calculate the equilibrium constant for each of the following reactions at 298 K. [Solved]

GO VOLS!

Use the equation: E^0 = 0.059/n x log K
a) Cu ====> Cu^2 2e- -0.337
Ag e- === Ag 0.799 not 0.0799
-0.337 0.799 = 0.462
To write the balanced equation, the silver half reaction is multiplied by 2 so that the number of electrons
gained equals the number lost. Thus, n in the total number of electrons transferred, which is 2

0.462 = 0.059/2 x log K ; log K = 15.6 K = 10^15.6 = 3.98 x 10^15
b) n = 3 E^o = 1.61 0.32
c) n 4

Fern gave a good explanation. Your plugged in values should be:
A: 4.2 * 10^15
B: 2.8 * 10^65
C: 7.6 * 10^39

0.462 = 0.059/2 x log K ; log K = 15.6 K = 10^15.6 = 3.98 x 10^15
b) n = 3 E^o = 1.61 0.32
c) n 4

Source(s): Mastering Chem Answers Youre welcome.

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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