Using, E°cell = 0.0592/n logK
Where, n = 2 (because two electrons are lost or gained)
K = 9.95×10^3
E°cell = 0.0592/2 ×log(9.95×10^3)
= 0.0592×3.99/2
= 0.118 V
Calculate the standard potential, Eº, for this reaction from its equilibrium constant at 298 K. X(s)...
Calculate the standard potential, E®, for this reaction from its equilibrium constant at 298 K. X(s) +Y4+ (aq) = X4+ (aq) + Y(s) K = 7.92 x 10-8 E =
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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...
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The equilibrium constant, K. for a redox reaction is related to the standard potential, E, by the equation Fe(s) + Ni+ (aq) +Fe?+ (aq) + NI(s) FE In K = Express your answer numerically. View Available Hints) where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mole). R (the gas constant) is equal to 8.314 J/(mol-K). and T is the Kelvin temperature. ΟΙ ΑΣΦ h ? KK- Submit Previous Answers *...
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6. At 298 K, the equilibrium constant for the below reaction is 4.17 x 103. What is the concentration of Clat equilibrium? Pb2+ (aq) + 2 C1" (aq) = PbCl2 (5) (a) 20.2 M (b) 0.078 M (c) 10.1 M (d) 0.039 M
> No idea why someone gave this a thumbs down, it worked for me!
crystal hammond Fri, Dec 3, 2021 5:49 PM