This is all the information that is provided in the question. The oxidation of H

ID: 870456 • Letter: T

Question

This is all the information that is provided in the question.

The oxidation of HNO2 to NO3- in acid solution is a 4-electron transfer described by the following equation:

2HNO2(aq) + O2(g) = 2NO3-(aq) + 2H+(aq) Eocell = 0.289 V

If the reaction at 298 K is in equilibrium with air (PO2 = 0.10 atm) at pH 6.10, what is the ratio of [NO3-] to [HNO2]?

This is other information (not provided by the question) What other information do I need? And What is the answer?

NO3-(aq) + H2O(l) + 2e- = NO2-(aq) + 2OH-(aq) n=2 Eo(V) = 0.01

HNO2(l) + H+(aq) + e- = NO(g) + H2O(l) n=1 Eo(V) = 1.00

Hint: log(K) = nEocell / 0.0592V After solving for K, use the equilibrium constant expression to solve for the ratio [HNO2] to [NO3-]

pH = -log[H+]

Thus, [H+] = 10-pH = 10-6.1 = 7.943*10-7 M

2HNO2(aq) + O2(g) <---------> 2NO3-(aq) + 2H+(aq) ; E0cell = 0.289 V

Now, as per Nernst Equation:-

E0cell = (0.059/n)*logKeq ; where n = number of electron transfer taking place in the overall balanced reaction = 4

Keq = {[NO3-]2*[H+]2}/{[O2]*[HNO2]2}

Thus, 0.289 = (0.059/4)*logKeq

or, logKeq = 19.593

or, Keq = 1019.593 = 3.92*1019

Thus, 3.92*1019 = {[NO3-]2*[H+]2}/{[O2]*[HNO2]2}

or, 3.92*1019 = {[NO3-]/[HNO2]}2*{(7.943*10-7)2/0.1} =

or, 3.92*1019 = {[NO3-]/[HNO2]}2*(6.31*10-12)

or, 6.213*1030 = {[NO3-]/[HNO2]}2

or, {[NO3-]/[HNO2]}= 2.49*1015

or, [HNO2]/[NO3-] = 4.012*10-16

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