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I want to figure out a way how to determine the products of a redox-reaction by looking at the documented Standard electrode potential (for example here). But with my "way" I have some trouble/contradictions with common redox reaction, in my case the reduction of $\ce{MnO4-}.

Here is my procedure for the reduction of $\ce{MnO4-}$:

  1. Collect all available reactions with their Standard Potential:* \begin{align} \ce{MnO4- + 2H2O + 4e- &-> MnO2 + 4OH-}& E(\text{Red})&= \pu{0.83 V}\tag{a}\label{a}\\ \ce{MnO4- + H+ + e- &-> HMnO4^-} & E(\text{Red}) &= \pu{0.72 V}\tag{b}\label{b}\\ \ce{MnO4- + 8H+ + 5e- &-> Mn^2+}& E(\text{Red}) &= \pu{1.23 V}\tag{c}\label{c}\\ \ce{MnO4- + 4H+ + 3e- &-> MnO2 + 2H2O}& E(\text{Red}) &=\pu{1.6 V}\tag{d}\label{d} \end{align}

  2. Then figure out the reaction with the highest reduction potential, because according to $E = E(\text{Red}) - E(\text{Ox})$, $E$ is maximal with the biggest $E(\text{Red})$ and the reaction is going to happen with the highest $E$.

    In my example, $\eqref{d}$ has the highest potential, so according to my theory $\ce{MnO2}$ is formed in the reduction of $\ce{MnO4-}$ (in combination with for example the oxidation of $\ce{I-}$. But in all textbooks, internet sites, the reduction $\eqref{c}$ is named - not reaction $\eqref{d}$ which I had determined.

I am aware that these potentials are pH- and concentration dependent, I fit the $E(\text{Red})$-data with the Nernst-Equation to $\mathrm{pH} = 3$. Nevertheless, $E(\text{Red})$ of $\eqref{d}$ is always higher in $E(\text{Red})$ than $\eqref{c}$. According to my calculations (graph of $\eqref{d}$) which is dependent on the $\mathrm{pH}$ is always higher than $\eqref{c}$).

I have drawn a diagram which shows the $E(\text{Red})$ potential of the two reaction $\eqref{c}$ and $\eqref{d}$ (see legend). Reaction $\eqref{d}$ has always the higher standard potential - is the better oxidation reaction. Why does nevertheless reaction $\eqref{d}$ happen in acid solutions?

Functions calculated with Nernst-Equation: \begin{align} E(\text{pH}) &= 1.507 - \frac{0.059}{5} \cdot 8 \text{pH}\tag{c'}\\ E(\text{pH}) &= 1.7 - \frac{0.059}{3} \cdot 4 \text{pH}\tag{d'} \end{align} (other concentrations are assumed as equal)

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  • $\begingroup$ Please check some unbalance equations and corresponding $\mathrm{E^o}$ (red) values in equations (a)-(d). $\endgroup$ – Mathew Mahindaratne Feb 14 at 1:40
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What you have listed are standard redox potentials. You need two of them to determine a reaction. As I'm not sure what reaction you intend, I'll give an example below, which you can use to work out your own reaction. The diagram helps to sort out what can happen.

redox diagram

Note that a reducing agent is an electron donor, and an oxidising agent an electron acceptor and $$\ce{oxidised form + electrons <=> reduced form}.$$

For example:
The oxidiser $\ce{Sn^{2+}}$ reacts completely with reducer $\ce{Cr^{2+}}$ (at standard redox potentials). \begin{align} E^\circ &= \pu{-0.136 V}:& \ce{Sn^2+ + 2e- &<=> Sn}\\ E^\circ &= \pu{-0.41 V}:& \ce{Cr^3+ + e^- &<=> Cr^2+}\\ && \ce{2Cr^2+ + Sn^2+ &<=>> 2Cr^3+ + Sn v} \end{align} With energy $$E= (\pu{-0.136 V}) - (\pu{-0.41 V}) = \pu{0.274 V}$$ which corresponds to a negative free energy and hence is spontaneous: \begin{align} \Delta G &= -nFE= (-2\times 96485\times 0.274) \pu{J};& (\pu{1J} &= \pu{1C*V}). \end{align}

Although the reverse reaction, $\ce{Cr^{3+}}$ to oxidise $\ce{Sn}$ metal could occur, this happens to a negligible extent.

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  • $\begingroup$ First thank you for your answer, I understand your point, but if I look at following reaction: 2I<sup>-</sup> -> I<sub>2</sub> + 2e<sup>-</sup> E(Red) = -0.536 V with c) $\Delta E =$ E(Red) - E(Ox) = 1.23 V + 0.536 V = 1,766 V with d) $\Delta E =$ E(Red) - E(Ox) = 1.6V + 0.536 V = 2,136 V => reaction d) (in redox with the I<sup>-</sup> oxidation) will happen according to the "theory" - but in reality, there is no brown, dirty solution - clear, weak pink (indicates c)!) $\endgroup$ – Nilsfrank99 Sep 26 '16 at 17:48
  • $\begingroup$ OK, bad example. Its brown from the I<sub>2</sub>. But all reactions in the internet and books claim the formation of Mn<sup>2+</sup>... $\endgroup$ – Nilsfrank99 Sep 26 '16 at 18:19

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