Calculate the equilibrium constant for each of the reactions at 25 °C Standard Electrode Potentials at 25 °C Reduction Half - Reaction E°(V) Pb2+ (aq) + 2e -- > Pb(s) -0.13 Mg2+ (aq) + 2e--> Mg(s) -2.37 Br2(1) + 2e--> 2Br-(aq) 1.09 Cl2(g) + 2e-- > 2Cl-(aq ) 1.36 MnO2(s) + 4H+(aq) + 2e--> Mn2+(aq) + 2H2 O(l) 1.21 Cu2+ (aq) + 2e--> Cu(s) 0.16 Part B Br2(l )+2Cl-(aq) -> 2Br-(aq) + Cl2(g)
Calculate the equilibrium constant for each of the reactions at 25 °C Standard Electrode Potentials at 25 °C Reduction Half - Reaction E°(V) Pb2+ (aq) + 2e -- > Pb(s) -0.13 Mg2+ (aq) + 2e--> Mg(s) -2.37 Br2(1) + 2e--> 2Br-(aq) 1.09 Cl2(g) + 2e-- > 2Cl-(aq ) 1.36 MnO2(s) + 4H+(aq) + 2e--> Mn2+(aq) + 2H2 O(l) 1.21 Cu2+ (aq) + 2e--> Cu(s) 0.16 Part B Br2(l )+2Cl-(aq) -> 2Br-(aq) + Cl2(g)
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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
Transcribed Image Text:Calculate the equilibrium constant for each of the
reactions at 25 °C Standard Electrode Potentials at
25 C Reduction Half - Reaction E°(V) Pb2+ (aq) + 2e
--> Pb(s) -0.13 Mg2+ (aq) + 2e--> Mg(s) -2.37
Br2(1) 2e 2Br-(aq) 1.09 Cl2(g) + 2e--> 2Cl-(aq
>
) 1.36 MnO2(s) + 4H+(aq) + 2e- - > Mn2+ (aq) + 2H2
O(l) 1.21 Cu2+ (aq) + 2e-- > Cu(s) 0.16 Part B Br2(l
)+2Cl(aq) -> 2Br-(aq) + Cl2(g)
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