The standard electrode potentials for the following redox couples are given as: >$E_0(\ce{Fe^{2+}/Fe}) = x~\mathrm{V}$ and $ E_0(\ce{Fe^{3+}/Fe^{2+}}) = y~\mathrm{V}$
What will be potential $E_0(\ce{Fe^{3+}/Fe})$ (in $\mathrm{V}$)?

My approach: $\ce{Fe^{3+} + e- + Fe^{2+} + 2e- -> Fe^{2+} + Fe}$

So $E_0(\ce{Fe^{3+}/Fe})=(x+y)~\mathrm{V}$


You can write down the three half reactions:

\begin{aligned} \ce{Fe^{2+} + 2e- &-> Fe} & E_1&= x\,\mathrm{V}\\ \ce{Fe^{3+} + e- &-> Fe^{2+}} & E_2&= y\,\mathrm{V}\\ \ce{Fe^{3+} + 3e- &-> Fe} & E_3&= z\,\mathrm{V} \end{aligned}

The third half-reaction is simply the sum of the first two. However, we cannot just sum $E_1 + E_2$ in order to obtain $E_3$. The problem is with the meaning of the potentials, and their relation to the stoichiometry.

An half reaction with potential $E$ in volts means that the elecrons have some energy assigned to them. The meaning of volts is joules per coulomb. The charge of one mole of electrons is 96500 C (this is Faraday's constant: 1 F = 95600C/mol).

Take a look at the first half reaction. It transfers two moles of electron per mole of reaction. This means the electrons carry

  • $x$ joules per coulomb, or
  • $96500x$ joules per mole of electrons, or
  • $2\times 96500x$ joules per mole of reaction.

In the same fashion, the second half reaction has the electrons carrying

  • $y$ joules per coulomb,
  • $96500y$ joules per mole of electrons,
  • the same $96500y$ joules per mole of reaction.

The third half reaction is the sum of the first two, so its electrons carry

  • $2\times 96500x + 96500y$ joules per mole of reaction, which is
  • $3\times 96500 z$ joules per mole of reaction, or
  • $96500z$ joules per mole of electrons, or
  • $z$ joules per coulomb.

Now you can simplify this to $2x + y = 3z$ and calculate the potential you want.

  • $\begingroup$ You should add that Electrode potentials of two reactions cannot be added because they are not state functions, i.e. dependent on the path. On the other hand $\ce{\Delta G}$ i.e. the energy carried by electrons, is a state function, and can be added. $\endgroup$ – Shoubhik R Maiti Dec 15 '17 at 11:23

As correct as eflaschuks approach is, I found it rather confusing, because much text and few mathematics. So I am adding them here.

First of all, let us write down the Nernst Equation for an arbitrary reaction $\ce{Ox + z\cdot e- <=> Red}$: \begin{aligned} E=E^\ominus(\ce{Red/Ox}) - z^{-1}\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Red}}{\ce{Ox}} \end{aligned}

For our given system that boils down to the following reaction and their potentials: \begin{aligned} \ce{Fe^{2+} + $2\cdot$ e- &-> Fe}& E_1 &= E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}& \text{(1)}\\ \ce{Fe^{3+} + $1\cdot$ e- &-> Fe^{2+}}& E_2 &= E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}& \text{(2)}\\ \ce{Fe^{3+} + $3\cdot$ e- &-> Fe}& E_3 &= E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}}& \text{(3)}\\ \end{aligned}

Now you are looking for the standard potential $E^\ominus(\ce{Fe/Fe^{3+}})$, so you have to consider the potential $E_3$, which can be build from $E_1$ and $E_2$: $$E_3 = a\cdot E_1 + b\cdot E_2$$

Substituting $(1)$, $(2)$ and $(3)$ yields:

$$\begin{multline} E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} = \\ a\cdot\left(E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) \\+ b\cdot\left(E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right) \end{multline}$$

In order to obtain the standard potential the logarithmic terms have to cancel. Therfore we can split these equations: \begin{aligned} &&E^\ominus(\ce{Fe/Fe^{3+}}) &= a\cdot E^\ominus(\ce{Fe/Fe^{2+}}) + b\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) \\ &&\frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &= a\cdot\left(\frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) + b\cdot\left(\frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right)\\ \therefore&& \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &= \frac{3a}{2} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}} + 3b \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}} \end{aligned}

In order for the $\ce{Fe^{2+}}$ terms to cancel it is fairly obvious, that $$a=\frac23,\\ b=\frac13,$$ therfore resulting in \begin{aligned} &&E^\ominus(\ce{Fe/Fe^{3+}}) &= \frac{2}{3}\cdot E^\ominus(\ce{Fe/Fe^{2+}}) + \frac{1}{3}\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}). \end{aligned}

You can also check that against experimental data (from wikipedia): \begin{array}{llrlr}\hline \text{halfreaction} &&&& \text{potential}\\\hline \ce{Fe^{2+} + $2\cdot$ e- &-> Fe} & E_1 &=& -0.44\mathrm{V} \\ \ce{Fe^{3+} + $1\cdot$ e- &-> Fe^{2+}}& E_2 &=& 0.77\mathrm{V}\\ \ce{Fe^{3+} + $3\cdot$ e- &-> Fe} & E_3 &=& -0.04\mathrm{V}\\\hline \end{array}


Your approach is a correct one.

Explanation: (I call the given two reduction half cell reactions as 1 and 2 respectively)

The required half cell reaction $\ce{Fe^{3+} -> Fe}$ could be obtained by adding reaction 1 and 2 .Thus the net $\ce{E0}$ value would be the sum of $\ce{E0}$ values of reaction 1 and 2 (as both are reduction half cell reactions.)

Hence net $\ce{E0(Fe^{3+}/Fe^{2+})} = (x + y) \, \text{V}$

  • $\begingroup$ Are you sure? I am quite sceptical about my approach. $\endgroup$ – Rudstar May 18 '14 at 6:15
  • 3
    $\begingroup$ Why should not the standard half reaction potentials for $\ce{Fe^{3+} -> Fe^0}$ be $\frac{2x+y}{3}$, as per calculating potentials between oxidation numbers in a Latimer diagram? $\endgroup$ – Yoda May 18 '14 at 16:04

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