I am referring to the following three reactions. Two of these reactions are imaginary ($\ce{M}$ and $\ce{X}$ are imaginary).

$$\begin{alignat}{3} \ce{X- + 2e- \;&<=> X^3-} \qquad &&E^\circ_{\ce{X^3-}/\ce{X-}}=0.9\ \mathrm{V}\qquad &&&\text{(a)}\\ \ce{2H+ + MO^2+ + e- \;&<=> M^3+ + H2O} \qquad &&E^\circ_{\ce{MO^2+}/\ce{M^3+}}=0.5\ \mathrm{V} \qquad &&&\text{(b)}\\ \ce{8H+ + MnO4- + 5e- \;&<=> Mn^2+ + 4H2O} \qquad &&E^\circ_{\ce{MnO4-}/\ce{Mn^2+}}=1.5\ \mathrm{V} \qquad &&&\text{(c)} \end{alignat}$$

Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution. (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)

To find out possible reactions, following steps are taken.

Equations $\text{(a)}$ and $\text{(c)}$;

$$ \begin{align} &{-}\text{(a)} \times 5 + \text{(c)} \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO4- -> 2Mn^2+ + 5X- + 8H2O} \qquad E^\circ_1\qquad\text{(1)} \\ \\ &E^\circ_1 = 1.5\ \mathrm{V} + (-0.9\ \mathrm{V}) = 0.6\ \mathrm{V} \\ &\Delta G^\circ_1 = -nE^\circ_1F = -10 \times 0.6 \times F = -6F \end{align} $$

Equations $\text{(b)}$ and $\text{(c)}$;

$$ \begin{align} &{-}\text{(b)} \times 5 + \text{(c)}: \\ \ce{&5M^3+ + H2O + MnO4- -> Mn^2+ + 5MO^2+ + 2H+} \qquad E^\circ_2 \qquad(2)\\ \\ &E^\circ_2 = 1.5\ \mathrm{V} + (-0.5\ \mathrm{V}) = 1.0\ \mathrm{V} \\ &\Delta G^\circ_2 = -nE^\circ_2F = -5 \times 1.0 \times F = -5F \end{align}$$

My questions:

  1. Are there any errors in above calculations? If so, please suggest corrections.
  2. According to $E^\circ$ values, reaction $(2)$ is more feasible than reaction $(1)$. Am I correct?
  3. According to $\Delta G^\circ$ values, reaction $(1)$ is more feasible than reaction $(2)$. Am I correct?
  4. Is it useless to use $E$ values to predict the feasibilities of redox reactions? Should we always resort to $\Delta G$?
  5. Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict?
  6. If not how can these results be explained?
  • $\begingroup$ Standard potentials are written for reduction reactions and you wrote the first one for an oxidation reaction. $\endgroup$
    – Papul
    Jun 12, 2015 at 16:17
  • $\begingroup$ (a) is written as a reduction. $\endgroup$
    – user16096
    Jun 13, 2015 at 13:47

2 Answers 2


$\Delta G = -n {\mathcal F} \Delta {\mathcal E} $

$\mathcal F$ is Faraday's constant, which physicists currently believe has been constant throughout the universe for all time. $n$ is the number of electrons transfered in an electrochemical reaction. For a given reaction, $n$ is also constant in all places and times.

Thus, $\Delta G$ determines $\Delta {\mathcal E}$ and vice versa.

Are there any errors in above calculations?

Yes, there must be.

According to E∘ values, reaction (2) is more feasible than reaction (1). Am I correct ? According to ΔG∘ values, reaction (1) is more feasible than reaction (2). Am I correct ?

Both of these cannot be correct.

Is it useless to use E values to predict the feasibilities of redox reactions ?

No, it is just as good and exactly as informative as $\Delta G$.

Should we always resort to ΔG ?

No. Feel free to resort to $\Delta G$ if you like units like kJ/mol instead of volts, but other than personal preference there is no reason to prefer one over another.

Is it possible for predictions based on E values and ΔG values to contradict ?


If not how can these results be explained ?

UPDATE: Arithmetic error.

