Question

PAGE
→ Calculate the equilibrium constant for the
reaction at 298K
4 Br + O2 + 4H+
2 Br2
+2H2O
Given that Ecell = 0.16V​

Answer 1

Question:

Calculate the equilibrium constant for the

following cell reaction

4Br + 02 + 4H+ → 2Br2 + 2H2O

Given that Ecell = 0.16 V.

Solution:

The given equation can be split in two half cell as follows :

Oxidation:

 4Br- → 2Br2 + 4 e-

 Reduction :

 O2 + 4H+ + 4e- → 2H2O

 So, there is total transfer of 4 electrons during the cell reaction.

We know that :

 Putting the standard value of R, T and F in given equation we get

• E° cell = 0.0591/n log Kc.......(1) 

There is transfer of 4 electrons during the reaction so n would be 4.

Putting the value of n and Eº cell in equation  (1) we get, 

 0.16 = 0.0591/4 x log Kc

 log Kc = 0.16 x 4/0.0591

 log Kc = 10.83

 Kc = antilog of 10.83

 Kc = 6.76 x 10^10