Question

For the half cell reaction
NO2 +2OH ----NO3 +H2O +2e-
E°cell is +0.425.if the overall potential is 0.6250.what is the pH of the solution at:(1) 25°C (2)47°C.if the concentrations of the nitrite ion and nitrate ions are 0.10 and 1.97×10^-10 mol/dm^3 respectively.

Answer 1

Solution.

a) T = 25 °C

$$E=Eo + \frac{R \times T}{n \times F} \times ln(\frac{[Ox]}{[Red]})$$

$$E=Eo + \frac{R \times T}{n \times F} \times ln(\frac{[NO3^{-}]}{[NO2^{-}][OH^{-}]^2})$$

F = 96500

n = 2

T = 25 + 273 = 298

R = 8.31

$$[OH^{-}] = 0.001830$$

$$pOH = -lg[OH^{-}]$$

$$pH = 14-pOH$$

pOH = 2.74

pH = 11.26

b) T = 47 °C

$$E=Eo + \frac{R \times T}{n \times F} \times ln(\frac{[Ox]}{[Red]})$$

$$E=Eo + \frac{R \times T}{n \times F} \times ln(\frac{[NO3^{-}]}{[NO2^{-}][OH^{-}]^2})$$

F = 96500

n = 2

T = 47 + 273 = 320

R = 8.31

$$[OH^{-}] = 0,003128$$

$$pOH = -lg[OH^{-}]$$

$$pH = 14-pOH$$

pOH = 2.50

pH = 11.50

Answer:

a) pH = 11.26

b) pH = 11.50