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Setup: A magnesium electrode dipped in a magnesium sulfate solution and an iron electrode dipped into iron (III) sulfate solution

Am I right in thinking that you need to first identify the redox reactions to determine to determine overall voltage of the cell at standard conditions?

I have identified Mg to be oxidized to Mg(2+) at the anode, but I am not sure which is reduced at the cathode. Would it be the sulfate ions?

If it is indeed the sulfate ions, I have calculated the standard electrode potential to be 2.54 volts (is this correct)?

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  • $\begingroup$ @Mithoron Oxidize to sulfurous acid? $\endgroup$ – user264985 Oct 11 '15 at 19:59
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    $\begingroup$ Sulfate can indeed be reduced by magnesium as magnesium is very strongly reducing, and you would get $E^\circ = +2.54\mathrm{~V}$ as you said, but there's probably a stronger oxidising agent than sulfate in that $\ce{Fe2(SO4)3}$ solution. $\endgroup$ – orthocresol♦ Oct 11 '15 at 20:50
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[...] I am not sure which is reduced at the cathode. Would it be the sulfate ions?

orthocresol has already pointed out in his comment that there's more in $\ce{Fe2(SO4)3}$ than just the sulfate.

Take a look at the cation:

$$\begin{align} \ce{Fe^3+& + e- &<=>&Fe^2+\quad &+0.77V} \\ \ce{Fe^3+& + 3e- &<=>&Fe(s)\quad &-0.04V} \end{align}$$

(Standard electrode potentials are cited from wikipedia.)

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