I have some challenging homework of balancing redox equations! I get really frustrated from it and I have a quiz 2 days later! Please help me out!

*When dealing with acidic/basic conditions, we write half reactions as a compound so that we can balance O by addiong H2O and balance H by add H+.

Balance the following equations using half-reactions and oxidation numbers:

1)H2O2 + Cr2O7 2-  ->  Cr 3+ + O2 + H2O (ACIDIC condition)
oxidation numbers are +1,-1,+6,-2,->,+3,0,+1,-2 respectively

Half reactions:(not balanced)
Cr2O7 2- -> 2Cr 3+
H2O2 -> O2 + H2O

Half reactions: (charge and atoms balanced)
14H+ + 6e + Cr2O7 2- -> 2Cr 3+ +7H2O
H2O2 -> O2 + H2O  (The teacher told us to do it like this) For this one, how do I balance the charges and atoms? By adding H2O and H+ or just balance by coefficients?

I try it this way:
2H2O2 -> O2 + 2H2O (The H and O atoms are balanced and their charges are the same, so I have no way to add electrons) Then how can I do? By adding H2O and H+ instead of adding coefficinets?


2) Cu + HNO3 -> Cu(NO3)2 + NO +H2O (ACIDIC condition)
oxidation numbers are: 0,+1,+5,-2,->,+2,+5,-2,+2,-2,+1,-2 respectively

Half reactions:
Cu -> Cu(NO3)2 <-this bugs me, how can I balance the N atoms
3H+ + 3e + HNO3 -> NO +2H2O


3)KIO3 + KI + HCL -> KCL + I2 + H2O (ACIDIC condition)

Half reactions:
KIO3 -> I2 (How can I blance the K atoms with no K on the product side?)
KI -> I2 (same problem)


4)This one I am just not sure if I am right!
OF2 + I- -> F- + I3 -   (ACIDIC condition)
oxidation nubmers are: +2,-1,-1,-1,-1/3 respectively

Half reactions:
2H+ + 4e + OF2 -> 2F- + H2O  (I put an H2O on the product side, beucase it is in acidic solution (the teacher told me that) and it is the only way to balance the O atoms)
2(3I- -> I3 - + 2e)

Add the half reactions:

2H+ + OF2 + 6I- -> 2F- + 2I3 - + H2O
(suprisely the charge and atoms are balanced )

I feel sorry to have that many questions! (Hopefully we are done with balanceing redox equations) Please help me! Thank you a million times!

1) 14H+ + 6e + Cr2O7 2- -> 2Cr 3+ +7H2O (gain electron)--equation 1
2H2O2 -> O2 + 2H2O -2e (loss electron) ---equation2

Then equation 2 x3, then give
6H2O2 -> 3O2 + 6H2O -6e

add two equations
14H+ + Cr2O7 2- + 6H2O2 -> 3O2 +6H2O + 2Cr 3+ +7H2O
the final equation is:
14H+ + Cr2O7 2- + 6H2O2 -> 3O2 +13H2O + 2Cr 3+

2) The half reaction is:
Cu --> Cu 2+ -2e (the nitrate is no change in oxidation state in the Cu(NO3))

3) KIO3 -> I2 + 5e ( only I reduced not K)
KI -> I2 -e (also only I oxidized)

So dont remember which reactants are under reduction and oxidation reaction.


Good luck!

But the charges are not balanced! For example
1)2H2O2 -> O2 + 2H2O -2e
Electric Charge of left side=0
Electric Charge of right side=2-

The charge of the final equation is not balanced too:
14H+ + Cr2O7 2- + 6H2O2 -> 3O2 +13H2O + 2Cr 3+
Sum of Electric Charges of left side=12+
Sum of Electric Charges of right side=6+

3) KIO3 -> I2 + 5e
Electric Charge of left side=0
Electric Charge of right side=5-

 

No, this has the charges unbalanced:
14H+ + Cr2O7 2- + 6H2O2 -> 3O2 +13H2O + 2Cr 3+
because this reaction is not a half reaction (no 2e)
2H2O2 -> O2 + 2H2O -2e (loss electron) ---equation2

This way:
14H+ + 6e + Cr2O7 2- -> 2Cr 3+ +7H2O  (Cr reduced)
3 (H2O2 -> 02 +  2H+  + 2e)  (O- is oxidized)
Add:
14H+ + Cr2O72-  + 3H2O2 -> 2Cr3+  + 3O2  + 6H+  + 7H2O
Now, simplify by subtracting off the 6H+ :
8H+  + Cr2O72-  + 3H2O2 -> 2Cr3+  + 3O2  +  7H2O

But O is reduced, right?
The oxidation number of O in H2O2 is -1, the oxidation number of O in H2O is -2


The equation that is to be balanced:
H2O2 + Cr2O7 2-  ->  Cr 3+ + O2 + H2O (ACIDIC condition)

for the other one:
You look at IO3- and realize I is being reduced to I2. Water will be formed on the right because there's O in IO3. H+ will be needed on the left to balance.
Then look at KI and realize I- will oxidize to the same I2. The K in both cases is just part of the salt and can be included early or excluded and added late. I'll include them here to show that you just add them when needed to balance.

10e + 12H+ + 2K+  + 2IO3- ->  I2 + 2K+  + 6H2O
5(2KI -> I2 + 2K+  + 2e)

12H+ + 2KIO3 + 10KI -> 6I2 + 6H2O + 12K+
Now just simplify (divide by 2) and add in the cl.

Thanks!
For question number 2, Cu + HNO3 -> Cu(NO3)2 + NO +H2O (ACIDIC condition)
oxidation numbers are: 0,+1,+5,-2,->,+2,+5,-2,+2,-2,+1,-2 respectively

Half reactions:
Cu -> Cu(NO3)2 (how can I balance the N atoms)
3H+ + 3e + HNO3 -> NO +2H2O

how can I balance the N atoms in the first half reactions?

Did you miss the post three above on KIO3?
balancing NO3- is similar.

IC, it has K+ ions

For question 3:
KIO3 + KI + HCL -> KCL + I2 + H2O (ACIDIC condition)

You said that:
"12H+ + 2KIO3 + 10KI -> 6I2 + 6H2O + 12K+
Now just simplify (divide by 2) and add in the Cl. "

Would the final answer be KIO3 + 5KI +6HCl -> 6KCl + 3I2 + 3H2O?
(I hope it is.. )

yes

yes

Thanks! I did it myself and got the right answer

For this one:
Fe(OH)2 + O2 -> Fe(OH)3 + OH-   (BASIC solution)

I worked on it and got 2H2O + 4Fe(OH)2 + O2 -> 4Fe(OH)3
Steps:
4(H2O + Fe(OH)2 -> Fe(OH)3 + e + H+
2H+ + 4e + O2 -> 2OH-

Add them:
4H2O + 4Fe(OH)2 + O2 -> 4Fe(OH)3 + 2H+ + 2OH-
4H2O + 4Fe(OH)2 + O2 -> 4Fe(OH)3 + 2H2O
2H2O + 4Fe(OH)2 + O2 -> 4Fe(OH)3

But is it possible that there is no OH- on the product side for equation that is basic condition?

Thank you!