Question

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

Answer 1

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

Explanation:

OXIDATION

`Fe(s) + 3/2Cl_2(g) rarr FeCl_3(s)`

COMPLEXATION

`FeCl_3(s) + 6H_2O(l) rarr [Fe(OH_2)_6]^(3+)(Cl^-)_3`

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

Answer 2

`2"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.

Your equation is:

`"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+)`
`"Cl"_2 → "Cl"^(-)`

2: Balance all atoms other than `"H"` and `"O"`.

`"Fe" → ["Fe"("H"_2"O")_6]^(3+)`
`"Cl"_2 → color(red)(2)"Cl"^(-)`

3: Balance `"O"`.

`"Fe"color(white)(l) + color(red)("6H"_2"O") → ["Fe"("H"_2"O")_6]^(3+)`
`"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"^(-))`
`"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"Fe" + 3"Cl"_2 + "12H"_2"O" → 2["Fe"("H"_2"O")_6]^(3+) + "6Cl"^(-)`

8: Check mass balance.

On the left: `"2 Fe"; "6 Cl"; "24 H"; "12 O"`
On the right: `"2 Fe"; "24 H"; "12 O"; "6 Cl"`

9: Check charge balance

On the left: `0`
On the right: `"6+ + 6-" = 0`

The balanced equation is

`2"Fe" + 3"Cl"_2 + "12H"_2"O" → 2["Fe"("H"_2"O")_6]^(3+) + "6Cl"^(-)`