F2(g)+2e-→2F-(aq)2.866O3(g)+2H+(aq)+2e-→O2(g)+H2O(l)2.076Co3+(aq)+e-→Co2+(aq)1.92H2O2(aq)+2H+(aq)+2e-→2H2O(l)1.776Ce4+(aq)+e-→Ce3+(aq)1.72Au+(aq)+e-→Au(s)1.692PbO2(s)+4H+(aq)+SO42-(aq)+2e-→PbSO4(s)+2H2O(l)1.691MnO4-(aq)+8H+(aq)+5e-→Mn2+(aq)+4H2O(l)1.507Au3+(aq)+3e-→Au(s)1.498Cr2O72-(aq)+14H+(aq)+6e-→2Cr3+(aq)+7H2O(l)1.36Cl2(g)+2e-→2Cl-(aq)1.358O2(g)+4H+(aq)+4e-→2H2O(l)1.229MnO2(s)+4H+(aq)+2e-→Mn2+(aq)+2H2O(l)1.2242IO3-(aq)+12H+(aq)+10e-→I2(s)+6H2O(l)1.195Br2(l)+2e-→2Br-(aq)1.066VO2+(aq)+2H+(aq)+e-→VO2+(aq)+H2O(l)0.991HNO2(aq)+H+(aq)+e-→NO(g)+H2O(l)0.983NO3-(aq)+4H+(aq)+3e-→NO(g)+2H2O(l)0.9572Hg2+(aq)+2e-→Hg22+(aq)0.92Hg2+(aq)+2e-→Hg(l)0.851Ag+(aq)+e-→Ag(s)0.800Hg22+(aq)+2e-→2Hg(l)0.797Fe3+(aq)+e-→Fe2+(aq)0.771O2(g)+2H+(aq)+2e-→H2O2(aq)0.695MnO4-(aq)+2H2O(l)+3e-→MnO2(s)+4OH-(aq)0.595I2(s)+2e-→2I-(aq)0.536Cu+(aq)+e-→Cu(s)0.521O2(g)+2H2O(l)+4e-→4OH-(aq)0.401Cu2+(aq)+2e-→Cu(s)0.342HSO4-(aq)+3H+(aq)+2e-→H2SO3(aq)+H2O(l)0.172SO42-(aq)+4H+(aq)+2e-→SO2(g)+2H2O(l)0.172Cu2+(aq)+e-→Cu+(aq)0.153Sn4+(aq)+2e-→Sn2+(aq)0.1512H+(aq)+2e-→H2(g)0.000Fe3+(aq)+3e-→Fe(s)-0.037Pb2+(aq)+2e-→Pb(s)-0.126CrO42-(aq)+4H2O(l)+3e-→Cr(OH)3(s)+5OH-(aq)-0.13Sn2+(aq)+2e-→Sn(s)-0.138N2(g)+5H+(aq)+4e-→N2H5+(aq)-0.214Ni2+(aq)+2e-→Ni(s)-0.257Co2+(aq)+2e-→Co(s)-0.28PbSO4(s)+H+(aq)+2e-→Pb(s)+HSO4-(aq)-0.359Cd2+(aq)+2e-→Cd(s)-0.403Cr3+(aq)+e-→Cr2+(aq)-0.407Fe2+(aq)+2e-→Fe(s)-0.447Cr3+(aq)+3e-→Cr(s)-0.744Zn2+(aq)+2e-→Zn(s)-0.7622H2O(l)+2e-→H2(g)+2OH-(aq)-0.828Cr2+(aq)+2e-→Cr(s)-0.913N2(g)+4H2O(l)+4e-→4OH-(aq)+N2H4(aq)-1.16Mn2+(aq)+2e-→Mn(s)-1.185Al3+(aq)+3e-→Al(s)-1.676Be2+(aq)+2e-→Be(s)-1.847Sc3+(aq)+3e-→Sc(s)-2.077Mg2+(aq)+2e-→Mg(s)-2.372Na+(aq)+e-→Na(s)-2.71Ca2+(aq)+2e-→Ca(s)-2.868Sr2+(aq)+2e-→Sr(s)-2.899Ba2+(aq)+2e-→Ba(s)-2.912K+(aq)+e-→K(s)-2.931Li+(aq)+e-→Li(s)-3.040 A certain half-reaction has a standard reduction potential Ered=+0.69 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that mustprovide at least 0.60 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 anode of the cell. Is there a minimum standard reductionpotential that the half-reaction used atthe cathode of this cell can have?If so, check the "yes" box and calculatethe minimum. Round your answer to 2decimal places. If there is no lowerlimit, check the "no" box.Is there a maximum standard reductionpotential that the half-reaction used atthe cathode of this cell can have?If so, check the "yes" box and calculatethe maximum. Round your answer to 2decimal places. If there is no upperlimit, check the "no" box.By using the information in the ALEKSData tab, write a balanced equationdescribing a half reaction that could beused at the cathode of this cell.Note: write the half reaction as it wouldactually occur at the cathode.yes, there is a minimum.0no minimumO yes, there is a maximum.no maximum0Ered0Ered= Ov=☐v

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Answered: Is there a minimum standard reduction…

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter17: Electrochemistry

Section: Chapter Questions

Problem 137CWP: Consider a galvanic cell based on the following half-reactions: a. What is the standard potential...

