For the following redox reaction at 25 ºC, Eº cell = 2.24 V. Calculate the equilibrium constant, K. Mg(s) + Pb 2+ (aq) → Mg 2+ (aq) + Pb(s)
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For the following redox reaction at 25 ºC, Eº cell = 2.24 V. Calculate the equilibrium...
1 For the following redox reaction at 25 ºC, Eº cell = 0.74 V. Calculate the equilibrium constant, K. Cd(s) + Cu 2+ (aq) → Cd 2+ (aq) + Cu(s) (show work please) 2 Write a nuclear equation for the alpha decay of 238 92U. a. 238 92U → 0 +1 e + 238 91Pa b. 238 92U → 1 0n + 237 92U c. 238 92U → 4 2He + 234 90Th d. 238 92U → 0 -1 e + 238 93Np
cell = 0.74 V. Calculate the equilibrium constant, K. For the following redox reaction at 25 °C, E° Cd(s) + Cu 2+ (aq) → Cd 2+ (aq) + Cu(s)
6. Calculate the equilibrium constant (K) for the following redox reaction at 25 ?C. Ecell = -0.71 V 2 Al(s) + 3 Mg2+(aq) --> 2 Al3+(aq) + 3 Mg(s
The equilibrium constant, K, for a redox reaction is related to the standard potential, Eº, by the equation In K = nFE° RT where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e), R (the gas constant) is equal to 8.314 J/(mol · K), and T is the Kelvin temperature. Standard reduction potentials Reduction half-reaction E° (V) Ag+ (aq) + e +Ag(s) 0.80 Cu²+ (aq) + 2e + Cu(s) 0.34...
Consider the following information at 25 ºC: AgBr(s) + e- → Ag(s) + Br-(aq) Eº(AgBr(s),Ag(s)) = 0.07133 V Ag+(aq) + e- → Ag(s) Eº(Ag+(aq),Ag(s)) = 0.7996 V What is the value of the solubility product equilibrium constant Ksp(AgBr) of silver bromide? AgBr(s) ↔ Ag+(aq) + Br-(aq) Ksp(AgBr)
30) Use the tabulated half-cell potentials below to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. Pb2+(aq) + Cu(s) → Pb(s) + Cu2+(aq) Pb2+(aq) + 2e → Pb(s) Cu2+ (aq) +2e → Cu(s) E° = -0.13 V E = 0.34 V C) 7.9 x 1015 A) 7.9 x 10-8 D) 1.3 x 10-16 B) 8.9 x 107 E) 1.1 x 10-8
25) Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. 2 Al(s)+3 Mg2+(aq) A) 1.1 x 1072 B) 8.9 x 10-73 C) 1.1 x 10-72 D) 1.0 x 1024 E) 4.6 x 1031 2 Al3+(aq) +3 Mg(s)
Cell Potential and Equilibrium Standard reduction potentials The equilibrium constant, K, for a redox reaction is related to the standard cell potential, Ecel, by the equation Reduction half-reaction (V) Ag+ (aq) + e-→Ag(s) Cu2+ (aq) + 2e-→Cu(s) 0.34 Sn (a) 4e-Sn(s 0.15 2H' (aq) + 2e-→H2 (g) Ni2+ (aq) + 2e-→Ni(s)-0.26 Fe2+ (aq) + 2e-→Fe(s)-0.45 Zn2+ (aq) + 2e-→Zn(s)-0.76 Al3+ (aq) +3e-→Al(s) -1.66 Mg2+ (aq) + 2e-→Mg(s) -2.37 0.80 n FEcell where n is the number of moles of electrons...
Calculate the standard potential, Eº, for this reaction from its equilibrium constant at 298 K. X(s) + y2+(aq) = x2+(aq) + Y(s) K = 9.95 103 E = V
Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. 3 I2(s) + 2 Fe(s) → 2 Fe3+(aq) + 6 I(aq) A.8.9 × 10-18 B.1.1 × 1017 C.1.7 × 1029 D.2.4 × 1058 E.3.5 × 10-59