How would you go about writing the reaction between water and chromium metal under standard acid and base conditions?

I know you would have to look at the vertical lines at pH 0 and 14 but I don’t know how you would know which species to look at and whether or not you write the reduction or oxidation of water.

Also, at pH 14 how would you calculate the ${\Delta G}$ value for the reduction between water and chromium metal?

Pourbaix Diagram


1 Answer 1


The red line in your diagram corresponds to the redox potential of hydrogen, which depends on $\mathrm{pH}$:

$$\begin{alignat}{3} \ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=0)\\ \ce{2H2O + 2e- \;&<=> H2 + 2OH-}\quad\quad &&E^\circ = -0.828\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=14) \end{alignat}$$

The blue line in your diagram corresponds to the redox potential of oxygen, which depends on $\mathrm{pH}$, too:

$$\begin{alignat}{3} \ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=0)\\ \ce{O2 + 2H2O + 4e- \;&<=> 4OH-}\quad\quad &&E^\circ = +0.401\ \mathrm{V}\quad\quad &&&(\mathrm{pH}=14) \end{alignat}$$

Therefore, only areas between these lines are stable in aqueous solutions.

Your initial species is elemental chromium ($\ce{Cr}$). It is shown below the red line; therefore, it is unstable with respect to oxidation to $\ce{Cr^3+}$ at $\mathrm{pH}=0$

$$\begin{alignat}{2} \ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm{V}\\ \ce{Cr^3+ + e- \;&<=> Cr^2+}\quad\quad &&E^\circ = -0.408\ \mathrm{V}\\ \ce{Cr^2+ + 2e- \;&<=> Cr}\quad\quad &&E^\circ = -0.913\ \mathrm{V}\\\\ \ce{2Cr + 6H+ &-> 2Cr^3+ + 3H2} \end{alignat}$$

and to $\ce{Cr(OH)3}$ at $\mathrm{pH}=14$

$$\begin{alignat}{2} \ce{2H2O + 2e- \;&<=> H2 + 2OH-}\quad\quad &&E^\circ = -0.828\ \mathrm{V}\\ \ce{Cr(OH)3 + 3e- \;&<=> Cr + 3OH-}\quad\quad &&E^\circ = -1.33\ \mathrm{V}\\\\ \ce{2Cr + 6H2O &-> 2Cr(OH)3 + 3H2} \end{alignat}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.