The very first step in balancing redox reactions is to split up the reaction into one of oxidation and one of reduction. So.......
O2 + Sb ----> H202 + SbO2^1- (SbO2's charge is minus 1. I'm just not sure how to denote that.)
So splitting the top reaction into two half reactions, shouldn't it be:
O2 ----> H2O2
Sb -----> SbO2^1-
??
But when I start following the procedures in balancing redox reactions, I can't seem to get the answer in the book.
1) Are my two half reactions correct?
2) If they are (meaning I did something else wrong), could you show me, step by step?
Thank You! :)
O2 + Sb ----> H202 + SbO2^1- (SbO2's charge is minus 1. I'm just not sure how to denote that.)
So splitting the top reaction into two half reactions, shouldn't it be:
O2 ----> H2O2
Sb -----> SbO2^1-
??
But when I start following the procedures in balancing redox reactions, I can't seem to get the answer in the book.
1) Are my two half reactions correct?
2) If they are (meaning I did something else wrong), could you show me, step by step?
Thank You! :)
-
1st step, write 2 half reactions
2nd step, balance atoms other than O or H
3rd step, balance O by adding H2O to one side
4th step, balance H by adding H+ to one side
5th step balance charge by adding electrons, e-, to one side
6th step, balance electron transfer by multiplying entire equation by number.
O2 ----> H2O2
Balance H by adding 2 H+1 to left side
2 H+1 + O2 ----> H2O2
balance charge by adding 2 e- to left side
Eq . #1
2 e-+ 2 H+1 + O2 ----> H2O2
Sb -----> SbO2^1-
Balance O by adding 2 H2O to left side
2 H2O + Sb -----> SbO2^1-
Balance H by adding 4 H+1 to right side
2 H2O + Sb -----> SbO2^1- + 4 H+1
balance charge by adding 3 e- to right side
Eq . #2
2 H2O + Sb -----> SbO2^1- + 4 H+1 + 3 e-
Eq. #1 has 2 e- and Eq . #2 has 3 e-, so multiply Eq . #1
by 3 and Eq . #2 by 2
6 e-+ 6 H+1 + 3 O2 → 3 H2O2
4 H2O + 2 Sb → 2 SbO2^1- + 8 H+1 + 6 e-
Now add both equations, eliminating the 6 e-
6 H+1 + 3 O2 + 4 H2O + 2 Sb → 3 H2O2 + 2 SbO2^1- + 8 H+1
Now subtract 6 H+1 from both sides
3 O2 + 4 H2O + 2 Sb → 3 H2O2 + 2 SbO2^1- + 2 H+1
H+1 + SbO2^1- → HSbO2, so
3 O2 + 4 H2O + 2 Sb → 3 H2O2 + 2 HSbO2
This is the balanced equation
2nd step, balance atoms other than O or H
3rd step, balance O by adding H2O to one side
4th step, balance H by adding H+ to one side
5th step balance charge by adding electrons, e-, to one side
6th step, balance electron transfer by multiplying entire equation by number.
O2 ----> H2O2
Balance H by adding 2 H+1 to left side
2 H+1 + O2 ----> H2O2
balance charge by adding 2 e- to left side
Eq . #1
2 e-+ 2 H+1 + O2 ----> H2O2
Sb -----> SbO2^1-
Balance O by adding 2 H2O to left side
2 H2O + Sb -----> SbO2^1-
Balance H by adding 4 H+1 to right side
2 H2O + Sb -----> SbO2^1- + 4 H+1
balance charge by adding 3 e- to right side
Eq . #2
2 H2O + Sb -----> SbO2^1- + 4 H+1 + 3 e-
Eq. #1 has 2 e- and Eq . #2 has 3 e-, so multiply Eq . #1
by 3 and Eq . #2 by 2
6 e-+ 6 H+1 + 3 O2 → 3 H2O2
4 H2O + 2 Sb → 2 SbO2^1- + 8 H+1 + 6 e-
Now add both equations, eliminating the 6 e-
6 H+1 + 3 O2 + 4 H2O + 2 Sb → 3 H2O2 + 2 SbO2^1- + 8 H+1
Now subtract 6 H+1 from both sides
3 O2 + 4 H2O + 2 Sb → 3 H2O2 + 2 SbO2^1- + 2 H+1
H+1 + SbO2^1- → HSbO2, so
3 O2 + 4 H2O + 2 Sb → 3 H2O2 + 2 HSbO2
This is the balanced equation
-
This is not the correct answer to your question; the correct answer is
2Sb + 2OH + 2H2O + 3O2 ------> 2SbO2 + 3H2O2; this was basic solution
and not an acidic solution, and the step for solving it is different. If you would like
the proper steps, please feel free to leaving me a message.
2Sb + 2OH + 2H2O + 3O2 ------> 2SbO2 + 3H2O2; this was basic solution
and not an acidic solution, and the step for solving it is different. If you would like
the proper steps, please feel free to leaving me a message.
Report Abuse