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This question already has an answer here:

I mean by "combining" is to make a new half-reaction equation and not an overall equation for a reaction in whole.

For instance, I was trying to arrive at the following half-reaction:

$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

by combining

$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -> Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ= 1.51 \,\mathrm{V}$

and

$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H_2O(l)}\quad\quad E^\circ= 1.7\,\mathrm{V}$

using Hess's Law. My calculation yielded +0.19 V, and not +1.23 V as I would expect.

Why caused this? I know that there are half-reactions that can be constructed in this way (as above, by combining two or more other half-reactions) that result in the correct potential being calculated.

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marked as duplicate by orthocresol Apr 10 '17 at 22:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ I think there is a mistake in your middle equation. It is not balanced w/r/t H or O: $$\ce{MnO4- + 8H+ + 5e- -> Mn^{2+} + 4H2O}$$ $\endgroup$ – Ben Norris Apr 19 '15 at 19:20
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    $\begingroup$ Also, for folks who might expect that there is a mistake in the reduction potentials, here is a table with references. Hess's law should work here... $\endgroup$ – Ben Norris Apr 19 '15 at 19:23
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    $\begingroup$ I think you got -0.19V. At least, that is what I get. $\endgroup$ – LDC3 Aug 3 '15 at 0:17
  • $\begingroup$ @LDC3 - I also got -0.19 V, but for the sake of preserving the original content of the post, kept the answer the OP got in there. $\endgroup$ – Todd Minehardt Aug 3 '15 at 0:38
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In general, you cannot simply subtract electrode potentials like this to find the potential of another half-reaction. They have to be weighted by $n$ (the stoichiometric coefficient of $\ce{e-}$). In cases where your method works, it is only because $n$ is coincidentally the same for both half-reactions you are combining.

The foolproof way is always to convert electrode potentials to Gibbs free energy changes: $$\Delta G^\circ = -nFE^\circ$$

So $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ is

$$(-5)(96485 \text{ C mol}^{-1})(1.51 \text{ V}) = -728.4 \text{ kJ mol}^{-1}$$

and $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O}$ is

$$(-3)(96485 \text{ C mol}^{-1})(1.7 \text{ V}) = -492.1 \text{ kJ mol}^{-1}$$

Subtracting the second equation from the first, you will find that $\Delta G^\circ$ for the half-reaction of interest ($\ce{MnO2 + 4H+ + 2e- -> Mn2+ + 2H2O}$) is

$$(-728.4) - (-492.1) = -236.3 \text{ kJ mol}^{-1}$$

which gives:

$$E^\circ = -\frac{-236.3 \text{ kJ mol}^{-1}}{(2)(96485 \text{ C mol}^{-1})} = 1.22 \text{ V.}$$

I presume some rounding errors led to the slight discrepancy between my answer and the answer you want. You may leave out the Faraday constant in the calculations once you are more familiar with them; basically that means you work with the quantity $nE^\circ$ instead of $\Delta G^\circ$. The answer should be the same since they are directly proportional.

For more information you can refer to the redox chapters in any inorganic chemistry textbook (Shriver/Atkins, Housecroft, etc.). These calculations are dealt with under the sections discussing Latimer and Frost diagrams.

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