The choice of $n$ is arbitrary. Once you pick $n$ for a given half reaction, it determines the choice of $n$ for the other half reaction, but that's it. If you then consider a separate cell with separate reactions, you could write that new cell's equation with an arbitrary $n$. The $\textbf{mol}$ in $\Delta G$'s units of $\frac{\textrm{ kJ}}{\textbf {mol}}$ is $\textbf{mol}$ of reaction, and thus the value of $\Delta G$ depends on how you write the reaction. For example you could write equation (b) + (c) as

$$\ce{10M^3+ + 2 H2O + 2 MnO4- -> 2 Mn^2+ + 10 MO^2+ +4 H+}$$

Now, $n$ is twice as large, so $\Delta G$ is twice as large. But $\Delta \mathcal E$ is unchanged. This is because $\Delta \mathcal E$ has a basis of "$\textbf{mol}$" of electrons, not of moles of a reaction of arbitrary scale. In reality the "$\textbf{mol}$" of electrons is converted to a different units (Coulombs) to make things more confusing. (Remember, a volt is a joule per coulomb.)

So contrary to what I said above, the sign of $\Delta G$ but not the magnitude of $\Delta G$ is fixed, meaning that whether a reaction is absolutely spontaneous can determined by either $\Delta G$ or $\Delta \mathcal E$, but to compare the relative spontaneity of two reactions, either the $n$ values of the two reactions must be the same and then $\Delta G$ values can be used, or else without worrying about $n$ you could just look at $\Delta \mathcal E$.

  • $\begingroup$ Thanks for the answer. Can you please point where the arithmetic error is ? Could it be in the provided $E^{\circ}$ values of imaginary reactions ? $\endgroup$
    – user16096
    Jun 13, 2015 at 13:45
  • $\begingroup$ Doesn't $\Delta G$ has the unit kJ or J because, $\Delta G = -n (mol) \times E (JC^{-1}) \times F (C mol^{-1})$ ? $\endgroup$
    – user16096
    Jun 14, 2015 at 17:20

(Not too good with MathJax, so sorry about the lack of elegance. If someone could suggest a good tutorial, I'd appreciate it.)

Your methods are all correct, but it seems you are not comparing apples to apples. The $\Delta G$ of a particular reaction is going to scale with the amount of reactants you have. For example, the $\Delta G$ of $\ce{10N2 + 10H2 -> 10NH3}$ is going to be 10 times that of $\ce{N2 + H2 -> NH3}$. The same reasoning can be applied to your equations. So, in your first calculation of $-(a)\times5+(c)\times2$ the $\Delta G$ of $a$ and $c$ are 5 and 2 times greater than what they should, respectively. This is not a fair comparison.

What you need to do is to normalize your equations such that the same numbers of electrons are being transferred. You can do this with any integer, but it's probably easiest to just normalize to 1. In your first calculation, normalizing to an electron transfer of 1 leads to:

$$\ce{8/5H+ + 1/5MnO4- + 1/2X^3- -> 1/2X- + 1/5Mn^2+ + 4/5H2O}$$

$$E^\circ= 1.5\ \mathrm{V} + \left(-0.9\ \mathrm{V}\right) = 0.6\ \mathrm{V}$$ $$\Delta G^{\circ} =−nE^\circ F=-1\times0.6\times F=-0.6F$$

Applying the same normalization to the second equation yields:

$$E^\circ= 1.5\ \mathrm{V} + \left(-0.5\ \mathrm{V}\right) = 1.0\ \mathrm{V}\\ ΔG^\circ=−nE^\circ F=-1\times1.0\times F=-1.0F$$

The standard potential and the $\Delta G$ of the second overall reaction of half reactions b and c are in fact more negative than the first overall reaction. In case you're wondering why $E$ does not scale with the size of the reaction it's because $E$ is proportional to $G/n$. As you scale the size of the reaction, both $G$ and $n$ will grow or shrink in the exact same proportion and thus $E$ will always stay the same.

  • $\begingroup$ Welcome to Chemistry.SE. Our own in house tutorials are here and here. The second link has further reading suggested. $\endgroup$
    – Ben Norris
    Jul 19, 2015 at 10:29
  • $\begingroup$ Also, this page and this ‎one are good complements to what @Ben suggested. $\endgroup$
    – M.A.R.
    Jul 19, 2015 at 12:49
  • $\begingroup$ Yea, those links help a lot. Can't wait to make use of them next time. Thanks editors for cleaning up my garble. $\endgroup$
    – Zeruff
    Jul 20, 2015 at 1:54

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