Question

For the portion of the pE-pH diagram where the reduction of NO2 - to NH4 + is the relevant reaction, use the expression (NO2- + 8H+ + 6e- → NH4+ + 2H20 and pE=15.14-log([NH44]⅙ /[NO2- ]⅙ [H+]4/3])) to find the equation for the line that will separate the two species. Remember that the line will be where the species are equal in concentration. (Hint: your equation should be in the form y = mx + b where y = pE and x = pH)

Answer 1

The given expression can be written as follows.

pE = 15.14 - Log([NH₄⁺]^1/6/[NO₂^-]^1/6) - 4/3 * (-Log[H⁺])

i.e. pE = 15.14 - 1/6 * Log([NH₄⁺]/[NO₂^-]) - 4/3 pH

Where [NO₂^-] = [NH₄⁺], i.e. [NH₄⁺]/[NO₂^-] = 1, i.e. Log([NH₄⁺]/[NO₂^-]) = 0

Therefore, pE = -4/3 * pH + 15.14

The above equation is in the form of y = mx + c, where y = pE, x = pH

The slope for the line of pE versus pH, i.e. m = -4/3

And the y-intercept, i.e. c = 15.14