Consider the following half-reactions along with their respective standard reduction potential. Which of the species will react with Hg(spontaneously? E° = +1.22 V Мпоz() + 4H(ag) + 2e - Mna) + 2H,0) + so ag) E° = +0.61 V Hg,SO46) + 2e¯ → 2Hg() E° = -0.95 V Sn02(s) + 2H,0m + 4e¯ → Sn(3) + 40H(aq) E° = -1.48 V Cr(0H)3(3) + 3e → Cra + 30HTaq) O Sno2(3) O MnO2(s) O Cr(OH)3(3) O Sno2(s) and Cr(OH)3(3)
Consider the following half-reactions along with their respective standard reduction potential. Which of the species will react with Hg(spontaneously? E° = +1.22 V Мпоz() + 4H(ag) + 2e - Mna) + 2H,0) + so ag) E° = +0.61 V Hg,SO46) + 2e¯ → 2Hg() E° = -0.95 V Sn02(s) + 2H,0m + 4e¯ → Sn(3) + 40H(aq) E° = -1.48 V Cr(0H)3(3) + 3e → Cra + 30HTaq) O Sno2(3) O MnO2(s) O Cr(OH)3(3) O Sno2(s) and Cr(OH)3(3)
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![Consider the following half-reactions along with their respective standard reduction potential. Which of the species will react with Hg(spontaneously?
E° = +1.22 V
Мпоz() + 4H(ag)
+ 2e -
Mna) + 2H,0)
+ so ag)
E° = +0.61 V
Hg,SO46) + 2e¯ → 2Hg()
E° = -0.95 V
Sn02(s) + 2H,0m + 4e¯ → Sn(3) + 40H(aq)
E° = -1.48 V
Cr(0H)3(3)
+ 3e → Cra + 30HTaq)
O Sno2(3)
O MnO2(s)
O Cr(OH)3(3)
O Sno2(s)
and Cr(OH)3(3)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F156f4ac9-81e3-4b5e-a79f-b4c6427810ea%2Fd4064c83-2b9d-4b07-b43e-5cdbbb6ae00e%2F0vbqnfr_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the following half-reactions along with their respective standard reduction potential. Which of the species will react with Hg(spontaneously?
E° = +1.22 V
Мпоz() + 4H(ag)
+ 2e -
Mna) + 2H,0)
+ so ag)
E° = +0.61 V
Hg,SO46) + 2e¯ → 2Hg()
E° = -0.95 V
Sn02(s) + 2H,0m + 4e¯ → Sn(3) + 40H(aq)
E° = -1.48 V
Cr(0H)3(3)
+ 3e → Cra + 30HTaq)
O Sno2(3)
O MnO2(s)
O Cr(OH)3(3)
O Sno2(s)
and Cr(OH)3(3)
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