# How do I write the overall reaction for the reaction of hexaaquacobalt(II) chloride with sodium thiosulfate and ethylenediamine in the presence of "HCl" to form tris(ethylenediamine)cobalt(III) chloride and sodium bisulfate?

## Did I do the first step right? ["Co"("OH"_2)_6]"Cl"_2 + 3("en") -> ["Co"("en")_3]"Cl"_2 + 6"H"_2"O" Is sodium thiosulfate ${\text{Na"_2"S"_2"O}}_{7}$?

Nov 24, 2015

$N {a}_{2} {S}_{2} {O}_{7}$ exists (sodium pyrosulfate), but it's not very stable.

Yeah, the first step's fine. That happens since en is a stronger-field ligand than water. You need to oxidize cobalt(II) to cobalt(III), and you need to explain where you got your third chloride and your extra water at the end. So let's see...

$\left[C o {\left(O {H}_{2}\right)}_{6}\right] C {l}_{2} + 3 \left(e n\right) \to \left[C o {\left(e n\right)}_{3}\right] C {l}_{2} + 6 {H}_{2} O$

Here, you've added ethylenediamine to exchange with the aqua ligands.

Since water is now in solution, your $N {a}_{2} {S}_{2} {O}_{8}$ should dissolve just fine, leaving ${S}_{2} {O}_{8}^{2 -}$ and $2 N {a}^{+}$, where ${S}_{2} {O}_{8}^{2 -}$ as mentioned is an oxidant, so it gets reduced.

Let's write the reduction reaction for it (including chloride).

${S}_{2} {O}_{8}^{2 -} + 2 H C l + 2 {e}^{-} \to 2 H S {O}_{4}^{-} + 2 C {l}^{-}$

Persulfate hydrolyzes under acidic conditions to form bisulfate. Meanwhile, cobalt(II) is supposed to get oxidized to cobalt(III).

$2 \left(\left[C o {\left(e n\right)}_{3}\right] C {l}_{2} + C {l}^{-} \to \left[C o {\left(e n\right)}_{3}\right] C {l}_{3} + {e}^{-}\right)$

So, you've accounted for how you acquired a third chloride (where else would it come from?). And the remaining sodium ions react:

$2 N {a}^{+} + 2 H S {O}_{4}^{-} \to 2 N a H S {O}_{4}$

Let's see where we are... we need to double the quantities in the first step now to account for the two tris(ethylaminediamine)cobalt(II) chlorides we reacted in step 3.

Cancel out intermediates...

$\setminus m a t h b f \left(2\right) \left(\left[C o {\left(O {H}_{2}\right)}_{6}\right] C {l}_{2} + 3 \left(e n\right) \to \cancel{\left[C o {\left(e n\right)}_{3}\right] C {l}_{2}} + 6 {H}_{2} O\right)$

${S}_{2} {O}_{8}^{2 -} + 2 H C l + \cancel{2 {e}^{-}} \to 2 H S {O}_{4}^{-} + \cancel{2 C {l}^{-}}$

$\cancel{2 \left[C o {\left(e n\right)}_{3}\right] C {l}_{2}} + \cancel{2 C {l}^{-}} \to 2 \left[C o {\left(e n\right)}_{3}\right] C {l}_{3} + \cancel{2 {e}^{-}}$

and I think we have it. Adding in the sodium on both sides...

$\textcolor{b l u e}{2 \left[C o {\left(O {H}_{2}\right)}_{6}\right] C {l}_{2} + 6 \left(e n\right) + N {a}_{2} {S}_{2} {O}_{8} + 2 H C l \to}$
$\textcolor{b l u e}{2 \left[C o {\left(e n\right)}_{3}\right] C {l}_{3} + 12 {H}_{2} O + 2 N a H S {O}_{4}}$

As for the extra water in the product, I'm not sure. Maybe one of the water molecules on the product side comes near, giving:

$2 \left[C o {\left(O {H}_{2}\right)}_{6}\right] C {l}_{2} + 6 \left(e n\right) + N {a}_{2} {S}_{2} {O}_{8} + 2 H C l \to$
$2 \left[C o {\left(e n\right)}_{3}\right] C {l}_{3} \cdot {H}_{2} O + 10 {H}_{2} O + 2 N a H S {O}_{4}$

It wouldn't bind though, because cobalt has already occupied all six of its binding sites.