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When group 2 metals, says magnesium, is reacted with nitric acid, oxide of nitrogen is produced beside hydrogen gas. The standard electrode potential values agree with this:

$$ \begin{align} \ce{Mg^2+(aq) + 2 e- &<=> Mg(s)} &\quad E^\circ &= \pu{-2.38 V} \\ \ce{H+(aq) + e- &<=> 1/2 H2(g)} &\quad E^\circ &= \pu{0 V} \\ \ce{NO3-(aq) + 4 H+ + 3e- &<=> NO(g) + 2H2O(l)} &\quad E^\circ &= \pu{+0.96 V} \end{align} $$

Does sulfate ion from sulfuric acid is reduced too when reacted with group 2 metals? The standard electrode potential values suggest it is possible, but I can’t find any source from internet stating $\ce{SO2}$ is produced from reaction of sulfuric acid with, says, magnesium.

$$ \begin{align} \ce{Mg^2+(aq) + 2 e- &<=> Mg(s)} &\quad E^\circ &= \pu{-2.38 V} \\ \ce{SO4^2-(aq) + 4 H+ + 2 e- &<=> SO2(g) + 2 H2O(l)} &\quad E^\circ &= \pu{+0.17 V} \end{align} $$

If sulfate ion can’t be reduced by magnesium to sulfur dioxide, then what’s the reason?

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We should not extend the idea of electrode potentials for very high concentrations such as concentrated sulfuric acid which is on the order of 18 M. The interesting property of these acids (HNO3 or H2SO4) is that they behave as oxidizing agents or even as dehydrating agents (H2SO4) especially at high temperature and at high concentrations. This is not to be predicted by tabulated electrode potentials. These tabulated values are reliable for dilute solutions (~ 1 M or less).

So the short answer is that in dilute H2SO4, and Mg, we would see a classical reaction of Mg forming magnesium sulfate and hydrogen. Under concentrated conditions and heat, Mg, may be able to able to reduce it to SO2, just like copper.

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  • $\begingroup$ I am not clear yet. You said that these electrode potential value are only reliable for dilute solution, doesn’t mean that under dilute solution of sulphuric acid, Mg will react with it to form SO2? $\endgroup$ – JF9199 Apr 6 at 17:45
  • $\begingroup$ Practically not. Mg will form H2 in dilute sulfuric acid. You are forgetting another half cell 2H+ 2 e -> H2 Eo = 0.00 V. If there are multiple possible half cells, the one which has the highest tendency will proceed first. $\endgroup$ – M. Farooq Apr 6 at 18:28

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