Ag+(aq)+e-→Ag(s)0.800Al3+(aq)+3e-→Al(s)-1.676Au+(aq)+e-→Au(s)1.692Au3+(aq)+3e-→Au(s)1.498Ba2+(aq)+2e-→Ba(s)-2.912Be2+(aq)+2e-→Be(s)-1.847Br2(l)+2e-→2Br-(aq)1.066Ca2+(aq)+2e-→Ca(s)-2.868Cd2+(aq)+2e-→Cd(s)-0.403Ce4+(aq)+e-→Ce3+(aq)1.72Cl2(g)+2e-→2Cl-(aq)1.358Co2+(aq)+2e-→Co(s)-0.28Co3+(aq)+e-→Co2+(aq)1.92Cr2+(aq)+2e-→Cr(s)-0.913Cr3+(aq)+e-→Cr2+(aq)-0.407Cr3+(aq)+3e-→Cr(s)-0.744CrO42-(aq)+4H2O(l)+3e-→Cr(OH)3(s)+5OH-(aq)-0.13Cr2O72-(aq)+14H+(aq)+6e-→2Cr3+(aq)+7H2O(l)1.36Cu+(aq)+e-→Cu(s)0.521Cu2+(aq)+e-→Cu+(aq)0.153Cu2+(aq)+2e-→Cu(s)0.342F2(g)+2e-→2F-(aq)2.866Fe2+(aq)+2e-→Fe(s)-0.447Fe3+(aq)+e-→Fe2+(aq)0.771Fe3+(aq)+3e-→Fe(s)-0.0372H+(aq)+2e-→H2(g)0.0002H2O(l)+2e-→H2(g)+2OH-(aq)-0.828H2O2(aq)+2H+(aq)+2e-→2H2O(l)1.776Hg2+(aq)+2e-→Hg(l)0.8512Hg2+(aq)+2e-→Hg22+(aq)0.92Hg22+(aq)+2e-→2Hg(l)0.797I2(s)+2e-→2I-(aq)0.5362IO3-(aq)+12H+(aq)+10e-→I2(s)+6H2O(l)1.195K+(aq)+e-→K(s)-2.931Li+(aq)+e-→Li(s)-3.040Mg2+(aq)+2e-→Mg(s)-2.372Mn2+(aq)+2e-→Mn(s)-1.185MnO2(s)+4H+(aq)+2e-→Mn2+(aq)+2H2O(l)1.224MnO4-(aq)+8H+(aq)+5e-→Mn2+(aq)+4H2O(l)1.507MnO4-(aq)+2H2O(l)+3e-→MnO2(s)+4OH-(aq)0.595HNO2(aq)+H+(aq)+e-→NO(g)+H2O(l)0.983N2(g)+5H+(aq)+4e-→N2H5+(aq)-0.214N2(g)+4H2O(l)+4e-→4OH-(aq)+N2H4(aq)-1.16NO3-(aq)+4H+(aq)+3e-→NO(g)+2H2O(l)0.957Na+(aq)+e-→Na(s)-2.71Ni2+(aq)+2e-→Ni(s)-0.257O2(g)+2H+(aq)+2e-→H2O2(aq)0.695O2(g)+4H+(aq)+4e-→2H2O(l)1.229O2(g)+2H2O(l)+4e-→4OH-(aq)0.401O3(g)+2H+(aq)+2e-→O2(g)+H2O(l)2.076Pb2+(aq)+2e-→Pb(s)-0.126PbO2(s)+4H+(aq)+SO42-(aq)+2e-→PbSO4(s)+2H2O(l)1.691PbSO4(s)+H+(aq)+2e-→Pb(s)+HSO4-(aq)-0.359HSO4-(aq)+3H+(aq)+2e-→H2SO3(aq)+H2O(l)0.172SO42-(aq)+4H+(aq)+2e-→SO2(g)+2H2O(l)0.172Sc3+(aq)+3e-→Sc(s)-2.077Sn2+(aq)+2e-→Sn(s)-0.138Sn4+(aq)+2e-→Sn2+(aq)0.151Sr2+(aq)+2e-→Sr(s)-2.899VO2+(aq)+2H+(aq)+e-→VO2+(aq)+H2O(l)0.991Zn2+(aq)+2e-→Zn(s)-0.762 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.Oyes, there is a minimum.0no minimumyes, there is a maximum.no maximum0Ered = v0Ered=Ov A certain half-reaction has a standard reduction potential E+0.84 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that mustprovide at least 0.50 V of electrical power. The cell will operate under standard conditions.redNote for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell.=

