Main Menu Contents Grades Reservations Syllabus Course Conte

Main Menu Contents Grades Reservations Syllabus Course Contents.Exam III - Online retake, due 9am on Dec. 13 Timer Not Calculate the equilibrium constant, K, for a galvanic cell based on the following half reactions at 285 K: Mn® + 2 e Mn -1.18V Ag+ + e- Ag 0.808 V For the retake, enter your answer using the following format: 1.23e +45 L Submit Answer Tries 0/3 Post Discussion

Solution

from data table:

Eo(Mn2+/Mn(s)) = -1.18 V

Eo(Ag+/Ag(s)) = 0.808 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Ag+/Ag(s))

anode is (Mn2+/Mn(s))

The chemical reaction taking place is

2 Ag+(aq) + Mn(s) --> 2 Ag(s) + Mn2+(aq)

Eocell = Eocathode - Eoanode

= (0.808) - (-1.18)

= 1.988 V

here, number of electrons being transferred, n = 2

Eo = (2.303*R*T)/(n*F) log Kc

At 25 oC or 298 K, R*T/F = 0.0592

So, Eo = (0.0592/n)*log Kc

1.988 = (0.0592/2)*log Kc

log Kc = 67.1622

Kc = 1.45*10^67

Answer: 1.45e67