Balance Oxidation-Reduction Equations - Steps & Examples

  • Thread starter Thread starter gigi9
  • Start date Start date
  • Tags Tags
    Ap Chem
AI Thread Summary
Balancing oxidation-reduction equations involves identifying elements that appear in only one component on each side of the equation. The example of photosynthesis is used, starting with the unbalanced equation H2O + CO2 -> C6H12O6 + O2. To balance it, carbon is first addressed by adjusting the amount of CO2, followed by hydrogen using water, and finally oxygen is balanced by adding O2 on the right-hand side. The discussion emphasizes that redox reactions are more complex than standard stoichiometry, prompting a request for further examples to clarify the balancing process. The conversation highlights the importance of systematic adjustments to achieve balance in chemical equations.
gigi9
Messages
40
Reaction score
0
Someone please give me a good explanation on how to "Balance Oxidation- Reduction Equations." Plz show me the steps and an example of how to do this type of equations. Thanks a lot.
 
Physics news on Phys.org
gigi9,
photosynthesis might be a good example. Let's start from
H2O + CO2 -> C6H12O6 + O2.
First step, let's add what we have.
Left-hand side: 2H + 3O + 1C.
Right-hand side: 12H + 8O + 6C. So the formula is not balanced.

As a first step, we should look for an element that appears only in one component on each side. This is carbon. So, to balance carbon, we have to use 6 molecules of carbon dioxide:
H2O + 6CO2 -> C6H12O6 + O2.
Let's add what we have.
Left-hand side: 2H + 13O + 6C.
Right-hand side: 12H + 8O + 6C.

Is there another element which appears only once on each side? Yes, hydrogen! So we better use 6 molecules of water:
6H2O + 6CO2 -> C6H12O6 + O2.
Left-hand side: 12H + 18O + 6C.
Right-hand side: 12H + 8O + 6C.

We can balance this by adding more oxygen on the RHS:
6H2O + 6CO2 -> C6H12O6 + 6O2.

Now it's balanced.
 
Originally posted by arcnets
gigi9,
photosynthesis might be a good example. Let's start from
H2O + CO2 -> C6H12O6 + O2.
First step, let's add what we have.
Left-hand side: 2H + 3O + 1C.
Right-hand side: 12H + 8O + 6C. So the formula is not balanced.

As a first step, we should look for an element that appears only in one component on each side. This is carbon. So, to balance carbon, we have to use 6 molecules of carbon dioxide:
H2O + 6CO2 -> C6H12O6 + O2.
Let's add what we have.
Left-hand side: 2H + 13O + 6C.
Right-hand side: 12H + 8O + 6C.

Is there another element which appears only once on each side? Yes, hydrogen! So we better use 6 molecules of water:
6H2O + 6CO2 -> C6H12O6 + O2.
Left-hand side: 12H + 18O + 6C.
Right-hand side: 12H + 8O + 6C.

We can balance this by adding more oxygen on the RHS:
6H2O + 6CO2 -> C6H12O6 + 6O2.

Now it's balanced.

That's good arcnets, but gigi is talking about balancing redox reactions, which is a bit more complicated than regular stoichiometry.
 
why don't you post an example and we'll work from there gigi9.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top