Standard Electrode Potentials at 25 °C olumn to the appropriate blanks in the sentences on the right. Make plete before submitting your answer. Reduction Half-Reaction E (V) Au+3 (aq) + 3e + Au(s) 1.50 C2(g) + 2e + 2C1 (aq) 1.36 Rese NO (aq) + 4H*(aq) + 3e + NO(g) + 2H20(1) 0.96 + Ag(s) Ag (aq) +e Fe (aq) +e 0.80 Nitric acid (HNO3) can oxidize metals through the following reduction half-reaction: NO, (aq) + 4H*(aq) + 3e → NO(g) + 2H2O(1) E =0.96V. Because this half- in table, HNO3 can oxidize metal + Fe2+ (aq) 0.77 Cu2+ (aq) + 2e + Cu(s) 0.34 reaction is the reduction of 2H+(aq) + 2e - H2 (g) for example) that can't be oxidized by HCI. Fe+ (aq) + 3e + Fe(s) --0.036 more + Pb(s) + Ni(s) + Fe(s) + Zn(s) Pb2-(aq) + 2e -0.13 -0.23 Ni+ (aq) + 2e -0.45 Fe+ (aq) + 2e more -0.76 Zn2+(aq) + 2e

Chemistry: Principles and Practice
3rd Edition
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Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.59QE
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Chp 20 #14
I Review | Constants Periodic lable
Standard Electrode Potentials at 25 °C
plumn to the appropriate blanks in the sentences on the right. Make
plete before submitting your answer.
Reduction Half-Reaction
E (V)
Au+3(aq) + 3e
+ Au(s)
1.50
Cl2 (g) + 2e
+ 2C1 (aq)
1.36
Reset
NO, (aq) +4H*(aq) + 3e
- NO(g) + 2H2O(1) 0.96
Ag+ (aq) + e
Fe+ (aq) + e
→ Ag(s)
+ Fe2+ (aq)
0.80
Nitric acid (HNO3) can oxidize metals through the following reduction half-reaction:
0.77
NO, (aq) + 4H*(aq) + 3e NO(g) +2H2O(1) E =0.96V. Because this half-
Cu2+ (aq) + 2e
- Cu(s)
0.34
in table, HNO3 can oxidize metal
reaction is
the reduction of
2H+(aq) + 2e
+ H2 (g)
,for example) that can't be oxidized by HCl.
+ Fe(s)
-0.036
more
Fe (aq) + 3e
Pb2+(aq) + 2e
- Pb(s)
-0.13
Nit (aq) + 2e
- Ni(s)
-0.23
Fe2+ (aq) + 2e
+ Fe(s)
-0.45
more
Zn2+ (aq) + 2e
+ Zn(s)
-0.76
Transcribed Image Text:I Review | Constants Periodic lable Standard Electrode Potentials at 25 °C plumn to the appropriate blanks in the sentences on the right. Make plete before submitting your answer. Reduction Half-Reaction E (V) Au+3(aq) + 3e + Au(s) 1.50 Cl2 (g) + 2e + 2C1 (aq) 1.36 Reset NO, (aq) +4H*(aq) + 3e - NO(g) + 2H2O(1) 0.96 Ag+ (aq) + e Fe+ (aq) + e → Ag(s) + Fe2+ (aq) 0.80 Nitric acid (HNO3) can oxidize metals through the following reduction half-reaction: 0.77 NO, (aq) + 4H*(aq) + 3e NO(g) +2H2O(1) E =0.96V. Because this half- Cu2+ (aq) + 2e - Cu(s) 0.34 in table, HNO3 can oxidize metal reaction is the reduction of 2H+(aq) + 2e + H2 (g) ,for example) that can't be oxidized by HCl. + Fe(s) -0.036 more Fe (aq) + 3e Pb2+(aq) + 2e - Pb(s) -0.13 Nit (aq) + 2e - Ni(s) -0.23 Fe2+ (aq) + 2e + Fe(s) -0.45 more Zn2+ (aq) + 2e + Zn(s) -0.76
I Review | Constants | Periodic Table
Standard Electrode Potentials at 25 °C
Part B
Reduction Half-Reaction
E° (V)
Au+ (aq) + 3e
In HNO3?
→ Au(s)
+ 2C1-(aq)
1.50
Cl2 (g) + 2e
Match the words in the left column to the appropriate blanks in the sentences on the right. Make
certain each sentence is complete before submitting your answer.
1.36
NO, (aq) + 4H (aq) + 3e
→ NO(g) + 2H20(1) |0.96
Ag+ (aq) + e
+ Ag(s)
0.80
Fet (aq) +e
+ Fe?+ (aq)
0.77
Cu2+(aq) + 2e
→ Cu(s)
0.34
zinc
Nitric acid (HN03) can oxidize metals through the following re
2H+ (aq) + 2e
+ H2 (g)
NO, (aq) + 4H+(aq) + 3e NO(g) + 2H20(1) E
Fet (aq) + 3e
+ Fe(s)
-0.036
reaction is
the reduction of
in 1
iron
Pb2+(aq) + 2e
→ Pb(s)
for example) that can't be oxidized by HCl.
-0.13
above (and therefore more
Ni+ (aq) + 2e
+ Ni(s)
-0.23
positive)
Fe2+ (aq) + 2e
+ Fe(s)
-0.45
Cl2
Zn2+ (aq) + 2e
+ Zn(s)
-0.76
below (and therefore more
negative)
copper
Transcribed Image Text:I Review | Constants | Periodic Table Standard Electrode Potentials at 25 °C Part B Reduction Half-Reaction E° (V) Au+ (aq) + 3e In HNO3? → Au(s) + 2C1-(aq) 1.50 Cl2 (g) + 2e Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. 1.36 NO, (aq) + 4H (aq) + 3e → NO(g) + 2H20(1) |0.96 Ag+ (aq) + e + Ag(s) 0.80 Fet (aq) +e + Fe?+ (aq) 0.77 Cu2+(aq) + 2e → Cu(s) 0.34 zinc Nitric acid (HN03) can oxidize metals through the following re 2H+ (aq) + 2e + H2 (g) NO, (aq) + 4H+(aq) + 3e NO(g) + 2H20(1) E Fet (aq) + 3e + Fe(s) -0.036 reaction is the reduction of in 1 iron Pb2+(aq) + 2e → Pb(s) for example) that can't be oxidized by HCl. -0.13 above (and therefore more Ni+ (aq) + 2e + Ni(s) -0.23 positive) Fe2+ (aq) + 2e + Fe(s) -0.45 Cl2 Zn2+ (aq) + 2e + Zn(s) -0.76 below (and therefore more negative) copper
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