How Do You Balance Oxidation-Reduction Equations in Acidic and Basic Conditions?

[Nelo]

Homework Statement

Several problems.

1) Balance these oxidation-reduction equations in both acidic and basic conditions. In each equation,
underline the oxidizing agent. (I have highlighted them. Remember, the oxidizing agent is the species that
is reduced and gains electrons.)

a)Cr2o7-+ C2H5OH → Cr3+ + CO
b) HNO3 + P -> No + h3po4.
c) C2h4 + MNO4- -> Mn2+ + CO2

Homework Equations

No equations needed

The Attempt at a Solution

I just seem to not understand what the hell is going on. We talked about this breifly but it will be on the exam ( on tuesday) Here is what I got from what she taught...

a) balance all elements except hydrogen and oxygen
b) Balance oxygen using H2o
c) Balance hydrogen using H+
d) balance the charge of rctnt side with charge of product side by adding electrons to either reactant or product
e) Multiply 1 or both balanced half reactions by integer to equalize number of electrons transferred in the two half reactions.


So..#1 in acidic :: Cr2O7-+ C2H5OH → Cr3+ + CO

Cr2O7 ---> Cr3+ First i found charge of Cr in Cr2O7 (which is 6) . since difference of 6 and 3 is 3. and it went from +6 to +3 it gained 3 electrons and was reduced. so the half rctn would be

14H+ + Cr2O7 + 3e ----> Cr3+ + 7H2O

Second half reaction.. : H2C2O4 --> 2 CO2 + 2 H+ + 2 e- ( I don't even understand why that's 2 electrons...?)

Thats basically all i understand .. and that's just the beginning part. Can anyone help me to get to the right steps of this??

 

[Borek]

14H+ + Cr2O7 + 3e ----> Cr3+ + 7H2O

This is not balanced.

Second half reaction.. : H2C2O4 --> 2 CO2 + 2 H+ + 2 e- ( I don't even understand why that's 2 electrons...?)

2 electrons to balance charge. Properly balanced reaction (or half reaction) must be balanced in terms of atoms AND charge. There is no charge on the left (total of zero), so you need zero on the right. After balancing atoms you have 2H⁺ there - that means charge of +2, to neutralize it you need to add two electrons.

When balancing half reactions you don't use oxidation numbers, they aren't needed. First balance atoms, then balance charge adding electrons on one of the sides.

See:

balancing equations with half reactions method

balancing equations with oxidation numbers method

 

[Nelo]

Are you always balancing the oxidation one by multiplying it by set amount of value (# of electrons ) . ?

 

[Nelo]

Also, to completely balance that in terms of electrons you would multiply the dicromate half reaction by 2 , and the CO2 half reaction by 3 ? giving 6 electrons? thus cancelling

 

[Nelo]

I still don't understand it.

Cr2O7 2- + C2H5OH -> Cr3+ + CO2.

1) 14H+ + Cr2O7-2 ----> 2Cr 3+ + 7H2O .
the only thing left is the electrons

The overall charge on the left side is (14) (-2) = 12. overall charge on right is (3) (0) *unless the coefficnat multiplies with the 3+ giving 6 on the right, thus needing 6 electrons to be stable. (resulting in multplying the other half rctn by 3)

 

[Nelo]

I did a separate problem. Seem to have gotten the right answer. Please tell me if you notice something off.

Problem : As2O3 + NO3- ---> H3AsO4 + NO

What i did : 4H+ + No3- ---> NO + 2H2o +3e

2nd half : 5H2o + As2o3 + 4e -----> 2H3AsO4 + 4H+

Since we need electrons to balance, multiply top half by 4, and bottom half by 3, then combine and simplify.

Giving : 3 As2O3 + 4 NO3- + 7H2o +4H+ ----> 6H3AsO4 + 4NO . I'm going to try doing basic solution now.

 

[Borek]

1) 14H+ + Cr2O7-2 ----> 2Cr 3+ + 7H2O .
the only thing left is the electrons

The overall charge on the left side is (14) (-2) = 12. overall charge on right is (3) (0) *unless the coefficnat multiplies with the 3+ giving 6 on the right, thus needing 6 electrons to be stable. (resulting in multplying the other half rctn by 3)

Charge of 14H⁺ means 14*+1 - you did that correctly. Why do you have any doubts what is the charge of 2Cr³⁺? 2*+3 it is, we ara talking about total charge, not a single ion cherge.

What i did : 4H+ + No3- ---> NO + 2H2o +3e

What is charge on the left? What is charge on the right? Are they identical?

 

[Nelo]

C2H4 + MnO4-  Mn2+ + CO2

How do you solve this one?

I get to 8H+ + MnO4- ----> Mn 2+ + 4H2O
(7 charge) (2 charge) needs 5 electrons.

2nd half being 4H2O + C2H4 ----> 2CO2 + 12H+
needs 12 electrons.

This doesn't make sense.

 

[Borek]

Yes, it needs 12 electrons.