wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Determine the standard equilibrium constant of the following reaction at 298 K.
2MnO4+6H++5H2C2O42Mn2++8H2O+10CO2 (Ecell=2.0 V)

Open in App
Solution

Anode : 5H2C2O410CO2+10e
Cathode : 2MnO4+10eMn2+
n=2e, ΔG=RTlnKf=nFE0
logKf=20.059×E0
logKf=67.8
Kf=1067.8

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Watch What You Eat! - Food Poisoning and Preservation
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon