Question

The maximum allowed nitrite (NO2-) in drinking water is listed as 1 mg/L; for nitrate (NO3-) it is 10 mg/L. The standard half-cell potential for the nitrate <-> nitrite reaction is Eo = 0.881 V. What is the pE of a water body whose pH is 5.4 that is just at the limits of allowed nitrite and nitrate?

Answer 1

for Nitrite (NO2₋) =1mg/L
for Nitrate (NO3₋)=10mg/L
The Nitarire ion reaction --> NO2₋+H2O⟶2H⁺+2e⁻+NO3₋
Here, the Half cell potencial Eo=0.881V
pE=pE^o_{$-log [Reduced]/[Oxidized]$}and pE naught = 14.9

$pE= pE naught - pH -log[NO2-\wedge 1/2/NO3-\wedge 1/2]$

$pE= 14.9 - 5.4 -log[1\wedge 1/2/10-\wedge 1/2]$

$pE= 10$