Question

Ag+(aq)+e-→Ag(s)0.800Al3+(aq)+3e-→Al(s)-1.676Au+(aq)+e-→Au(s)1.692Au3+(aq)+3e-→Au(s)1.498Ba2+(aq)+2e-→Ba(s)-2.912Be2+(aq)+2e-→Be(s)-1.847Br2(l)+2e-→2Br-(aq)1.066Ca2+(aq)+2e-→Ca(s)-2.868Cd2+(aq)+2e-→Cd(s)-0.403Ce4+(aq)+e-→Ce3+(aq)1.72Cl2(g)+2e-→2Cl-(aq)1.358Co2+(aq)+2e-→Co(s)-0.28Co3+(aq)+e-→Co2+(aq)1.92Cr2+(aq)+2e-→Cr(s)-0.913Cr3+(aq)+e-→Cr2+(aq)-0.407Cr3+(aq)+3e-→Cr(s)-0.744CrO42-(aq)+4H2O(l)+3e-→Cr(OH)3(s)+5OH-(aq)-0.13Cr2O72-(aq)+14H+(aq)+6e-→2Cr3+(aq)+7H2O(l)1.36Cu+(aq)+e-→Cu(s)0.521Cu2+(aq)+e-→Cu+(aq)0.153Cu2+(aq)+2e-→Cu(s)0.342F2(g)+2e-→2F-(aq)2.866Fe2+(aq)+2e-→Fe(s)-0.447Fe3+(aq)+e-→Fe2+(aq)0.771Fe3+(aq)+3e-→Fe(s)-0.0372H+(aq)+2e-→H2(g)0.0002H2O(l)+2e-→H2(g)+2OH-(aq)-0.828H2O2(aq)+2H+(aq)+2e-→2H2O(l)1.776Hg2+(aq)+2e-→Hg(l)0.8512Hg2+(aq)+2e-→Hg22+(aq)0.92Hg22+(aq)+2e-→2Hg(l)0.797I2(s)+2e-→2I-(aq)0.5362IO3-(aq)+12H+(aq)+10e-→I2(s)+6H2O(l)1.195K+(aq)+e-→K(s)-2.931Li+(aq)+e-→Li(s)-3.040Mg2+(aq)+2e-→Mg(s)-2.372Mn2+(aq)+2e-→Mn(s)-1.185MnO2(s)+4H+(aq)+2e-→Mn2+(aq)+2H2O(l)1.224MnO4-(aq)+8H+(aq)+5e-→Mn2+(aq)+4H2O(l)1.507MnO4-(aq)+2H2O(l)+3e-→MnO2(s)+4OH-(aq)0.595HNO2(aq)+H+(aq)+e-→NO(g)+H2O(l)0.983N2(g)+5H+(aq)+4e-→N2H5+(aq)-0.214N2(g)+4H2O(l)+4e-→4OH-(aq)+N2H4(aq)-1.16NO3-(aq)+4H+(aq)+3e-→NO(g)+2H2O(l)0.957Na+(aq)+e-→Na(s)-2.71Ni2+(aq)+2e-→Ni(s)-0.257O2(g)+2H+(aq)+2e-→H2O2(aq)0.695O2(g)+4H+(aq)+4e-→2H2O(l)1.229O2(g)+2H2O(l)+4e-→4OH-(aq)0.401O3(g)+2H+(aq)+2e-→O2(g)+H2O(l)2.076Pb2+(aq)+2e-→Pb(s)-0.126PbO2(s)+4H+(aq)+SO42-(aq)+2e-→PbSO4(s)+2H2O(l)1.691PbSO4(s)+H+(aq)+2e-→Pb(s)+HSO4-(aq)-0.359HSO4-(aq)+3H+(aq)+2e-→H2SO3(aq)+H2O(l)0.172SO42-(aq)+4H+(aq)+2e-→SO2(g)+2H2O(l)0.172Sc3+(aq)+3e-→Sc(s)-2.077Sn2+(aq)+2e-→Sn(s)-0.138Sn4+(aq)+2e-→Sn2+(aq)0.151Sr2+(aq)+2e-→Sr(s)-2.899VO2+(aq)+2H+(aq)+e-→VO2+(aq)+H2O(l)0.991Zn2+(aq)+2e-→Zn(s)-0.762 Is there a minimum standard reductionpotential that the half-reaction used atthe cathode of this cell can have?If so, check the &#34;yes&#34; box and calculatethe minimum. Round your answer to 2decimal places. If there is no lowerlimit, check the &#34;no&#34; box.Is there a maximum standard reductionpotential that the half-reaction used atthe cathode of this cell can have?If so, check the &#34;yes&#34; box and calculatethe maximum. Round your answer to 2decimal places. If there is no upperlimit, check the &#34;no&#34; 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.Oyes, there is a minimum.0no minimumyes, there is a maximum.no maximum0Ered = v0Ered=Ov A certain half-reaction has a standard reduction potential E+0.84 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that mustprovide at least 0.50 V of electrical power. The cell will operate under standard conditions.redNote for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell.=

