Predict the cell potential for the following rxn when the P O2(g)= 2.50 atm, [H+]= 0.1M, [Br] 0.26M?

Question

O2(g)+4H+(aq)+4Br(aq)-->2H2O(l)+2Br2(l)

E^0 of cell=0.152V

Answer 1

nernst:

E = Eo - (0.0592/n)log Q

E = 0.152  - (0.0592/4 mol e-) log prod / reactant

E = 0.152  - (0.0148) log 1 / [O2] [H+]^4 [Br-]^4

E = 0.152  - (0.0148) log 1 / [2.5] [0.1]^4 [0.26]^4

E = 0.152  - (0.0148) log 1 / 1.14e-6

E = 0.152  - (0.0148) log 875319

E = 0.152  - (0.0148) 5.942

E = 0.152  - 0.088

your answer is: E = 0.064 volts