Refer to this table of standard reduction potentials to answer the questions. Reduction half-reaction F2(g) +2e2F-(aq) Potential (V) ▸ View Available Hint(s) Reset Help +2.87 Na(s) Br₂(1) H2(g) O2(g) O2(g)+4H+(aq) + 4e¯¯→2H₂O(1) +1.23 Br2 (1)+2e2Br (aq) +1.07 +0.80 Anode Cathode Not produced Ag++e→Ag(s) 2H2O(1)+2e→H2(g) +20H(aq) -0.83 Na+ (aq) +eNa(s) -2.71 Note that at [H+]=10-7 M, the electrode potential for O2(g)+4H+(aq) +4e+2H2O(1) under these nonstandard conditions is E= +0.82 V. At [OH] 107 M, the electrode potential for 2H2O(1)+2e→H2(g) +20H(aq) under these nonstandard conditions is E = -0.41 V. Submit Previous Answers × Incorrect; Try Again; 2 attempts remaining You sorted 2 out of 4 items incorrectly. Compare the potentials for two species that could be oxidized at the anode. For example, the oxidation of water has a potential of -0.82 V, since the conditions are nonstandard and potential for the oxidation of a species is the opposite that for the reduction. The species with the less negative potential will be oxidized at the anode. You may want to review Hint 1. Identify possible oxidation reactions.
Refer to this table of standard reduction potentials to answer the questions. Reduction half-reaction F2(g) +2e2F-(aq) Potential (V) ▸ View Available Hint(s) Reset Help +2.87 Na(s) Br₂(1) H2(g) O2(g) O2(g)+4H+(aq) + 4e¯¯→2H₂O(1) +1.23 Br2 (1)+2e2Br (aq) +1.07 +0.80 Anode Cathode Not produced Ag++e→Ag(s) 2H2O(1)+2e→H2(g) +20H(aq) -0.83 Na+ (aq) +eNa(s) -2.71 Note that at [H+]=10-7 M, the electrode potential for O2(g)+4H+(aq) +4e+2H2O(1) under these nonstandard conditions is E= +0.82 V. At [OH] 107 M, the electrode potential for 2H2O(1)+2e→H2(g) +20H(aq) under these nonstandard conditions is E = -0.41 V. Submit Previous Answers × Incorrect; Try Again; 2 attempts remaining You sorted 2 out of 4 items incorrectly. Compare the potentials for two species that could be oxidized at the anode. For example, the oxidation of water has a potential of -0.82 V, since the conditions are nonstandard and potential for the oxidation of a species is the opposite that for the reduction. The species with the less negative potential will be oxidized at the anode. You may want to review Hint 1. Identify possible oxidation reactions.
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter17: Electrochemistry
Section: Chapter Questions
Problem 77AP
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Please correct answer and don't use hand rating
![Refer to this table of standard reduction potentials to
answer the questions.
Reduction half-reaction
F2(g) +2e2F-(aq)
Potential
(V)
▸ View Available Hint(s)
Reset
Help
+2.87
Na(s)
Br₂(1)
H2(g) O2(g)
O2(g)+4H+(aq) + 4e¯¯→2H₂O(1)
+1.23
Br2 (1)+2e2Br (aq)
+1.07
+0.80
Anode
Cathode
Not produced
Ag++e→Ag(s)
2H2O(1)+2e→H2(g) +20H(aq) -0.83
Na+ (aq) +eNa(s)
-2.71
Note that at [H+]=10-7 M, the electrode potential
for O2(g)+4H+(aq) +4e+2H2O(1) under
these nonstandard conditions is E= +0.82 V.
At [OH] 107 M, the electrode potential for
2H2O(1)+2e→H2(g) +20H(aq) under
these nonstandard conditions is E = -0.41 V.
Submit
Previous Answers
× Incorrect; Try Again; 2 attempts remaining
You sorted 2 out of 4 items incorrectly. Compare the potentials for two species that could be oxidized at the
anode. For example, the oxidation of water has a potential of -0.82 V, since the conditions are
nonstandard and potential for the oxidation of a species is the opposite that for the reduction. The species
with the less negative potential will be oxidized at the anode.
You may want to review Hint 1. Identify possible oxidation reactions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F781b2bf0-e3a1-4f4d-a802-819a67abd4f8%2F71e52e26-1660-4aa2-99ef-aba5b4813ed1%2Fovyaf6w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Refer to this table of standard reduction potentials to
answer the questions.
Reduction half-reaction
F2(g) +2e2F-(aq)
Potential
(V)
▸ View Available Hint(s)
Reset
Help
+2.87
Na(s)
Br₂(1)
H2(g) O2(g)
O2(g)+4H+(aq) + 4e¯¯→2H₂O(1)
+1.23
Br2 (1)+2e2Br (aq)
+1.07
+0.80
Anode
Cathode
Not produced
Ag++e→Ag(s)
2H2O(1)+2e→H2(g) +20H(aq) -0.83
Na+ (aq) +eNa(s)
-2.71
Note that at [H+]=10-7 M, the electrode potential
for O2(g)+4H+(aq) +4e+2H2O(1) under
these nonstandard conditions is E= +0.82 V.
At [OH] 107 M, the electrode potential for
2H2O(1)+2e→H2(g) +20H(aq) under
these nonstandard conditions is E = -0.41 V.
Submit
Previous Answers
× Incorrect; Try Again; 2 attempts remaining
You sorted 2 out of 4 items incorrectly. Compare the potentials for two species that could be oxidized at the
anode. For example, the oxidation of water has a potential of -0.82 V, since the conditions are
nonstandard and potential for the oxidation of a species is the opposite that for the reduction. The species
with the less negative potential will be oxidized at the anode.
You may want to review Hint 1. Identify possible oxidation reactions.
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