[therealman]

Hi

I am trying to balance this equation:

Fe²⁺+ Cr₂O₇^2-+ I^-+ H⁺  Fe³⁺+ Cr³⁺+ I₂ + H₂O

I hope someone out threre can help me with this, and correct me if I have done somthing wrong.

[UG]

The preferred way is to seperate each one into oxidation and reductions reactions, for example, your very large equation can be simplified into 3 smaller ones, called 'half-reactions':

Oxidation: Fe²⁺  Fe³⁺
Reduction: Cr₂O₇^2-  Cr³⁺
Oxidation: I^-  I₂

Here's an example on how to balance half reactions:

VO⁺² (aq) V⁺³ (aq)

1. Balance all of the atoms which are not oxygen or hydrogen. There is no need to here.
2. Balance oxygen by 'adding water'

VO⁺² (aq) V⁺³ (aq) + H₂O (l)

3. Balance hydrogen by adding H⁺

VO⁺² (aq) + 2H⁺ (aq) V⁺³ (aq) + H₂O (l)

4. Balance the charges on each side

VO⁺² (aq) + 2H⁺ (aq) + e^- V⁺³ (aq) + H₂O (l)

See if you can balance all of them. Note: this is not the last step in balancing the entire equation.

[AWK]

Hi

I am trying to balance this equation:

Fe²⁺+ Cr₂O₇^2-+ I^-+ H⁺  Fe³⁺+ Cr³⁺+ I₂ + H₂O

I hope someone out threre can help me with this, and correct me if I have done somthing wrong.

These are two independent redox reactions. You should balance them separately.

[DesertRose]

Hi, your equation is not quite balanced. the mistake that you're making is that if you want to balance your equation like an ionic equation, before you insert the charges and the state symbols, you have to actually balance it like you would in an normal equation. Then when you are finished, you insert the state symbols. After, you break up each molecule to give you individual elements (making sure to add the SAME state symbol as previously, and it's charge). Lastly you cancel out each element on both side of the reaction (reactants and products) that has the same state, and number of elements, which will leave you with the answer.
If however, you wish to balance your equation alone, all you have to do, is make sure that you have the same number of each element on both sides, and that'll give you your answer.
Hope it helps!