A copper penny dropped into a solution of nitric acid produces a mixture of nitr

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^-_3(aq) implies 2NO(g) + 3Cu^2+(aq) + 4H_2O(l) Using the provided standard potentials, calculate E degree_cell for this reaction. Cu^2+(aq) + 2e^- implies Cu(s) E degree = 0.3419 V 4H^+(aq) + NO^-_3(aq) + 3e^- implies NO(g) + 2H_2O(l) E degree = 0.96 V What is the value of E degree_rxn at 264.00 K when [H^+] = 0.100 M, [NO^-_3] = 0.02500 M, [Cu^2+] = 0.02500 M, and the partial pressure of NO is 0.001000 atm?

Explanation / Answer

Cooper ios losing two electrons, so it's oxidizing, while NO3 is being reduced. (gaining electrons). In general,when the reaction is balanced, we have 6 electrons transfered so E° is:
E° = E° red - E°ox
E° = 0.94 - 0.3419 = 0.5981 V

With this we can calculate the E cell:
E = E° - 0.059/n logQ

let's calculate concentration of NO:
C = p/RT
C = 0.001 / 0.0821*298 = 4.09x10-5 M

Now let's calculate Q:
Q = (0.025)3 * 4.09x10-5 / 0.0252 * 0.18
Q = 102.25

E = 0.5981 - 0.059/6 log102.25
E = 0.5783 V

Hope this helps

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