Question

Question 4 (1 point) Given the following reaction, what is the AG for the spontaneous reaction in KJ/mol? Faraday's constant 96500 Coulomb/mol (96.5 kJ/V mol) Round your answer to the nearest whole number (i.e. if you calculated 52.02, input your answer as 52). IF YOUR ANSWER IS A NEGATIVE VALUE MAKE SURE TO INCLUDE THE NEGATIVE SIGN. IF IT IS POSITIVE YOU CAN LEAVE IT AS IS. O2+ 2Ht+2e NO3+2H*+ 2e NO2+ E'o 0.295 V H2O2 H20 E'o 0.421 V

Answer 1

O2 + 2 H+ + 2e- --------------- H2O2         E0 = 0.295V

NO3- + 2 H+ + 2e- ------------ NO2- + H2O            E0 = 0.421V

according to IUPAC convention , More SRP electrode acts as cathode and less SRP electode is anode

Hence, NO3- ion solution is a cathode

oxidation reaction at anode                    H2O2 ---------------------- O2 + 2H+ + 2e-

reduction reaction at cathode                 NO3- + 2 H+ + 2e- -------------- NO2 + H2O

                                           -------------------------------------------------------------------------------------

                                                         H2O2 + NO3-   ----------------- O2 + NO2- + H2O

                                            -------------------------------------------------------------------------------------

E0 cell = E0 cathode - E0 anode

E0cell = 0.421 - 0.295

E0cell = 0.126V

number of electrons transferred = n= 2

F= 96500 coloumbs

Delat G0 = - nFE0cell

Delat G0 = - 2 x 96500 x 0.126

Delat G0 = - 24318 J

Delat G0 = 24.318 KJ

Delat G0 = - 24 KJ