posted by RZeal on .
3HClO2(aq) +2Cr3+(aq) + 4H2O(l) -> 3HClO(aq) + (Cr2O7)2–(aq) + 8H+(aq)
At pH 0.00, with [Cr2O72–] = 0.80 M, [HClO2] = 0.15 M, and [HClO] = 0.20 M, the cell voltage is found to be 0.15 V. Calculate the concentration of [Cr3+] in the cell
Ecell = 0.31
Not sure how to really work this one out, more complicated than other problems like this
You have conflicting information. The problem says cell voltage is 0.15 v and you say Ecell = 0.31. I suspect one of those numbers is Eocell. If that is the case, then
Ecell = Eocell -(0.05916/6)*log Q and log Q (HClO)^3*(Cr2O7^2-)*(1)^8/(HClO2)^3*(Cr^3+)^2
The easy to is to solve for log Q, then Q, then solve for Cr^3+
Sorry about the confusion
Ecell = 0.15 and Eocell = 0.31
Just plug the numbers into my equation above and solve for [Cr^3+]
Thanks I was able to solve it!