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.) H₂O2 + 2H+ + 2e → 2H₂O Cr₂O72- + 14H+ + 6e €° = a. b. € V 2H+ + 2e → H₂ A1³+ + 3e → Al = + V + € = 1.78 V 2 Cr³+ + 7H₂O = 1.33 V € = 0.00 V E = -1.66 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.) H₂O2 + 2H+ + 2e → 2H₂O Cr₂O72- + 14H+ + 6e €° = a. b. € V 2H+ + 2e → H₂ A1³+ + 3e → Al = + V + € = 1.78 V 2 Cr³+ + 7H₂O = 1.33 V € = 0.00 V E = -1.66 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
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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![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.)
H₂O2 + 2H+ + 2e → 2H₂O
€ = 1.78 V
Cr₂O7²- + 14H+ + 6e¯ →2 Cr³+ + 7H₂O €° = 1.33 V
a.
b.
€ =
V
€ =
+
2H+ + 2e → H₂
→ Al
A1³+ +3e
V
€ = 0.00 V
€ = -1.66 V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb31ba3b8-2953-4b95-bdbf-2db1a69955fa%2Fd7af4f8f-5784-433e-aa7f-2e9420d8d0fc%2Fiwyp2cjs_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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.)
H₂O2 + 2H+ + 2e → 2H₂O
€ = 1.78 V
Cr₂O7²- + 14H+ + 6e¯ →2 Cr³+ + 7H₂O €° = 1.33 V
a.
b.
€ =
V
€ =
+
2H+ + 2e → H₂
→ Al
A1³+ +3e
V
€ = 0.00 V
€ = -1.66 V
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