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