What is produced at each electrode in the electrolysis of an aqueous solution of both nabr and agf?

Explanation:

reduction potential of Ag+/Ag = 0.80 V

the oxidation potential of Br-/Br2 = -1.066 V

reduction potential of Na+/Na = 2.71 V

reduction potential of 2H+/H2 = 0.00 V

oxidation potential of 2F-/F2 = -2.87 V

oxidation potential of 2H2O↔  O2 + 4H+  + 4e- = -1.23 V

So,  at cathode reduction occurs so, here Ag(s) will form because it has high reduction potential.

However, at anode oxidation occurs and , here the substance with high oxidation potential can oxidises easily so bromine will form.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j