# How would you balance the following equation: Fe + Cl2 + H2O → [Fe(H2O)6]3+ + Cl-?

Jan 8, 2016

You are oxidizing iron, and then making a coordination complex. I will give you a reaction scheme.

#### Explanation:

OXIDATION

$F e \left(s\right) + \frac{3}{2} C {l}_{2} \left(g\right) \rightarrow F e C {l}_{3} \left(s\right)$

COMPLEXATION

$F e C {l}_{3} \left(s\right) + 6 {H}_{2} O \left(l\right) \rightarrow {\left[F e {\left(O {H}_{2}\right)}_{6}\right]}^{3 +} {\left(C {l}^{-}\right)}_{3}$

I don't know whether this is completely what you want.

Jan 21, 2016

$2 {\text{Fe" + "3Cl"_2 + "12H"_2"O" → 2["Fe"("H"_2"O")_6]^(3+) + "6Cl}}^{-}$

#### Explanation:

A Socratic answer here shows how to balance redox equations.

${\text{Fe"color(white)(l) + "Cl"_2 + "H"_2"O" → ["Fe"("H"_2"O")_6]^(3+) + "Cl}}^{-}$

SOLUTION:

1: The two half-reactions are

"Fe" → ["Fe"("H"_2"O")_6]^(3+)
${\text{Cl"_2 → "Cl}}^{-}$

2: Balance all atoms other than $\text{H}$ and $\text{O}$.

"Fe" → ["Fe"("H"_2"O")_6]^(3+)
${\text{Cl"_2 → color(red)(2)"Cl}}^{-}$

3: Balance $\text{O}$.

"Fe"color(white)(l) + color(red)("6H"_2"O") → ["Fe"("H"_2"O")_6]^(3+)
${\text{Cl"_2 → 2"Cl}}^{-}$

4: Balance H.

Done.

5: Balance charge.

"Fe"color(white)(l)+ "6H"_2"O" → ["Fe"("H"_2"O")_6]^(3+) + color(red)("3e"^(-))
${\text{Cl"_2 + color(red)("2e"^(-)) → 2"Cl}}^{-}$

6: Equalize electrons transferred.

color(red)(2 ×) {"Fe" + "6H"_2"O" → ["Fe"("H"_2"O")_6]^(3+) + + "3e"^(-)}
color(red)(3 ×){"Cl"_2 + "2e"^(-) → 2"Cl"^(-)}

7: Add the two half-reactions.

$2 {\text{Fe" + 3"Cl"_2 + "12H"_2"O" → 2["Fe"("H"_2"O")_6]^(3+) + "6Cl}}^{-}$

8: Check mass balance.

On the left: $\text{2 Fe"; "6 Cl"; "24 H"; "12 O}$
On the right: $\text{2 Fe"; "24 H"; "12 O"; "6 Cl}$

9: Check charge balance

On the left: $0$
On the right: $\text{6+ + 6-} = 0$

The balanced equation is

$2 {\text{Fe" + 3"Cl"_2 + "12H"_2"O" → 2["Fe"("H"_2"O")_6]^(3+) + "6Cl}}^{-}$