MISSED THIS? Watch KCV 19.4; Read Section 19.4. You can click on the Review link to access the section in your eText. Consider the table of the standard electrode potentials at 25 °C: Reduction half-reaction Au3+ (aq) +3e → Au(s) Cr2O72(aq) + 14H+ (aq) + 6e →2Cr3+ (aq) + 7H₂O(1) 2H+ (aq) + 2e → H2(g) Cr3+(aq) +3e→ Cr(s) E° (V) 1.50 1.33 0 -0.73 Part A Identify whether Au or Cr or both or neither will dissolve in 1 M H2Cr2O7 Only Cr will dissolve. Only Au will dissolve. Both Au and Cr will dissolve. Neither Au nor Cr will dissolve. Submit Previous Answers Part B Correct H2Cr2O7 oxidizes metals through the following reduction half-reaction: Cr2O72 (aq) +14H+ (aq) + 6e → 2Cr3+ (aq) + 7H₂O (1) Because this half-reaction is above the reduction of H+ in the table, H₂ Cr2O7 can oxidize s metals (for example, copper or silver) that cannot be oxidized by HCl. In general, the metals reduction half-reactions listed below the reduction of Cr2O72- to Cr3+ in the table dissolve H2Cr2O7, whereas the metals listed above it do not. Therefore, Au (which has a reduction potential of 1.50 V) will not be oxidized, but Cr (which has a reduction potential of -0.73 V) be oxidized by both HCl and H2Cr2O7. Enter a balanced redox equation for the reaction that occurs. Express your answer as a balanced net ionic equation including phases. ΑΣΦ ? A chemical reaction does not occur for this question.
MISSED THIS? Watch KCV 19.4; Read Section 19.4. You can click on the Review link to access the section in your eText. Consider the table of the standard electrode potentials at 25 °C: Reduction half-reaction Au3+ (aq) +3e → Au(s) Cr2O72(aq) + 14H+ (aq) + 6e →2Cr3+ (aq) + 7H₂O(1) 2H+ (aq) + 2e → H2(g) Cr3+(aq) +3e→ Cr(s) E° (V) 1.50 1.33 0 -0.73 Part A Identify whether Au or Cr or both or neither will dissolve in 1 M H2Cr2O7 Only Cr will dissolve. Only Au will dissolve. Both Au and Cr will dissolve. Neither Au nor Cr will dissolve. Submit Previous Answers Part B Correct H2Cr2O7 oxidizes metals through the following reduction half-reaction: Cr2O72 (aq) +14H+ (aq) + 6e → 2Cr3+ (aq) + 7H₂O (1) Because this half-reaction is above the reduction of H+ in the table, H₂ Cr2O7 can oxidize s metals (for example, copper or silver) that cannot be oxidized by HCl. In general, the metals reduction half-reactions listed below the reduction of Cr2O72- to Cr3+ in the table dissolve H2Cr2O7, whereas the metals listed above it do not. Therefore, Au (which has a reduction potential of 1.50 V) will not be oxidized, but Cr (which has a reduction potential of -0.73 V) be oxidized by both HCl and H2Cr2O7. Enter a balanced redox equation for the reaction that occurs. Express your answer as a balanced net ionic equation including phases. ΑΣΦ ? A chemical reaction does not occur for this question.
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 120AE
Related questions
Question
Transcribed Image Text:MISSED THIS? Watch KCV 19.4; Read Section 19.4. You can click on the
Review link to access the section in your eText.
Consider the table of the standard electrode potentials at 25 °C:
Reduction half-reaction
Au3+ (aq) +3e → Au(s)
Cr2O72(aq) + 14H+ (aq) + 6e →2Cr3+ (aq) + 7H₂O(1)
2H+ (aq) + 2e → H2(g)
Cr3+(aq) +3e→ Cr(s)
E° (V)
1.50
1.33
0
-0.73
Part A
Identify whether Au or Cr or both or neither will dissolve in 1 M H2Cr2O7
Only Cr will dissolve.
Only Au will dissolve.
Both Au and Cr will dissolve.
Neither Au nor Cr will dissolve.
Submit
Previous Answers
Part B
Correct
H2Cr2O7 oxidizes metals through the following reduction half-reaction:
Cr2O72 (aq) +14H+ (aq) + 6e → 2Cr3+ (aq) + 7H₂O (1)
Because this half-reaction is above the reduction of H+ in the table, H₂ Cr2O7 can oxidize s
metals (for example, copper or silver) that cannot be oxidized by HCl. In general, the metals
reduction half-reactions listed below the reduction of Cr2O72- to Cr3+ in the table dissolve
H2Cr2O7, whereas the metals listed above it do not. Therefore, Au (which has a reduction
potential of 1.50 V) will not be oxidized, but Cr (which has a reduction potential of -0.73 V)
be oxidized by both HCl and H2Cr2O7.
Enter a balanced redox equation for the reaction that occurs.
Express your answer as a balanced net ionic equation including phases.
ΑΣΦ
?
A chemical reaction does not occur for this question.
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