Question

For the reaction: NO2 -1 (aq) + Cr2O7 -2 (aq) ® Cr+3(aq) + NO3 -1 (aq) at pH = 11.00

a. If pH = 11.00 and all other species are present in 0.47 M concentrations at 25 ?C, what is the cell potential?

Answer 1

Balanced ewquation is

3NO2^- (aq) + Cr2O7^2- (aq) + 8H+ (aq) ----> 2Cr3+ (aq) + 3NO3^- (aq) + 4H2O (l)

Q cell = [Cr3+] ^2 [ NO3-]^3 / [ NO2-]^3 [ Cr2O7^2-] [H+]^ 8 , given pH = 11 , hence [H+] = 10^ -11 M

now Q cell = ( 0.47^2) ( 0.47^3) / ( 0.47^3) ( 10^ -11)^ 8

= 2.2 x 10^ 87

Eo cell = Eo ( reduction half Cr2O7/Cr3+) - Eo ( oxidatio half NO3^- NO2^-)

= 1.33 - 0.83 = 0.5 V ( values of Eo taken from table )

now Ecell = Eocell - ( 0.059 /n) log Q

where n = number of electrons transferred per reaction = 6 ( since Cr2O7^2- has Cr in +6 state and 2Cr6+ accepts 6e- to become 2Cr3+)

now Ecell = 0.5 - ( 0.059/6) log ( 2.2 x 10^ 87)

= 0.36 V