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### Topic: [Help]Balancing redox equations  (Read 45559 times)

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#### 777888

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##### [Help]Balancing redox equations
« on: October 18, 2004, 10:56:09 AM »
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)

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!
« Last Edit: October 18, 2004, 11:07:55 AM by 777888 »

#### kevins

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##### Re:[Help]Balancing redox equations
« Reply #1 on: October 18, 2004, 12:11:56 PM »

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

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!

#### 777888

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##### Re:[Help]Balancing redox equations
« Reply #2 on: October 18, 2004, 12:42:15 PM »

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

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-

« Last Edit: October 18, 2004, 12:47:10 PM by 777888 »

#### Demotivator

• Guest
##### Re:[Help]Balancing redox equations
« Reply #3 on: October 18, 2004, 12:46:41 PM »
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)
14H+ + Cr2O72-  + 3H2O2 -> 2Cr3+  + 3O2  + 6H+  + 7H2O
Now, simplify by subtracting off the 6H+ :
8H+  + Cr2O72-  + 3H2O2 -> 2Cr3+  + 3O2  +  7H2O

#### 777888

• Guest
##### Re:[Help]Balancing redox equations
« Reply #4 on: October 18, 2004, 12:52:02 PM »
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)
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)

#### Demotivator

• Guest
##### Re:[Help]Balancing redox equations
« Reply #5 on: October 18, 2004, 01:04:17 PM »
How do you know O from H2O2 is reduced? I can say the  Cr7+  oxidizes the O- to O2, while the O- reduces the Cr7+. The O in H2O could come from Cr2O7.

#### Demotivator

• Guest
##### Re:[Help]Balancing redox equations
« Reply #6 on: October 18, 2004, 01:40:20 PM »
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.

#### kevins

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##### Re:[Help]Balancing redox equations
« Reply #7 on: October 18, 2004, 01:43:36 PM »
777888

I am sorry I get something wrong and thanks Demotivator to correct.

The redox equation is not a easy understand topic for some student.

1. Try to identify which reactants are under redox reaction.
2. Set up a partial equation for the oxidising agent
3. Set up a partial equation for the reducing agent
4. The redox reaction is e transfer process.
i.e. no of e loss = no of e gain

e.g. the reduction of potassium permangamente by iron (II) sulfate in the present of dilute sulfuric acid.

the first partial equation (reduction) is
MnO4 - ->Mn 2+

to balance atomically, 8H+ is required.
MnO4 - + 8H+ -> Mn 2+ + 4H2O

and to balance it electrically 5e is needed on the left-hand side
MnO4 - + 8H+  + 5e -> Mn 2+ + 4H2O

The second equation (oxidation) is
Fe 2+ -> Fe 3+

To balance this electrically 1e must be added to the right-hand side or subtracted from the left-hand side
Fe 2+ -> Fe 3+ + e

Now the gain and loss of e must be equal. Therefore two partial equations must apply in the ratio of 1 to 5.
so the final equation is

MnO4 - + 8H+  + 5Fe 2+ -> Mn 2+ + 5Fe 3+ + 4H2O

Good luck!

#### 777888

• Guest
##### Re:[Help]Balancing redox equations
« Reply #8 on: October 18, 2004, 06:03:58 PM »
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?

#### Demotivator

• Guest
##### Re:[Help]Balancing redox equations
« Reply #9 on: October 18, 2004, 08:22:30 PM »
Did you miss the post three above on KIO3?
balancing NO3- is similar.
« Last Edit: October 18, 2004, 08:24:52 PM by Demotivator »

#### 777888

• Guest
##### Re:[Help]Balancing redox equations
« Reply #10 on: October 18, 2004, 09:21:23 PM »
Did you miss the post three above on KIO3?
balancing NO3- is similar.
IC, it has K+ ions
« Last Edit: October 18, 2004, 09:34:15 PM by 777888 »

#### Demotivator

• Guest
##### Re:[Help]Balancing redox equations
« Reply #11 on: October 18, 2004, 09:49:12 PM »
well either this way
Cu + 2NO3-  -> Cu(NO3)2  + 2e
or
Cu  ->  Cu2+  +  2e
Then fill in the NO3 to balance after the equations are added .
« Last Edit: October 18, 2004, 10:00:46 PM by Demotivator »

#### 777888

• Guest
##### Re:[Help]Balancing redox equations
« Reply #12 on: October 18, 2004, 10:18:50 PM »
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.. )
« Last Edit: October 18, 2004, 10:19:30 PM by 777888 »

#### Demotivator

• Guest
##### Re:[Help]Balancing redox equations
« Reply #13 on: October 18, 2004, 10:21:12 PM »
yes

#### 777888

• Guest
##### Re:[Help]Balancing redox equations
« Reply #14 on: October 19, 2004, 10:31:37 AM »
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-