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!
Redox reactions
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- ChenBeier
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Re: Redox reactions
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
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
Re: Redox reactions
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).
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
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Re: Redox reactions
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 .
I-/I2 is positiv. 0,535 V
NO2-/ NO3- is also positiv 0,94 V
Elements are put in by definition as 1 .
Re: Redox reactions
Ok, but why did you write:
and no E2 = 0,535 - 0,059/2 *log( 1/(0,1)^2) ?E2 = 0,535 + 0,059/2 *log( 1/(0,1)^2)
- ChenBeier
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Re: Redox reactions
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.
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.