A certain half-reaction has a standard reduction potential Ed=+1.34 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 1.20 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 celI. Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? Ov yes, there is a minimum. red 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. no minimum Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? yes, there is a maximum. red dh Data 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. no maximum E° (V) Half-Reaction F2 (g) + 2e → 2F¯ (aq) 2.866 Co3+ (aq) + e -→ Co2+ (aq) 1.92 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. H2O2 (aq) + 2H+ (aq) + 2e 2H2O (I) 1.776 Au+ (aq) + e Au (s) 1.692 Mn2+ (aq) + 4H2O (I) MnO4 (aq) + 8H+ (aq) + 5e- 1.507 Note: write the half reaction as it would actually occur at the anode. Au3+ (aq) + 3e Au (s) 1.498 2C1 (aq) Cl2 (g) + 2e- 1.35827 O2 (g) + 4H+ (aq) + 4e → 2H20 (I) 1.229 Mn2+ (aq) + 2H2O (I) MnO2 (s) + 4H+ (aq) + 2e¯ → 1.224 2103 (aq) + 12H+ (aq) + 10e I2 (s) + 6H2O (I) 1.195 Br2 (1) + 2e- → 2Br (aq) 1.066 Vo2+ (aq) + 2H+ (aq) + e-→ vO2+ (aq) + H20 (1) 0.991 Explanation Check NO (g) + H20 (I) HNO2 (aq) + H+ (aq) + e 0.983 dard reduction ction used at Ov yes, there is a minimum. %3D red have? ala nh Data and calculate - answer to 2 no lower 2H+ (aq) + 2e- H2 (g) 0.000 no minimum Fe3+ (aq) + 3e Fe (s) -0.037 Pb2+ (aq) + 2e Pb (s) -0.1262 CrO42- (aq) + 4H,0 (1) + 3e → Cr(OH)3 (s) + 50H (aq) -0.13 dard reduction ction used at have? yes, there is a Sn2+ (aq) + 2e Sn (s) -0.1375 Ni2+ (aq) + 2e→ Ni (s) -0.257 and calculate r answer to 2 no upper Co2+ (aq) + 2e Co (s) -0.28 no maximum PBSO4 (s) + H+ (aq) + 2e Pb (s) + HSO4 (aq) -0.3588 Cr+ (aq) + e Cr2+ (aq) -0.407 Fe2+ (aq) + 2e Fe (s) -0.447 n the ALEKS equation chat could be cell. Cr3+ (aq) + 3e Cr (s) -0.744 Zn2+ (aq) + 2e Zn (s) -0.7618 on as it would e. 2H20 (1) + 2e→ H2 (g) + 2OH- (aq) -0.8277 Cr2+ (aq) + 2e Cr (s) =0.913 N2 (g) + 4H2O (I) + 4e → 40H- (aq) + N2H4 (aq) -1.16 Mn2+ (aq) + 2e Mn (s) -1.185 Al3+ (aq) + 3e- Al (s) -1.676 Sc3+ (aq) + 3e Sc (s) -2.077 Mg2+ (aq) + 2e Mg (s) -2.372 Na+ (aq) + e Na (s) -2.71 Check Ca2+ (aq) + 2e Ca (s) -2.868 Ba2+ (ag) + 2e Ba (s) -2.912

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Write out the half reaction as it would actuall Accur at the anode. And is there a min or a max standard reduction. 

A certain half-reaction has a standard reduction potential Ed=+1.34 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that
must provide at least 1.20 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 celI.
Is there a minimum standard reduction
potential that the half-reaction used at
the anode of this cell can have?
Ov
yes, there is a minimum.
red
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.
no minimum
Is there a maximum standard reduction
potential that the half-reaction used at
the anode of this cell can have?
yes, there is a maximum.
red
dh Data
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.
no maximum
E° (V)
Half-Reaction
F2 (g) + 2e → 2F¯ (aq)
2.866
Co3+ (aq) + e
-→ Co2+ (aq)
1.92
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.
H2O2 (aq) + 2H+ (aq) + 2e
2H2O (I)
1.776
Au+ (aq) + e Au (s)
1.692
Mn2+ (aq) + 4H2O (I)
MnO4 (aq) + 8H+ (aq) + 5e-
1.507
Note: write the half reaction as it would
actually occur at the anode.
