2H+ + 2e →H₂ La³+ + 3e¯→La €° = V + + + = 0.00 V ɛ° = −2.37 V +
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
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 135CWP: Consider a galvanic cell based on the following half-reactions: a. What is the standard potential...
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![Give the balanced cell equation and determine ɛ° for the galvanic cells based on the following half-reactions.
(Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
H₂O2 + 2H+ + 2e¯ → 2H₂O
€
2-
O
Cr₂O7²- + 14H+ + 6e¯ →2 Cr³+ + 7H₂O €°
=
V
a.
€°
+
+
b. 2H+ + 2e →H₂
La³+ + 3e¯ →La ɛ°
2.3+
ɛ° =
V
+
E
+
O
+
= 0.00 V
= -2.37 V
= 1.78 V
1.33 V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e049020-7449-404d-9d0e-5966c3ff89ef%2Fa021c8ba-5f7c-4a9f-aaf6-49cc6620a55c%2Frktlvq_processed.png&w=3840&q=75)
Transcribed Image Text:Give the balanced cell equation and determine ɛ° for the galvanic cells based on the following half-reactions.
(Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
H₂O2 + 2H+ + 2e¯ → 2H₂O
€
2-
O
Cr₂O7²- + 14H+ + 6e¯ →2 Cr³+ + 7H₂O €°
=
V
a.
€°
+
+
b. 2H+ + 2e →H₂
La³+ + 3e¯ →La ɛ°
2.3+
ɛ° =
V
+
E
+
O
+
= 0.00 V
= -2.37 V
= 1.78 V
1.33 V
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