# What will happen when a metal reacts with an acid which is neither dilute nor concentrated? [closed]

We have always studied about metals reacting with dilute and concentrated acids, but I do not understand what will happen if the same metal reacts with an acid that is neither concentrated nor dilute?

• I'm voting to close this question as off-topic because it's vague. Oct 16, 2017 at 14:33

This question is a bit vague, but briefly, the reason some acids behave differently at high and low concentration is because changing the activities of components in the solution changes the redox potential of the various possible reactions.

Let's take the reaction of iron with sulfuric acid as an example. This is not necessarily the only factor and I'll make a few assumptions that are likely quite inaccurate to simplify the problem, but it will illustrate the situation:

Let's assume that the half-reaction with iron is always the same—iron is being oxidized:

$\ce{Fe_{(s)}<=> Fe^2+_{(aq)} +2e-}\ E^\circ=+0.44\ \mathrm{V}$

Now, if you look at a table of standard reduction potentials, you'll see that there are a couple of potential half-reactions that could be reduced to drive the iron oxidation:

$\ce{2H+_{(aq)} +2e- <=> H2_{(g)}}\ E^\circ=0\ \mathrm{V}$

$\ce{SO4^{2-}_{(aq)} +4H+_{(aq)} + 2e- <=> SO2_{(g)} +2H2O_{(l)}}\ E^\circ=+0.17\ \mathrm{V}$

If all these reagents were at unity activity, then the second reaction would be more favoured and we would see mostly $\ce{SO2}$, but the activities of the substances in solution are also important. By the Nernst equation, $E=E^\circ -\frac{RT}{zF}\ln Q$, we see that the potential of a complete cell or a half-cell reaction is dependant on the reaction quotient $Q$. Because we're assuming that only the one reaction happens to the iron, we can just look at the reduction half-cell. If we assume that the sulfuric acid dissociates completely to $\ce{SO4^2-}$, then $[\ce{H+}]\approx 2\cdot[\ce{SO4^2-}]$. If we assume the gaseous products of the reaction don't affect the equilibrium, for our two reactions, $Q$ is $\frac{1}{[\ce{H+}]^2}=\frac{1}{(2[\ce{H2SO4}])^2}$ and $\frac{1}{[\ce{H+}]^4[\ce{SO4^{2-}}]}=\frac{1}{(2\cdot[\ce{H2SO4}])^4[\ce{H2SO4}]}=\frac{1}{16\cdot[\ce{H2SO4}]^5}$, respectively.

Now, if we substitute these into the Nernst equation for different concentrations of acid, we can see which reaction is more favourable: In dilute acid (1 mM), we get −160 mV for hydrogen formation and −238 mV for sulfur dioxide. The picture changes as the acid gets more concentrated. In 1 M acid, the hydrogen formation half-cell has a potential of 18 mV, but sulfur dioxide is now up to 206 mV.

So you can see how the concentration of acid changes which reaction is more favourable. To answer your question about what happens in the middle, one need only do the calculation for a moderate concentration and find that at some point, the two half-reactions become close in potential and we may see a significant mixture of both reactions occurring.

This is a very simplified picture and the Nernst equation is only technically valid under conditions of no current flow, meaning that other factors may influence what is observed experimentally, but I hope it's apparent that there's not really a solid line between dilute and concentrated. Many different reactions can occur simultaneously, even if a single reaction is dominant under certain conditions.

Some metals react explosively with water (Alkali metals, Group 1, sodium, potassium), and water is neutral. Some metals do not react even with very strong and consentraded acids (platinum).

Some acids are strong, some are weak. Even vitamine C is an acid, but so also are sulferic acid and nitric acid. Consentration also plays a role in this picture.

Magnesium reacts slowly with water, and is oxidized. Gold does not react. Iron reacts. and so on. The nobler the metal, the harder it is for an acid to oxidize it, and the stronger and / or more consentrated the acid will have to be to break down the metal. Alkali metals are 'weak' enough to be oxidized by clean air.

So this question is far from simple to give a short answer to.