A certain half-reaction has a standard reduction potential E=-1.25 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that red must provide at least 0.80 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer reguires this half-reaction to happen at the cathode of the cell. dh Data Ag+ (aq) + e Ag (s) 0.7996 Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? O yes, there is a minimum. Ov %3D red Al3+ (aq) + 3e-- Al (s) -1.676 Au+ (aq) + e- Au (s) 1.692 If so, check the "yes" box and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, check the "no" box. Au3+ (aq) + 3e Au (s) 1.498 O no minimum Ba2+ (aq) + 2e- Ba (s) -2.912 Brz (1) + 2e - - 2Br (aq) 1.066 Ca2+ (aq) + 2e - Ca (s) -2.868 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? = Ov Cl2 (9) + 2e → 2CI- (aq) O yes, there is a maximum. E red 1.35827 %3! Co2+ (aq) + 2e - Co (s) Co3+ (aq) + e Co2+ (aq) Cr2+ (aq) + 2e→ Cr (s) Cr3+ (aq) + 3e Cr (s) Cr+ (aq) + e Cr2+ (aq) Cro42- (aq) + 4H20 (1) + 3e - Cr(OH)3 (s) + 50H (aq) -0.28 If so, check the "yes" box and calculate the maximum. Round your answer to 2 decimal places. If there is no upper limit, check the "no" box. 1.92 O no maximum -0.913 -0.744 -0.407 -0.13 By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the anode of this cell. Cu2+ (aq) + 2e- Cu (s) 0.3419 Cu2+ (ag) + e - Cu* (aq) 0.153 Note: write the half reaction as it would actually occur at the anode. Cu* (ag) + e - Cu (s) 0.521 F2 (9) + 2e 2F (aq) 2.866 Fe2+ (ag) + 2e- Fe (s) -0.447 Explanation Check Fe3+ (aq) + e - Fe2+ (aq) 0.771 1. Fe+ (aq) + 3e Fe (s) -0.037 MacBook Air 20 FI F12 D00 F4 F10 FS F6 F7 F8 F9 FI F2 F3
A certain half-reaction has a standard reduction potential E=-1.25 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that red must provide at least 0.80 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer reguires this half-reaction to happen at the cathode of the cell. dh Data Ag+ (aq) + e Ag (s) 0.7996 Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? O yes, there is a minimum. Ov %3D red Al3+ (aq) + 3e-- Al (s) -1.676 Au+ (aq) + e- Au (s) 1.692 If so, check the "yes" box and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, check the "no" box. Au3+ (aq) + 3e Au (s) 1.498 O no minimum Ba2+ (aq) + 2e- Ba (s) -2.912 Brz (1) + 2e - - 2Br (aq) 1.066 Ca2+ (aq) + 2e - Ca (s) -2.868 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? = Ov Cl2 (9) + 2e → 2CI- (aq) O yes, there is a maximum. E red 1.35827 %3! Co2+ (aq) + 2e - Co (s) Co3+ (aq) + e Co2+ (aq) Cr2+ (aq) + 2e→ Cr (s) Cr3+ (aq) + 3e Cr (s) Cr+ (aq) + e Cr2+ (aq) Cro42- (aq) + 4H20 (1) + 3e - Cr(OH)3 (s) + 50H (aq) -0.28 If so, check the "yes" box and calculate the maximum. Round your answer to 2 decimal places. If there is no upper limit, check the "no" box. 1.92 O no maximum -0.913 -0.744 -0.407 -0.13 By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the anode of this cell. Cu2+ (aq) + 2e- Cu (s) 0.3419 Cu2+ (ag) + e - Cu* (aq) 0.153 Note: write the half reaction as it would actually occur at the anode. Cu* (ag) + e - Cu (s) 0.521 F2 (9) + 2e 2F (aq) 2.866 Fe2+ (ag) + 2e- Fe (s) -0.447 Explanation Check Fe3+ (aq) + e - Fe2+ (aq) 0.771 1. Fe+ (aq) + 3e Fe (s) -0.037 MacBook Air 20 FI F12 D00 F4 F10 FS F6 F7 F8 F9 FI F2 F3
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...
Related questions
Question
![A certain half-reaction has a standard reduction potential E =-1.25 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that
must provide at least 0.80 V of electrical power. The cell will operate under standard conditions.
Note for advanced students: assume the engineer requires this half-reaction to happen at the cathode of the cell.
h Data
Ag+ (aq) + e Ag (s)
0.7996
Is there a minimum standard reduction
potential that the half-reaction used at
the anode of this cell can have?
yes, there is a minimum.
Ered
= v
!!
Al3+ (aq) + 3e-- Al (s)
-1.676
Au+ (aq) + e Au (s)
1.692
If so, check the "yes" box and calculate
the minimum. Round your answer to 2
decimal places. If there is no lower
limit, check the "no" box.
Au3+ (aq) + 3e - Au (s)
1.498
O no minimum
Ba2+ (ag) + 2e - Ba (s)
-2.912
Brz (1) + 2e- 2Br (aq)
1.066
Ca2+ (ag) + 2e - Ca (s)
-2.868
Is there a maximum standard reduction
potential that the half-reaction used at
the anode of this cell can have?
E= Ov
Cl2 (g) + 2e - 2CI- (aq)
Co2+ (aq) + 2e- - Co (s)
O yes, there is a maximum.
1.35827
red
-0.28
If so, check the "yes" box and calculate
the maximum. Round your answer to 2
decimal places. If there is no upper
limit, check the "no" box.
Co3+ (aq) + e- Co2+ (aq)
1.92
O no maximum
Cr2+ (aq) + 2e→ Cr (s)
Cr+ (aq) + 3e - Cr (s)
Cr+ (aq) + e - Cr2+ (aq)
-0.913
-0.744
-0.407
Cro42- (aq) + 4H20 (1) + 3e -
Cr(OH)3 (s) + 5OH (aq)
-0.13
By using the information in the ALEKS
Data tab, write a balanced equation
describing a half reaction that could be
used at the anode of this cell.
Cu2+ (aq) + 2e - Cu (s)
0.3419
Cu2+ (ag) + e - Cut (ag)
0.153
Note: write the half reaction as it would
actually occur at the anode.
Cu* (ag) + e→ Cu (s)
0.521
F2 (g) + 2e 2F (aq)
2.866
Fe2+ (aq) + 2e Fe (s)
-0.447
Explanation
Check
Fe3+ (aq) + e-- Fe2+ (aq)
0.771
Fe+ (aq) + 3e Fe (s)
-0.037
MacBook Air
20
esc
000
000 FA
►►
F12
F8
F9
F10
FII
F1
F2
F3
F5
F6
F7
24
%
&
*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd61b6772-1f69-4fa3-8553-04a65cf6b22e%2F870ad5f8-22b0-4fb1-a295-c4e590eceff3%2F7y7f0zl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A certain half-reaction has a standard reduction potential E =-1.25 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that
must provide at least 0.80 V of electrical power. The cell will operate under standard conditions.
Note for advanced students: assume the engineer requires this half-reaction to happen at the cathode of the cell.
h Data
Ag+ (aq) + e Ag (s)
0.7996
Is there a minimum standard reduction
potential that the half-reaction used at
the anode of this cell can have?
yes, there is a minimum.
Ered
= v
!!
Al3+ (aq) + 3e-- Al (s)
-1.676
Au+ (aq) + e Au (s)
1.692
If so, check the "yes" box and calculate
the minimum. Round your answer to 2
decimal places. If there is no lower
limit, check the "no" box.
Au3+ (aq) + 3e - Au (s)
1.498
O no minimum
Ba2+ (ag) + 2e - Ba (s)
-2.912
Brz (1) + 2e- 2Br (aq)
1.066
Ca2+ (ag) + 2e - Ca (s)
-2.868
Is there a maximum standard reduction
potential that the half-reaction used at
the anode of this cell can have?
E= Ov
Cl2 (g) + 2e - 2CI- (aq)
Co2+ (aq) + 2e- - Co (s)
O yes, there is a maximum.
1.35827
red
-0.28
If so, check the "yes" box and calculate
the maximum. Round your answer to 2
decimal places. If there is no upper
limit, check the "no" box.
Co3+ (aq) + e- Co2+ (aq)
1.92
O no maximum
Cr2+ (aq) + 2e→ Cr (s)
Cr+ (aq) + 3e - Cr (s)
Cr+ (aq) + e - Cr2+ (aq)
-0.913
-0.744
-0.407
Cro42- (aq) + 4H20 (1) + 3e -
Cr(OH)3 (s) + 5OH (aq)
-0.13
By using the information in the ALEKS
Data tab, write a balanced equation
describing a half reaction that could be
used at the anode of this cell.