Au3+ (aq) + 3e Au (s)
1.498
2C1 (aq)
Cl2 (g) + 2e-
1.35827
O2 (g) + 4H+ (aq) + 4e →
2H20 (I)
1.229
Mn2+ (aq) + 2H2O (I)
MnO2 (s) + 4H+ (aq) + 2e¯ →
1.224
2103 (aq) + 12H+ (aq) + 10e
I2 (s) + 6H2O (I)
1.195
Br2 (1) + 2e- → 2Br (aq)
1.066
Vo2+ (aq) + 2H+ (aq) + e-→ vO2+ (aq) + H20 (1)
0.991
Explanation
Check
NO (g) + H20 (I)
HNO2 (aq) + H+ (aq) + e
0.983
Transcribed Image Text:A certain half-reaction has a standard reduction potential Ed=+1.34 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 1.20 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 celI. Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? Ov yes, there is a minimum. red 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. no minimum Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? yes, there is a maximum. red dh Data 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. no maximum E° (V) Half-Reaction F2 (g) + 2e → 2F¯ (aq) 2.866 Co3+ (aq) + e -→ Co2+ (aq) 1.92 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. H2O2 (aq) + 2H+ (aq) + 2e 2H2O (I) 1.776 Au+ (aq) + e Au (s) 1.692 Mn2+ (aq) + 4H2O (I) MnO4 (aq) + 8H+ (aq) + 5e- 1.507 Note: write the half reaction as it would actually occur at the anode. Au3+ (aq) + 3e Au (s) 1.498 2C1 (aq) Cl2 (g) + 2e- 1.35827 O2 (g) + 4H+ (aq) + 4e → 2H20 (I) 1.229 Mn2+ (aq) + 2H2O (I) MnO2 (s) + 4H+ (aq) + 2e¯ → 1.224 2103 (aq) + 12H+ (aq) + 10e I2 (s) + 6H2O (I) 1.195 Br2 (1) + 2e- → 2Br (aq) 1.066 Vo2+ (aq) + 2H+ (aq) + e-→ vO2+ (aq) + H20 (1) 0.991 Explanation Check NO (g) + H20 (I) HNO2 (aq) + H+ (aq) + e 0.983
dard reduction
ction used at
Ov
yes, there is a minimum.
%3D
red
have?
ala
nh Data
and calculate
- answer to 2
no lower
2H+ (aq) + 2e- H2 (g)
0.000
no minimum
Fe3+ (aq) + 3e
Fe (s)
-0.037
Pb2+ (aq) + 2e
Pb (s)
-0.1262
CrO42- (aq) + 4H,0 (1) + 3e
→ Cr(OH)3 (s) + 50H (aq)
-0.13
dard reduction
ction used at
have?
yes, there is a
Sn2+ (aq) + 2e
Sn (s)
-0.1375
Ni2+ (aq) + 2e→ Ni (s)
-0.257
and calculate
r answer to 2
no upper
Co2+ (aq) + 2e
Co (s)
-0.28
no maximum
PBSO4 (s) + H+ (aq) + 2e
Pb (s) + HSO4 (aq)
-0.3588
Cr+ (aq) + e
Cr2+ (aq)
-0.407
Fe2+ (aq) + 2e
Fe (s)
-0.447
n the ALEKS
equation
chat could be
cell.
Cr3+ (aq) + 3e
Cr (s)
-0.744
Zn2+ (aq) + 2e
Zn (s)
-0.7618
on as it would
e.
2H20 (1) + 2e→ H2 (g) + 2OH- (aq)
-0.8277
Cr2+ (aq) + 2e
Cr (s)
=0.913
N2 (g) + 4H2O (I) + 4e → 40H- (aq) + N2H4 (aq)
-1.16
Mn2+ (aq) + 2e
Mn (s)
-1.185
Al3+ (aq) + 3e- Al (s)
-1.676
Sc3+ (aq) + 3e
Sc (s)
-2.077
Mg2+ (aq) + 2e
Mg (s)
-2.372
Na+ (aq) + e
Na (s)
-2.71
Check
Ca2+ (aq) + 2e
Ca (s)
-2.868
Ba2+ (ag) + 2e
Ba (s)
-2.912
Transcribed Image Text:dard reduction ction used at Ov yes, there is a minimum. %3D red have? ala nh Data and calculate - answer to 2 no lower 2H+ (aq) + 2e- H2 (g) 0.000 no minimum Fe3+ (aq) + 3e Fe (s) -0.037 Pb2+ (aq) + 2e Pb (s) -0.1262 CrO42- (aq) + 4H,0 (1) + 3e → Cr(OH)3 (s) + 50H (aq) -0.13 dard reduction ction used at have? yes, there is a Sn2+ (aq) + 2e Sn (s) -0.1375 Ni2+ (aq) + 2e→ Ni (s) -0.257 and calculate r answer to 2 no upper Co2+ (aq) + 2e Co (s) -0.28 no maximum PBSO4 (s) + H+ (aq) + 2e Pb (s) + HSO4 (aq) -0.3588 Cr+ (aq) + e Cr2+ (aq) -0.407 Fe2+ (aq) + 2e Fe (s) -0.447 n the ALEKS equation chat could be cell. Cr3+ (aq) + 3e Cr (s) -0.744 Zn2+ (aq) + 2e Zn (s) -0.7618 on as it would e. 2H20 (1) + 2e→ H2 (g) + 2OH- (aq) -0.8277 Cr2+ (aq) + 2e Cr (s) =0.913 N2 (g) + 4H2O (I) + 4e → 40H- (aq) + N2H4 (aq) -1.16 Mn2+ (aq) + 2e Mn (s) -1.185 Al3+ (aq) + 3e- Al (s) -1.676 Sc3+ (aq) + 3e Sc (s) -2.077 Mg2+ (aq) + 2e Mg (s) -2.372 Na+ (aq) + e Na (s) -2.71 Check Ca2+ (aq) + 2e Ca (s) -2.868 Ba2+ (ag) + 2e Ba (s) -2.912
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