What are the products of [Ti(H2O)6]Cl3 + H2SO4 + K2Cr2O7 --->

Question

I am slightly lost on how to proceed with this equation. I know potassium dichromate is a strong oxidizing agent but I am unsure of how to deal with it along with the titanium complex. I was wondering if I could get tips on how to handle these types of equations.

Answer 1

I know potassium dichromate is a strong oxidizing agent

You have already identified one partner and its role in a redox reaction. Good start!

So which component will be oxidized to what?

Can we assume that the reaction is carried out in water? If so, let's have a look at the ions in solution.

We'll have $\ce{H3O+}$, $\ce{K+}$, $\ce{Cr2O7^{2-}}$, and "that complex".

Let's have a look at $\ce{[Ti(H2O)6]Cl3}$.

Apparently, it's a complex of titanium with 6 neutral molecules of water as the ligands and 3 $\ce{Cl-}$ as the counter ions. So, what is the oxidation state of titanium in the complex when it is neutral with 3 $\ce{Cl-}$ as the counter ions?

Let's look again: $\ce{[Ti(H2O)6]^{3+}}$. OK, it is obviously $\ce{Ti(III)}$ and it seems that we have found the victim of our stong oxidant in the system.

We remember that most titanium compounds in nature are $\ce{Ti(IV)}$ oxides and conclude that we will have to balance our redox reaction to account for a $\ce{Ti(III) -> Ti(IV)}$ oxidation.

We could now continue to conclude that the reduction of $\ce{Cr(VI)}$ to $\ce{Cr(III)}$ is accompanied by $\ce{[Ti(H2O)6]^{3+} -> [Ti(H2O)6]^{4+}}$ but then we remember that $\ce{Ti(IV)}$ - tiny and higly charged - is a hard Lewis acid and will not exist as a naked cation, barely covered with just six neutral ligands in a hexaquo complex!

Instead, the likely $\ce{Ti(IV)}$ species is the titanyl cation, $\ce{TiO}^{2+}.$

Time to collect the puzzle pieces and write the redox reaction.