K = 10^8.78 = 6.0 x 10^8

0.0592 / 2 log K = 0.26

0.0296 log K = 0.26

log K = 8.78

E = E^o (0.0592 / n) (log Q)

At equilibrium E = 0

K = 10^8.78 = 6.0 x 10^8

0.0592 / 2 log K = 0.26

0.0296 log K = 0.26

log K = 8.78

K = 10^8.78 = 6.0 x 10^8

Sn 2 2e- Sn(s) Eo = -0.14

Cd 2 2e- == Cd(s) Eo = -0.40

Sn 2 2e- == Sn -0.14

Cd === Cd 2 2e- 0.40

Cd(s) Sn 2 == Sn(s) Cd 2 E^o = 0.26

n = 2 since two electrons are transferred.

0.0592 / 2 log K = 0.26

0.0296 log K = 0.26

log K = 8.78

Sn 2 2e- == Sn -0.14

Cd === Cd 2 2e- 0.40

Cd(s) Sn 2 == Sn(s) Cd 2 E^o = 0.26

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