Redox reactions

[Neon]

Please help:

The standard electrode potential for the reduction of nitrate(V) ions to nitrate(III) ions is 0.940 V. The standard electrode potential for the reduction of iodine to iodide ions is 0.535 V. If we have a solution that has the same concentration of nitrate(V) and nitrate(III) ions and a concentration of iodide ions of 0.1 M, what is the maximum pH to oxidize iodide ions if the pKa of HNO₂ is 3.15.

Thanks in advance!

[ChenBeier]

We have two equations

HNO₂ + H₂O => HNO₃ + 2 H+ + 2e-

2 I- => I₂ + 2 e-

Nernst equation

E₁ = E0 + 0,059/2*log ( cHNO₃ * c^2 H+ /c HNO₂)

E₂ = E0 + 0,059/2 *log (cI₂/c^2 I-)

First equation

cHNO₃ = cHNO₂

E₁ = 0,94 +0,059/2 * log(c^2H+)

pH = -log cH+

E₁ = 0,94 - 0,059*pH

Second equation

E₂ = 0,535 + 0,059/2 *log( 1/(0,1)^2)

E₂ = 0,594 V

E₁ has to be E₂

E₁ = E₂

0,594 V = 0,94V - 0,059 *pH

- 0,346 = -0,059 *pH

pH = 5,86

[Neon]

Thanks, but I have some questions:

1. Why did you use plus and not minus in the second Nernst equation (for iodine).
2. Why did you assume that the activity of elemental iodine is 1? Iodine could also dissolve a little. When I was solving this, I thought as if I wanted to calculate the voltage during a redox titration before the addition of the titrant, and thus assumed the spontaneous dissociation of the iodide ion into iodine (e.g. 10^-6 M).

[ChenBeier]

Redox Potentials are measured to hydrogen and considered as oxidation.
I-/I₂ is positiv. 0,535 V
NO₂-/ NO₃- is also positiv 0,94 V

Elements are put in by definition as 1 .

[Neon]

Ok, but why did you write:

E₂ = 0,535 + 0,059/2 *log( 1/(0,1)^2)

and no E₂ = 0,535 - 0,059/2 *log( 1/(0,1)^2) ?

[ChenBeier]

Because in the logarithmen I used cox/cRed. Not the opposit.

Log cox/ cred = - log cred/cox

log/cI₂/c^2I- = - logc^2I-/ cI₂

https://en.m.wikipedia.org/wiki/Nernst_equation

You are right it should be a minus like in the Literature but also the reduced form as numerator and the oxidised form as denominator in both equations.

But mathematically the result is the same.