Calculate E° cll from the tabulated standard reduction potentials for each of the following reactions in aqueous solution. Then calculate AG° and K at 25°C from E° cell' F = 9.65 x 1o J/V•mol e R = 8.314 J/mol•K Acidic solution Standard reduction potential, E (Volts) Cu2" (ag) + e → Cu'(aq) Cu'(aq) + e - Cu(s) Mno, (ag) + 8H"(aq) + 5e Mn?"(ag) + 4H,0 Fe"(aq) + e – Fe2"(aq) 0.153 0.521 1.507 0.771 Basic solution Zn(OH) (s) + 2e Zn(s) + 40H (aq) Mno, (aq) + 2H,0 + 3e MnO,(s) + 4OH(aq) -1.22 0,588 a Mn04 + 5FE2+= Mn2+ + 5FE3+ (acidic solution) E°cell = AG° = kJ/mol K =

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Chapter19: Electrochemistry
Section: Chapter Questions
Problem 19.80QP: Calculate the standard cell potential of the cell corresponding to the oxidation of oxalic acid,...
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Calculate E°
cell
from the tabulated standard reduction potentials for each of the following reactions in aqueous solution. Then calculate AG° and K at 25°C from E°
cell'
4
F = 9.65 x 10 J/V•mol e
R = 8.314 J/mol•K
Acidic solution
Standard reduction potential, E (Volts)
Cu?" (aq) + e → Cu*(aq)
Cu'(ag) + e Cu(s)
Mno, (ag) + 8H"(ag) + 5e Mn2"(ag) + 4H,0
0.153
0.521
1.507
Fe"(aq) + e Fe?"(aq)
0.771
Basic solution
-1.22
Zn(OH)-(s) + 2e Zn(s) + 4OH(aq)
Mno, (aq) + 2H,O + 3e → Mno,(s) + 40H (aq)
0,588
Mn04 + 5FE2+= Mn2+ + 5FE3+ (acidic solution)
E°cell =
AG° =
kJ/mol
K =
Transcribed Image Text:Calculate E° cell from the tabulated standard reduction potentials for each of the following reactions in aqueous solution. Then calculate AG° and K at 25°C from E° cell' 4 F = 9.65 x 10 J/V•mol e R = 8.314 J/mol•K Acidic solution Standard reduction potential, E (Volts) Cu?" (aq) + e → Cu*(aq) Cu'(ag) + e Cu(s) Mno, (ag) + 8H"(ag) + 5e Mn2"(ag) + 4H,0 0.153 0.521 1.507 Fe"(aq) + e Fe?"(aq) 0.771 Basic solution -1.22 Zn(OH)-(s) + 2e Zn(s) + 4OH(aq) Mno, (aq) + 2H,O + 3e → Mno,(s) + 40H (aq) 0,588 Mn04 + 5FE2+= Mn2+ + 5FE3+ (acidic solution) E°cell = AG° = kJ/mol K =
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