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F₂(g)+2e^-→2F^-(aq)	2.866
O₃(g)+2H⁺(aq)+2e^-→O₂(g)+H₂O(l)	2.076
Co³⁺(aq)+e^-→Co²⁺(aq)	1.92
H₂O₂(aq)+2H⁺(aq)+2e^-→2H₂O(l)	1.776
Ce⁴⁺(aq)+e^-→Ce³⁺(aq)	1.72
Au⁺(aq)+e^-→Au(s)	1.692
PbO₂(s)+4H⁺(aq)+SO₄^2-(aq)+2e^-→PbSO₄(s)+2H₂O(l)	1.691
MnO₄^-(aq)+8H⁺(aq)+5e^-→Mn²⁺(aq)+4H₂O(l)	1.507
Au³⁺(aq)+3e^-→Au(s)	1.498
Cr₂O₇^2-(aq)+14H⁺(aq)+6e^-→2Cr³⁺(aq)+7H₂O(l)	1.36
Cl₂(g)+2e^-→2Cl^-(aq)	1.358
O₂(g)+4H⁺(aq)+4e^-→2H₂O(l)	1.229
MnO₂(s)+4H⁺(aq)+2e^-→Mn²⁺(aq)+2H₂O(l)	1.224
2IO₃^-(aq)+12H⁺(aq)+10e^-→I₂(s)+6H₂O(l)	1.195
Br₂(l)+2e^-→2Br^-(aq)	1.066
VO₂⁺(aq)+2H⁺(aq)+e^-→VO²⁺(aq)+H₂O(l)	0.991
HNO₂(aq)+H⁺(aq)+e^-→NO(g)+H₂O(l)	0.983

NO₃^-(aq)+4H⁺(aq)+3e^-→NO(g)+2H₂O(l)	0.957
2Hg²⁺(aq)+2e^-→Hg₂²⁺(aq)	0.92
Hg²⁺(aq)+2e^-→Hg(l)	0.851
Ag⁺(aq)+e^-→Ag(s)	0.800
Hg₂²⁺(aq)+2e^-→2Hg(l)	0.797
Fe³⁺(aq)+e^-→Fe²⁺(aq)	0.771
O₂(g)+2H⁺(aq)+2e^-→H₂O₂(aq)	0.695
MnO₄^-(aq)+2H₂O(l)+3e^-→MnO₂(s)+4OH^-(aq)	0.595
I₂(s)+2e^-→2I^-(aq)	0.536
Cu⁺(aq)+e^-→Cu(s)	0.521
O₂(g)+2H₂O(l)+4e^-→4OH^-(aq)	0.401
Cu²⁺(aq)+2e^-→Cu(s)	0.342
HSO₄^-(aq)+3H⁺(aq)+2e^-→H₂SO₃(aq)+H₂O(l)	0.172
SO₄^2-(aq)+4H⁺(aq)+2e^-→SO₂(g)+2H₂O(l)	0.172
Cu²⁺(aq)+e^-→Cu⁺(aq)	0.153
Sn⁴⁺(aq)+2e^-→Sn²⁺(aq)	0.151
2H⁺(aq)+2e^-→H₂(g)	0.000
Fe³⁺(aq)+3e^-→Fe(s)	-0.037
Pb²⁺(aq)+2e^-→Pb(s)	-0.126
CrO₄^2-(aq)+4H₂O(l)+3e^-→Cr(OH)₃(s)+5OH^-(aq)	-0.13
Sn²⁺(aq)+2e^-→Sn(s)	-0.138
N₂(g)+5H⁺(aq)+4e^-→N₂H₅⁺(aq)	-0.214
Ni²⁺(aq)+2e^-→Ni(s)	-0.257
Co²⁺(aq)+2e^-→Co(s)	-0.28
PbSO₄(s)+H⁺(aq)+2e^-→Pb(s)+HSO₄^-(aq)	-0.359
Cd²⁺(aq)+2e^-→Cd(s)	-0.403
Cr³⁺(aq)+e^-→Cr²⁺(aq)	-0.407
Fe²⁺(aq)+2e^-→Fe(s)	-0.447
Cr³⁺(aq)+3e^-→Cr(s)	-0.744
Zn²⁺(aq)+2e^-→Zn(s)	-0.762
2H₂O(l)+2e^-→H₂(g)+2OH^-(aq)	-0.828
Cr²⁺(aq)+2e^-→Cr(s)	-0.913
N₂(g)+4H₂O(l)+4e^-→4OH^-(aq)+N₂H₄(aq)	-1.16
Mn²⁺(aq)+2e^-→Mn(s)	-1.185
Al³⁺(aq)+3e^-→Al(s)	-1.676
Be²⁺(aq)+2e^-→Be(s)	-1.847
Sc³⁺(aq)+3e^-→Sc(s)	-2.077
Mg²⁺(aq)+2e^-→Mg(s)	-2.372
Na⁺(aq)+e^-→Na(s)	-2.71
Ca²⁺(aq)+2e^-→Ca(s)	-2.868
Sr²⁺(aq)+2e^-→Sr(s)	-2.899
Ba²⁺(aq)+2e^-→Ba(s)	-2.912
K⁺(aq)+e^-→K(s)	-2.931
Li⁺(aq)+e^-→Li(s)	-3.040

A certain half-reaction has a standard reduction potential Ered=+0.69 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that must
provide at least 0.60 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 anode of the cell.

Is there a minimum standard reduction
potential that the half-reaction used at
the cathode of this cell can have?
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.
Is there a maximum standard reduction
potential that the half-reaction used at
the cathode of this cell can have?
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.
By using the information in the ALEKS
Data tab, write a balanced equation
describing a half reaction that could be
used at the cathode of this cell.
Note: write the half reaction as it would
actually occur at the cathode.
yes, there is a minimum.
0
no minimum
O yes, there is a maximum.
no maximum
0
E
red
0
E
red
= Ov
=
☐v

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