Simple Redox Balancing, need confirmation

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In summary, balancing reactions involving hydrogen peroxide, oxygen, and water can be tricky due to the decomposition of hydrogen peroxide. While there are various approaches that can yield a formally correct result, they may not reflect the actual reaction. This can be seen in the case of the reaction between hydrogen peroxide and permanganate. Therefore, it is important to consider all possible reactions and their stoichiometry when balancing such reactions.
  • #1
AGNuke
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Question
Cr2O72- + H2O2 → Cr3+ + O2

Attempt
Cr2O72- + 3H2O2 + 8H+ → 2Cr3+ + 3O2 + 7H2O

Is my attempt correct. I think I had lost my grip in redox balancing.
 
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  • #2
Balancing reactions with hydrogen peroxide if oxygen and water are between products is always tricky, as these are products of the hydrogen peroxide decomposition. That means you can always combine your reaction with hydrogen decomposition reaction and you will still get something that is - from the formal point of view - correctly balanced, but it doesn't reflect any real reaction.

You can try to assume that oxygen produced has the sole source of hydrogen peroxide, that's the approach that yields correct results in the case of the reaction between hydrogen peroxide and permanganate. But just because it yields a correct result doesn't mean that's what is really happening - observed stoichiometry can agree with this approach just accidentally.

Compare http://pubs.acs.org/doi/abs/10.1021/ic00224a030
 
  • #3
Yeah. This questions seems to have infinite solutions. That's the base equation I got. There's a parallel disproportionation of Hydrogen Peroxide going on, so I guess I screwed up nonetheless.
 

1. What is Simple Redox Balancing?

Simple Redox Balancing is a method used in chemistry to balance chemical equations involving oxidation-reduction (redox) reactions. It involves balancing the number of atoms and the total charge on each side of the equation.

2. Why is balancing redox reactions important?

Balancing redox reactions is important because it allows us to accurately predict the products of a chemical reaction and determine the amount of each reactant needed. It also ensures that the law of conservation of mass and charge are upheld.

3. How do you know if a reaction is a redox reaction?

A redox reaction involves the transfer of electrons from one species to another. One way to identify if a reaction is a redox reaction is to check if the oxidation states of the elements involved are changing. If the oxidation state increases, it is being oxidized, and if it decreases, it is being reduced.

4. Can you give an example of a simple redox reaction and how to balance it?

One example of a simple redox reaction is the reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO). The unbalanced equation is:
Mg + O2 → MgO
To balance this equation, first, we need to determine the oxidation states of each element. Magnesium has an oxidation state of 0, while oxygen has an oxidation state of -2. We can write the oxidation states above each element:
Mg0 + O2-2 → MgO
Next, we need to balance the oxygen atoms by adding a coefficient of 2 in front of MgO:
Mg0 + O2-2 → 2MgO
Finally, we balance the magnesium atoms by adding a coefficient of 2 in front of Mg:
2Mg0 + O2-2 → 2MgO
The equation is now balanced, and the final step is to check if the total number of atoms and the total charge are the same on both sides of the equation.

5. Are there any shortcuts or tricks for balancing redox reactions?

Yes, there are a few shortcuts that can make balancing redox reactions easier. One technique is the half-reaction method, where you split the reaction into two half-reactions: one for the oxidation and one for the reduction. Another shortcut is the oxidation number method, where you assign oxidation numbers to each element and use them to balance the equation. It is also helpful to remember the rules for assigning oxidation numbers, such as the sum of oxidation numbers in a neutral compound is 0, and in a polyatomic ion, it equals the charge of the ion.

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