# If given a half reaction, how do you determine if an element can exist in a acidic conditions?

I was given the formulas,

\begin{alignat}{2} \ce{MnO4^2- \;&<=> MnO4- + e-}\quad &&E^\circ = -0.56\ \mathrm{V}\\ \ce{MnO2 + 2H2O \;&<=> MnO4^2- + 4H+ + 2e-}\quad &&E^\circ = -2.27\ \mathrm{V} \end{alignat}

I gathered you have to see if it’s non-spontaneous or spontaneous and compare it with the original reaction, but, when I asked my teacher on how to start, he said “multiply the first reaction by 2 and the second by −1.” I understand the ×2 to balance electrons, but why do you multiple it by −1? I don’t understand this question, any help would be appreciated. Thanks.

• Multiplying a reaction by -1 means that you reverse it; in other words, your reactants become your products and your products become your reactants. By reversing the second equation you then have 2 electrons in your reactants, so that when you add the reactions you start and end with the same number of electrons. – Derek Jun 5 '15 at 3:22

$\ce{2MnO_4^2- -> 2MnO_4^- + 2e^-}$ and
$\ce{MnO_4^2- + 4H^+ +2e^- -> MnO_2 + 2H2O}$
$\ce{3MnO_4^2- + 4H+ <=> 2MnO4- + MnO2 + 2 H2O}$.
The question is which way this goes. You can see that you have loads of $\ce{H+}$ on the left-hand side so you can expect that in acidic conditions the equilibrium will be shifted to the right hand side. You can quantify this by calculating the electrochemical potential from the Nernst equation for the two half cells.