Calculate the pH of the cathode compartment for the following reaction given Ecell = 3.01 V when [Cr3+] = 0.15 M, [Al3+] = 0.30 M, and [Cr2O72-] = 0.55 M.
2 Al(s) + Cr2O72-(aq) + 14 H+(aq) → 2 Al3+(aq) + 2 Cr3+(aq) + 7 H2O(l)
I'm only just starting with electrochemistry problems and I'm not sure what to do. First of all I don't know how to work out how many electrons are being transferred in the formal cell reaction as above.
Other than that, I know that
[tex]E = E° - \frac{R \cdot T}{F} \cdot log_e(Q)[/tex]
We can evaluate Q with the exception of [H+], which is just what we want. T I must assume is 298.15 K. However, how do we find E° - right now there are 2 variables, E° and [H+]. And what does it mean by pH in the "cathode compartment" - this is just the value of [H+] which contributes to Q, right?