Cu2+ (aq) + 2e - Cu (s)
0.3419
Cu2+ (ag) + e - Cut (ag)
0.153
Note: write the half reaction as it would
actually occur at the anode.
Cu* (ag) + e→ Cu (s)
0.521
F2 (g) + 2e 2F (aq)
2.866
Fe2+ (aq) + 2e Fe (s)
-0.447
Explanation
Check
Fe3+ (aq) + e-- Fe2+ (aq)
0.771
Fe+ (aq) + 3e Fe (s)
-0.037
MacBook Air
20
esc
000
000 FA
►►
F12
F8
F9
F10
FII
F1
F2
F3
F5
F6
F7
24
%
&
*
![2H (aq) + 2e H2 (g)
0.000
2H20 (1) + 2e → H2 (g) + 20H (aq)
-0.8277
H202 (aq) + 2H+ (aq) + 2e → 2H20 (1)
1.776
I2 (s) + 2e 21 (aq)
0.5355
2103 (aq) + 12H+ (aq) + 10e I2 (s) + 6H20 (1)
1.195
Mg2+ (aq) + 2e Mg (s)
-2.372
Mn2+ (aq) + 2e Mn (s)
-1.185
MnO2 () + 4H+ (aq) + 2e-
- Mn2+ (aq) + 2H20 (1)
1.224
MnO4 (aq) + 8H+ (aq) + 5e Mn2+ (aq) + 4H20 (1)
1.507
MnO4 (aq) + 2H20 (1) + 3e MnO2 (s) + 40H (aq)
0.595
HNO2 (aq) + H+ (aq) + e- NO (g) + H20 (1)
0.983
N2 (g) + 4H2O (1) + 4e- → 40H (aq) + N2H4 (aq)
-1.16
NO3 (aq) + 4H+ (aq) + 3e NO (g) + 2H20 (I)
0.957
Na+ (aq) + e Na (s)
-2.71
Ni2+ (ag) + 2e Ni (s)
-0.257
1.229
02 (g) + 4H+ (aq) + 4e- → 2H20 (1)
0.401
02 (g) + 2H20 (1) + 4e¬ → 40H (aq)
0.695
02 (g) + 2H+ (aq) + 2e → H2O2 (aq)
-0.1262
Pb2+ (ag) + 2e → Pb (s)
-0.3588
PBSO4 (s) + H (aq) + 2e Pb (s) + HSO4 (aq)
k Air
4)
F12
F11
II
F8
F9
F10
F7
%3D
delete
+ II](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd61b6772-1f69-4fa3-8553-04a65cf6b22e%2F870ad5f8-22b0-4fb1-a295-c4e590eceff3%2Fks1gi1r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2H (aq) + 2e H2 (g)
0.000
2H20 (1) + 2e → H2 (g) + 20H (aq)
-0.8277
H202 (aq) + 2H+ (aq) + 2e → 2H20 (1)
1.776
I2 (s) + 2e 21 (aq)
0.5355
2103 (aq) + 12H+ (aq) + 10e I2 (s) + 6H20 (1)
1.195
Mg2+ (aq) + 2e Mg (s)
-2.372
Mn2+ (aq) + 2e Mn (s)
-1.185
MnO2 () + 4H+ (aq) + 2e-
- Mn2+ (aq) + 2H20 (1)
1.224
MnO4 (aq) + 8H+ (aq) + 5e Mn2+ (aq) + 4H20 (1)
1.507
MnO4 (aq) + 2H20 (1) + 3e MnO2 (s) + 40H (aq)
0.595
HNO2 (aq) + H+ (aq) + e- NO (g) + H20 (1)
0.983
N2 (g) + 4H2O (1) + 4e- → 40H (aq) + N2H4 (aq)
-1.16
NO3 (aq) + 4H+ (aq) + 3e NO (g) + 2H20 (I)
0.957
Na+ (aq) + e Na (s)
-2.71
Ni2+ (ag) + 2e Ni (s)
-0.257
1.229
02 (g) + 4H+ (aq) + 4e- → 2H20 (1)
0.401
02 (g) + 2H20 (1) + 4e¬ → 40H (aq)
0.695
02 (g) + 2H+ (aq) + 2e → H2O2 (aq)
-0.1262
Pb2+ (ag) + 2e → Pb (s)
-0.3588
PBSO4 (s) + H (aq) + 2e Pb (s) + HSO4 (aq)
k Air
4)
F12
F11
II
F8
F9
F10
F7
%3D
delete
+ II
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