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How can I calculate the electrode potential from thermochemical data?enter image description here

My attempt: I know that the electrode potential is dependent on the enthalpies of: atomisation, first ionisation, second ionisation and hydration. So I summed up all to get $\Delta G$ and used the formula $\Delta G=-nFE$,but I am not getting the same value as given in the table.

Does the electrode potential depends on factors other than those given in the table, or there is something more to it than just addition?

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    $\begingroup$ You are summing $\Delta H$ values, how do you expect to get $\Delta G$? $\endgroup$ – orthocresol Apr 25 '16 at 15:12
  • $\begingroup$ @orthocresol But delta S is not mentioned in the table, how could I find it? $\endgroup$ – JM97 Apr 25 '16 at 15:33
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    $\begingroup$ You can't find it from the table. So the table isn't enough. $\endgroup$ – orthocresol Apr 25 '16 at 16:10
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As stated above by @orthocresol, these values of $\Delta H$ alone are not sufficient to find the electrode potential. To do so, you would need $\Delta G$ for each reaction, so you would either need to find a table of $\Delta G$ for these reactions, or you would need a table of $\Delta S$ along with the table given.

But your question is why do we care about those particular values to begin with. The reason for this is that those reactions summed together give the reduction reaction for that metal. For example, I will look at the reactions of iron.

The reduction reaction for iron: $$\ce{Fe(s) \to Fe^2+(aq) + 2e^-}$$

All the other reactions in the order they are listed in the original post: \begin{align} \ce{Fe(s) &\to Fe(g)} \\ \ce{Fe(g) &\to Fe^+(g) + e^-} \\ \ce{Fe^+(g) &\to Fe^2+(g) + e^-} \\ \ce{Fe^2+(g) &\to Fe^2+(aq)} \end{align}

If you add together these reactions, you should obtain the reduction reaction. Consequently, by Hess's law for Gibb's free energy, the sum of the $\Delta G$'s for these reactions will give you the $ΔG$ for the reduction reaction.

Then, it is just a matter of converting the $\Delta G$ to a reduction potential using the equation given in the original question:$$\Delta G^{\circ}=-nFE^\circ$$ where $n$ is the moles of electrons, $F$ is Faraday's constant, and $E^\circ$ is the standard reduction potential.

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