The standard reduction potential for the reaction: 2 D^+(aq) + 2 e^? rightarrow D_2(g) where D is deuterium, is ?0.0034V at 25 degree C. Calculate E degree and K at 25 degree C for the reaction: 2 H^+ (aq) + D_2(g) rightarrow 2 D^+(aq) + H_2(g)
Expert Answer
Reaction given:
2H+ + D2 —> 2D+ + H2
Here H+ is getting reduced and D2 is getting oxidized.
E0 = Eoxid + Ered = 0.0034 + 0 = 0.0034 V
Using relation:
E0 = (0.0591/n)*log K
Putting values we get:
0.0034 = (0.0591/2)*log K
Solving we get:
K = 100.115 = 1.303