Having trouble balancing redox equations. Teacher says I should use the method of half reactions and I'm not sure how. Any explanation or formula would be nice.
C6H12O2 --> C2H4O2
SO2+H2O+O2 --> H2SO4
C2H6O+O2 --> CO2 + H2O
I disagree that the half reaction method is easier in all circumstances; in fact, I think it is harder to do them this way for these three than straight trial and error. Let me illustrate on the last one. Here is what I do.
C2H6O + O2 ==> CO2 + H2O
I look and see 2C on the left so let's make it 2 on the right.
C2H6O + O2 ==> 2CO2 + H2O
That takes care of C. Now let's do H. I have 6 on the left, let's get 6 on the right.
C2H6O + O2 ==> 2CO2 +3H2O
We now know (or highly suspect, that the coefficients on the right are now "in stone". So let's count up O on the right and make that balance on the left. We have 3 + 4 = 7 O on the right. On the left we have 1 from C2H6O which leave 6 unaccounted for. So we add 3 for the O2.
C2H6O + 3O2 ==> 2CO2 + 3H2O
With a little practice these can be done in minimal time.
Let's do the second one this way and see if it's just as easy.
SO2+H2O+O2 --> H2SO4
I see 1 S on the left and 1 on the right along with 2 H on the left and 2 on the right so my first inclination is to leave that SO2 and H2O on the left and H2SO4 on the right alone. Let's count up O and see where we are. We have 4 O on the right and 3 from the SO2 + H2O (notice I don't count the O2 because that's the only one I can change). So we need one more O on the left. I can get that bv 1/2 O2 (1/2 * 2 = 1) so let's put that in.
SO2 + H2O + 1/2 O2 --> H2SO4
But of course we can't use fractions. How do we get rid of the 1/2. Just multiply everything by 2 which gives us
2SO2 + 2H2O + O2 ==> 2H2SO4.
Voila!
Is something missing from #1?
yes I apologize,
#1 is C6H12O6 --> C2H4O2
thankyou for all your help!
C6H12O6 ==> 3C2H4O2
Balancing redox equations can be challenging, but the method of half reactions is a helpful approach. It involves breaking down the redox equation into two half reactions—one for the oxidation half and another for the reduction half. Here are the steps to follow:
Step 1: Split the equation into half reactions.
Divide the redox equation into the oxidation half (the portion where atoms lose electrons) and the reduction half (the portion where atoms gain electrons).
For example, let's start with the first equation: C6H12O2 → C2H4O2
The oxidation half would be: C6H12O2 → 2C2H4O2 (Notice that carbon goes from +3 to +2 oxidation state)
The reduction half would be: 2H+ + 2e- → H2O (Notice that the hydrogen goes from +1 to 0 oxidation state)
Step 2: Balance the atoms other than oxygen and hydrogen.
In each half reaction, balance the atoms involved, except for oxygen and hydrogen. To accomplish this, add water (H2O) to balance the oxygen atoms and adjust the hydrogen ions (H+) to balance the hydrogen atoms.
Continuing with the first equation:
Oxidation half: C6H12O2 → 2C2H4O2 (already balanced)
Reduction half: 2H+ + 2e- → H2O (Notice that there is only one oxygen atom on the left and two on the right, so add another water molecule to balance: 2H+ + 2e- + H2O → 2H2O)
Step 3: Balance the charges by adding electrons.
To ensure that the charges are balanced, add electrons (e-) to one or both half reactions. The number of electrons added should be equal to the difference in charge between the reactants and products.
For our example:
Oxidation half: C6H12O2 → 2C2H4O2 + ?e- (Notice that carbon goes from +3 to +2, so we need to add 2e- to balance)
Reduction half: 2H+ + 2e- + H2O → 2H2O (already balanced)
Step 4: Balance the electrons in the half reactions.
Multiply each half reaction by a coefficient, if necessary, to ensure that the number of electrons is the same in both half reactions. This allows us to cancel out the electrons when we combine the two half reactions.
In our example, the oxidation half already has 2e-. Thus, we multiply the reduction half by 2 to balance the electrons:
Oxidation half: C6H12O2 → 2C2H4O2 + 2e-
Reduction half: 4H+ + 4e- + 2H2O → 4H2O
Step 5: Combine the two half reactions.
Add the two half reactions together to form the balanced redox equation. Ensure that the number of atoms and charges are balanced on both sides.
Final balanced equation: C6H12O2 + 4H+ + 4e- → 2C2H4O2 + 2H2O
This is the balanced redox equation formed by using the method of half reactions. Remember to follow these steps for each redox equation you encounter.