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!
To solve this problem, we can use the Nernst equation to relate the cell voltage (Ecell) to the concentrations of the reactants and products. The Nernst equation is given by:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- E°cell is the standard cell potential
- n is the number of electrons transferred in the balanced equation
- Q is the reaction quotient
In this case, the balanced equation shows that 6 electrons are transferred, so n = 6.
We can rearrange the Nernst equation to solve for Q:
Q = 10^((E°cell - Ecell) / (0.0592/n))
Given that E°cell = 0.31 and Ecell = 0.15, we can plug in the values to calculate Q:
Q = 10^((0.31 - 0.15) / (0.0592/6))
= 10^((0.16) / (0.00986))
≈ 8.02
Now, let's look at the reaction in the balanced equation. We start with 2Cr3+ and 3HClO2, and they combine to form (Cr2O7)2-, HClO, and H+. The stoichiometric coefficients indicate a 2:3 ratio between Cr3+ and Cr2O7)2-.
Using this information, we can calculate the concentration of [Cr3+]:
[Cr3+] = [Cr2O7)2-] / 2
= 0.80 M / 2
= 0.40 M
Therefore, the concentration of [Cr3+] in the cell is 0.40 M.