[huskywolf]

Hi,

Can you tell me if I balanced this equation correctly please?

MnO₄^- + Fe²⁺ + H⁺ => Fe³⁺ + Mn²⁺ + H₂O
=
MnO₄^- + Fe²⁺ + 8H⁺ => Fe³⁺ + Mn²⁺ + 4H₂O
Fe is oxidised, and Mn is reduced
? Thanks

[sinthreck]

Charge is not balanced.

LHS: -1+2+8 = +9
RHS: 3+2 = +5

I just had a go at this questions and came up with this result:

Mn0₄^- + 5Fe²⁺ + 8H⁺  Mn²⁺ + 5Fe³⁺ + 4H₂O

The charge is balanced (+17 on each side).

The only thing is, when I balanced it, I ignored the original H+ and H2O that was in the equation. I'm not sure you can do this.

[sinthreck]

Taken from my textbook:

Balancing redox reactions in aqueous solutions:

Step 1) Identify and separate half-reactions
Step 2) For each half-reaction
           - Balance elements (except H and O)
           - Balance O using H₂O
           - Balance H using H⁺
           - Balance charge using electrons.
Step 3) Equalise number of electrons in each half-reaction (i.e. multiply half reaction by a factor)
Step 4) Add half-reactions (cancel identical species)
Step 5) Check elements and charges balance


On second thoughts, I didn't really ignore the H₂O and H⁺ in the original reaction. My answer should be fine.

[huskywolf]

Great ,thanks for the help people