Question

A copper penny dropped into a solution of nitric acid produces a mixture of nitrogen oxides. The following reaction describes the formation of NO, one of the products. 3Cu(s) + 8H^+(aq) + 2NO^- (aq) rightarrow 2NO(g) + 3Cu^2+(aq) + 4H_2O(l) E degree _cell for this reaction is 0.620 V. What is the value of E degree _rxn at 208 K, when [H^+] = 0.100 M, [NO_3-] = 0.0350 M, [Cu^2+] = 0.0400 M, and the partial pressure of NO is 0.0018 atm?

Answer 1

Using Nernst equation to calculate E_cell at 208K

E_cell = E⁰_cell - RT/nF * lnQ

as this reaction involves a 6-electron change (i.e. Cu ==> Cu2+ + 2e- is a 2-electron change, so 3 of those (3Cu ==> 3Cu2+) is a 6-electron change)

therefore, E_cell = 0.62- RT/nF * lnQ

= 0.62 - 0.059/6 log ((P NO)^2 [Cu2+]^3) / ([H+]^8 [NO3-]^2))

= 0.62 - 0.0098 log ((0.0018)^2 (0.04)^3) / (0.100)^8 (0.0350)^2)) = 0.62 - 0.0098 log 19.0 = 0.61 V