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?
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
3 answers
yes I apologize,
#1 is C6H12O6 --> C2H4O2
thankyou for all your help!
#1 is C6H12O6 --> C2H4O2
thankyou for all your help!
C6H12O6 ==> 3C2H4O2
3 answers