Calculate the pH of the cathode compartment for the following reaction?

Question

Calculate the pH of the cathode compartment for the following reaction given ℰcell = 3.01 V when [Cr3+ ] = 0.15 M, [Al3+ ] = 0.31 M, and [Cr2O72− ] = 0.64 M. The reaction given is:

2 Al(s) + Cr2O72−(aq) + 14 H +(aq) → 2 Al3+(aq) + 2 Cr3+(aq) + 7 H2O(l)

Currently I am using the E = E - (0.0591/n) log(Q) equation but I do not know how I can utilize that to find the pH. What would the n be? Is the equation even correct and if it is, how do I move through the problem from here? Thank you

Answer 1

So, you are correct to use the modified Nernst equation. Now, you should ask yourself, what is the definition of pH? It is the negative log of the [H+], so if you can find [H+], it will be easy to find pH.

To find n, you have....

Al(s) -->Al³⁺ +3e-

Cr⁶⁺ + 3e- -->Cr³⁺

So n = 3 moles electrons

From the balanced equation

2 Al(s) + Cr₂O₇²⁻⁽aq) + 14 H⁺(aq) → 2 Al³⁺(aq) + 2 Cr³⁺(aq) + 7 H2O(l)

Q = [Al³⁺]²[Cr³⁺]² / [Cr₂O₇^2-][H⁺]¹⁴

Now, the difficult part is solving for [H+]. Once you have that, just take -log [H+] to get pH