A certain half-reaction has a standard reduction potential +0.14 V. An engineer proposes using this half-reactio...
A certain half-reaction has a standard reduction potential E = -0.45 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 1.00 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. 0-0 . " Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell...
A certain half-reaction has a standard reduction potential E Ted =-0.99 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least 0.90 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 O- Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?...
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
ol. Data Х -2.372 -1.185 1.224 1.507 0.595 0.983 -1.16 0.957 -2.71 -0.257 Mg2+ (aq) + 2e → Mg (s) Mn2+ (aq) + 2e - Mn (s) MnO2 (s) + 4H+ (aq) + 2e – Mn2+ (aq) + 2H20 (1) Mn04- (aq) + 8H+ (aq) + 5e – Mn2+ (aq) + 4H20 (1) Mn04- (aq) + 2H20 (1) + 3e - MnO2 (s) + 40H- (aq) HNO2 (aq) + H+ (aq) + - NO (9) + H20 (1) N2 (9)...
olo Data Х -0.8277 1.776 0.5355 1.195 -2.372 -1.185 1.224 1.507 0.595 0.983 -1.16 2H20 (1) + 2e - H2(g) + 20H(aq) H2O2 (aq) + 2H+ (aq) + 2e - 2H20 (1) 12 (s) + 2e - 21- (aq) 2103 (aq) + 12H+ (aq) + 10 → 12 (5) + 6H20 (1) Mg2+ (aq) + 2e - Mg(s) Mn2+ (aq) + 2e – Mn (s) MnO2 (s) + 4H+ (aq) + 2e – Mn2+ (aq) + 2H20 (1) Mn04- (aq)...
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction. Round your answer to 3 significant digits Mn (aq)+2H,O()+2Fe (aq)MnO, ()+4H (aq)+2Fe (aq) dhData Cu (aq) + e Cu (5) F2 (0)+2e2F (aq) Fe (aq) +2e Fe (s) Fe (aq) + eFe2 (aq) Fe (aq) + 3e Fe (s) 2.866 X ? -0.447 0.771 -e.037 2H (aq)+2e H (0) e.000 2H)O (I)+2e H2 (a) +20H(aq) -0.8277 1.776 H2O2 (aq)...
Try Again Your answer is wrong. In addition to checking your math, check that you used the right data and DID NOT round any intermediate calculations. 0.7996 -1.676 1.692 1.49B -2.912 1.066 -2.868 Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG" for the following redox reaction. clo Data Round your answer to 5 significant digits. Ag+ (aq) + e + Ag (s) H, (8)+20H(aq) + C1, (8) + 2H,0 (1)+201 (aq) A13+...
Please show all steps taken (prefer typed solution) Half-Reaction E ° (V) Ag+ (aq) + e− → Ag (s) 0.7996 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 Br2 (l) + 2e− → 2Br− (aq) 1.066 Ca2+ (aq) + 2e− → Ca (s) −2.868 Cl2 (g) + 2e− → 2Cl− (aq) 1.35827 Co2+ (aq) + 2e−...
ALEKS data table may not include the value, if not please just include the E value and I will go through the table and leave the answer in the comments A certain half-reaction has a standard reduction potential Ered- provide at least 0.50 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 0.53 V. An engineer proposes using this half-reaction...
Fill in the Blanks 0.34 Half Reaction E. (V) Half Reaction Erd® (V) F + 2e →2F 0+ 2H + 2e →H,02 0.68 Ag* + e → Ag Cu* + e → Cu 0.52 Co? + e Co? O2 + 2H,0 + 4e → 40H 0.40 H,O, + 2H + 2e → 2H,0 Cu2+ + 2e → Cu Cet+e → Ces Cu2+ + e → Cut 0.16 PbO, + 4H+ +50,2 +2e → PbSO, + 2H,0 2H* + 2e →...