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I am currently stuyding electrochemistry, and recently I've stumbled upon a problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate, but I obtained an unrealistic result and I became confused.

What I first did was write the half-reactions involved within the voltaic cell:

Anode (-): 5Fe2+(aq) ⇌ 5Fe3+(aq) + 5e- ε01= -0.77 V
Cathode (+): MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l) ε02= 1.5 V

Notice the fact that I obtained the first reaction, the oxidation, by reversing the equation for the reduction of Fe3+. By combining both half-cell reactions, one can obtain the equation for the whole process:

5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ⇌ 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) E0= 0.73 V

A positive standard potential will yield a negative value for the Gibbs free energy, so the reaction is product-favoured. The problem is concerned with calculating the concentration at equillibrium, corresponding to the equivalence point of our titration, so the overall Gibbs free energy will be 0, just as the non-standard potential, which we can calculate using the Nernst equation (considering 298.15 K of temperature and 1 atm of pressure). I'm also assuming that the reaction quotient is equivalent to the equilibrium constant, since we are at the equivalence point.

\begin{align*} E=E^{0}-\frac{0.0592}{5}\log{K} \end{align*} \begin{align*} E=0, E^{0}=\frac{0.0592}{5}\log{K} \end{align*} I then obtained the mathematical expression for the equilibrium constant: \begin{align*} K=10^{\frac{5E^{0}}{0.0592}} \end{align*} I then entered the values, and calculated K (substantially larger than 1): \begin{align*} K=4.523*10^{61} \end{align*} However, this is the hurdle I have yet to overcome. I didn't quite understand what activities are in the Nernst equation (afaik they are not equilibrium concentrations), so I assigned a formula for K based on the equilibrium reaction:

\begin{align*} K=\frac{[Fe^{3+}]^{5}[Mn^{2+}]}{[Fe^{2+}]^{5}[MnO4^{-}][H^{+}]^{8}} \end{align*}

The problem mentions the following concentrations for the titration:
Fe2+:0.1 N (I assumed it was 0.1/5=0.02 M)
H2SO4: 1N (probably 0.1M for H+)
KMnO4: 1N (0.2M MnO4- following the same steps)

In my conception, the activities of the reagents in the equilibrium constant were their initial concentrations, and the activities of the products were their equillibrium concentrations. I then made an ice table, considering a concentration of x. By replacing those values into the equilibrium constant, I obtained: \begin{align*} K=\frac{5x^{5}*x}{0.02^{5}*0.2*0.1^{8}} \end{align*} X was supposed to be the concentration of Fe^{2+}, however the answer I got is definitely not the right one. \begin{align*} x=6726501.235 \end{align*}

Where did I go wrong?

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  • $\begingroup$ When you computed $\frac{RT}{nF}$, did you copy over the correct number of zeros? $\endgroup$
    – Zhe
    Commented Apr 9, 2020 at 17:12
  • $\begingroup$ What do you mean? I always used 0.0592. As for the other concentrations, I copied those correctly as well. $\endgroup$
    – TheRelentlessNucleophile
    Commented Apr 9, 2020 at 17:54
  • $\begingroup$ $\frac{(8.3141\ \mathrm{J}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}) 298\ \mathrm{K}}{96485\ \mathrm{C}\ \mathrm{mol}^{-1}} \neq 0.0592\ \mathrm{J}\ \mathrm{mol}\ \mathrm{C}^{-1}$ $\endgroup$
    – Zhe
    Commented Apr 10, 2020 at 12:38
  • $\begingroup$ We actually are supposed to do more than that. The original Nernst equation uses the natural logarithm of the constant, as opposed to the generalized formula which uses the log function. It's just for the purpose of simplifying things. What you wrote would be roughly equal to 0.0257, which, when divided by loge, yields 0.05919 which can be safely approximated to 0.0592. More precise calculations would lead to about 0.059156973, but that is really "simplifiable" to 0.0592. You can also use 0.059 when doing rough calculations. $\endgroup$
    – TheRelentlessNucleophile
    Commented Apr 10, 2020 at 19:01

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problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate,

There are several things which may be problematic in your approach.

  1. For the stated problem you do not need to invoke Nernst equation. Simply use the balanced equation.

  2. When potentiometric titration curve are to be estimated, it is far more convenient would use the half-cell instead of a full reaction. I am assuming the redox electrode is dipping in iron solution.

\begin{align*} E=E^{0} -\frac{0.0592}{5}\log{[Fe^{2+}]/[Fe^{3+}]} \end{align*}

where $E^{0}$ is the iron half cell value. You should not calculate equilibrium constants from potentiometric titrations.

See page 17 for a full example Potentometric Titration clculation

  1. I do not know when people will stop teaching normality. It is obsolete (18th century concept). Your normality conversions are incorrect.

The problem mentions the following concentrations for the titration:
Fe2+:0.1 N (I assumed it was 0.1/5=0.02 M)
H2SO4: 1N (probably 0.1M for H+)
KMnO4: 1N (0.2M MnO4- following the same steps)

1 N sulfuric acid will be half the molarity =0.5 M

1 N permanganate will be 1/5th the molarity = 1/5 M

Check the concept of normality in your textbook.

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  • $\begingroup$ Apologies for misusing normal concentrations. However, how is applying the equation you've mentioned in your second point different from basically applying the Nernst equation for the anodic half cell? And why is it [Fe2+]/[Fe3+] and not the opposite, [Fe3+]/[Fe2+], since Fe3+ is the product obtained via the oxidation of Fe2+. Also, should i replace the values in the square brackets with the activities, or with the equilibrium concentrations? $\endgroup$
    – TheRelentlessNucleophile
    Commented Apr 9, 2020 at 15:10
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    $\begingroup$ No problem, normality is a part of Indian syllabus. When you use negative sign in Nernst equation before the log term, the reduced form comes in the numerator. I remember it by a word NERO = Negative sign reduced over oxidized. If you use a positive sign, PORE = Positive is oxidized over reduced. $\endgroup$
    – ACR
    Commented Apr 9, 2020 at 15:17
  • $\begingroup$ What I didn't understand, though, and the textbook failed to explain that clearly, is the difference between activities and molar concentrations. I never quite actually understood what are activities and how do they differ from molar concentrations, nor how are they used. When working with the Nernst equation, why are we allowed to compute the activities of the reactants and the equilibrium concentrations of the products? This is the stuff that really baffled me and I'd be highly appreciative if you could clarify it for me. $\endgroup$
    – TheRelentlessNucleophile
    Commented Apr 9, 2020 at 17:58
  • $\begingroup$ @user89030, please post this as a separate question so that it gets more attention. Also search the Chemistry stack exchange (previous question), I am sure it must have been addressed before. $\endgroup$
    – ACR
    Commented Apr 9, 2020 at 18:01

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