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

2 Answers
Jan 8, 2016

Answer:

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.

Jan 21, 2016

Answer:

#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"^(-)#