Accepted Answer

Answered: Image /qna-images/answer/ad036cbd-926b-4e6f-85cd-b0e7b48c262f.jpg

Ag⁺(aq)+e^-→Ag(s)	0.800
Al³⁺(aq)+3e^-→Al(s)	-1.676
Au⁺(aq)+e^-→Au(s)	1.692
Au³⁺(aq)+3e^-→Au(s)	1.498
Ba²⁺(aq)+2e^-→Ba(s)	-2.912
Be²⁺(aq)+2e^-→Be(s)	-1.847
Br₂(l)+2e^-→2Br^-(aq)	1.066
Ca²⁺(aq)+2e^-→Ca(s)	-2.868
Cd²⁺(aq)+2e^-→Cd(s)	-0.403
Ce⁴⁺(aq)+e^-→Ce³⁺(aq)	1.72
Cl₂(g)+2e^-→2Cl^-(aq)	1.358
Co²⁺(aq)+2e^-→Co(s)	-0.28
Co³⁺(aq)+e^-→Co²⁺(aq)	1.92
Cr²⁺(aq)+2e^-→Cr(s)	-0.913
Cr³⁺(aq)+e^-→Cr²⁺(aq)	-0.407
Cr³⁺(aq)+3e^-→Cr(s)	-0.744
CrO₄^2-(aq)+4H₂O(l)+3e^-→Cr(OH)₃(s)+5OH^-(aq)	-0.13
Cr₂O₇^2-(aq)+14H⁺(aq)+6e^-→2Cr³⁺(aq)+7H₂O(l)	1.36
Cu⁺(aq)+e^-→Cu(s)	0.521
Cu²⁺(aq)+e^-→Cu⁺(aq)	0.153
Cu²⁺(aq)+2e^-→Cu(s)	0.342
F₂(g)+2e^-→2F^-(aq)	2.866
Fe²⁺(aq)+2e^-→Fe(s)	-0.447
Fe³⁺(aq)+e^-→Fe²⁺(aq)	0.771
Fe³⁺(aq)+3e^-→Fe(s)	-0.037
2H⁺(aq)+2e^-→H₂(g)	0.000
2H₂O(l)+2e^-→H₂(g)+2OH^-(aq)	-0.828
H₂O₂(aq)+2H⁺(aq)+2e^-→2H₂O(l)	1.776
Hg²⁺(aq)+2e^-→Hg(l)	0.851
2Hg²⁺(aq)+2e^-→Hg₂²⁺(aq)	0.92
Hg₂²⁺(aq)+2e^-→2Hg(l)	0.797
I₂(s)+2e^-→2I^-(aq)	0.536
2IO₃^-(aq)+12H⁺(aq)+10e^-→I₂(s)+6H₂O(l)	1.195
K⁺(aq)+e^-→K(s)	-2.931
Li⁺(aq)+e^-→Li(s)	-3.040
Mg²⁺(aq)+2e^-→Mg(s)	-2.372
Mn²⁺(aq)+2e^-→Mn(s)	-1.185
MnO₂(s)+4H⁺(aq)+2e^-→Mn²⁺(aq)+2H₂O(l)	1.224
MnO₄^-(aq)+8H⁺(aq)+5e^-→Mn²⁺(aq)+4H₂O(l)	1.507
MnO₄^-(aq)+2H₂O(l)+3e^-→MnO₂(s)+4OH^-(aq)	0.595
HNO₂(aq)+H⁺(aq)+e^-→NO(g)+H₂O(l)	0.983
N₂(g)+5H⁺(aq)+4e^-→N₂H₅⁺(aq)	-0.214
N₂(g)+4H₂O(l)+4e^-→4OH^-(aq)+N₂H₄(aq)	-1.16
NO₃^-(aq)+4H⁺(aq)+3e^-→NO(g)+2H₂O(l)	0.957
Na⁺(aq)+e^-→Na(s)	-2.71
Ni²⁺(aq)+2e^-→Ni(s)	-0.257
O₂(g)+2H⁺(aq)+2e^-→H₂O₂(aq)	0.695
O₂(g)+4H⁺(aq)+4e^-→2H₂O(l)	1.229
O₂(g)+2H₂O(l)+4e^-→4OH^-(aq)	0.401
O₃(g)+2H⁺(aq)+2e^-→O₂(g)+H₂O(l)	2.076
Pb²⁺(aq)+2e^-→Pb(s)	-0.126
PbO₂(s)+4H⁺(aq)+SO₄^2-(aq)+2e^-→PbSO₄(s)+2H₂O(l)	1.691
PbSO₄(s)+H⁺(aq)+2e^-→Pb(s)+HSO₄^-(aq)	-0.359
HSO₄^-(aq)+3H⁺(aq)+2e^-→H₂SO₃(aq)+H₂O(l)	0.172
SO₄^2-(aq)+4H⁺(aq)+2e^-→SO₂(g)+2H₂O(l)	0.172
Sc³⁺(aq)+3e^-→Sc(s)	-2.077
Sn²⁺(aq)+2e^-→Sn(s)	-0.138
Sn⁴⁺(aq)+2e^-→Sn²⁺(aq)	0.151
Sr²⁺(aq)+2e^-→Sr(s)	-2.899
VO₂⁺(aq)+2H⁺(aq)+e^-→VO²⁺(aq)+H₂O(l)	0.991
Zn²⁺(aq)+2e^-→Zn(s)	-0.762

A certain half-reaction has a standard reduction potential "red = +0.84 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least 0.50 V of electrical power. The cell will operate under standard conditions. FO