If I am given the individual oxidation and reduction half-equations, how can I find the emf of the combination? For example:

$$ \begin{align} E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) &= \pu{0.77 V}\\ E^\circ(\ce{Fe^2+}/\ce{Fe^0}) &= \pu{-0.44V} \end{align} $$

How can I find the emf of this redox pair, i.e. $E^\circ(\ce{Fe^3+}/\ce{Fe^0})?$

  • 1
    $\begingroup$ Determine the Gibbs energy of reaction for each against the standard hydrogen electrode, add them up and determine the standard potential from the Gibbs energy of the summed reactions. $\endgroup$
    – Karsten
    May 18, 2019 at 4:05
  • 1
    $\begingroup$ You can’t combine the two in a cell because both are reduction half reactions. You could do a disproportionation or comproportionation, maybe. $\endgroup$
    – Karsten
    May 18, 2019 at 4:17

1 Answer 1


This is my first answer in stack exchange if some mistake comment on. $$E^\circ_\ce{Fe^3+/Fe} \neq E^\circ_\ce{Fe^3+/Fe^2+} + E^\circ_\ce{Fe^2+/Fe}$$


$$\Delta_r G^\circ_1 = \Delta_r G^\circ_2 + \Delta_r G^\circ_3$$

for the three reactions:

$$\ce{Fe^3+ + 3/2 H2 -> Fe + 3H+}\tag{1}$$

$$\ce{Fe^3+ + 1/2 H2 -> Fe^2+ + H+}\tag{2}$$

$$\ce{Fe^2+ + H2 -> Fe + 2H+}\tag{3}$$

With $$\Delta_r G^\circ = - n F E^\circ , $$ we can establish a relationship between the reduction potentials:

$$n_1 F E^\circ_1 = n_2 F E^\circ_2 + n_3 F E^\circ_3$$

Cancelling F and solving for $E^\circ_1$ allows us to calculate the reduction potential of the iron/iron(III) half reaction:

$$n_1 E^\circ_1 = n_2 E^\circ_2 + n_3 E^\circ_3$$

$$E^\circ_1 = \frac{n_2 E^\circ_2 + n_3 E^\circ_3}{n_1}$$

$$ = \frac{1 \cdot \pu{0.77 V} + 2 \cdot (\pu{-0.44 V})}{3}$$

$$ = \pu{-0.367 V}$$

answer to the question

  • $\begingroup$ Technically, a half reaction does not have a Gibbs energy of reaction. I think the numerical answer is correct, though, $\endgroup$
    – Karsten
    May 18, 2019 at 4:20
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    – andselisk
    May 18, 2019 at 5:48
  • $\begingroup$ Please format this using MathJax for easier readability. This page will get you started. Also, thank you for submitting an answer and, of course, welcome to the site! It's always awesome to see a new contributor answering questions! $\endgroup$ May 18, 2019 at 7:42

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