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

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Discussion Overview

The discussion revolves around balancing oxidation-reduction equations in both acidic and basic conditions, focusing on specific examples provided by participants. The scope includes homework problems and the application of balancing techniques in redox reactions.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant outlines a method for balancing redox equations, emphasizing the steps of balancing elements, oxygen, hydrogen, and charge using electrons.
  • Another participant challenges the balance of a half-reaction, stating that it must be balanced in terms of both atoms and charge, and questions the use of oxidation numbers.
  • There is a query about whether the oxidation half-reaction should be multiplied by a specific value to balance electrons.
  • Some participants suggest multiplying half-reactions to equalize the number of electrons transferred, with one proposing specific multipliers for the dichromate and CO2 half-reactions.
  • A participant presents a separate problem and claims to have reached the correct answer, inviting feedback on their method of balancing the reactions.
  • Another participant expresses confusion regarding the charge balance in their half-reaction calculations, particularly concerning the total charge on both sides of the equation.
  • One participant presents a new redox equation involving C2H4 and MnO4-, noting the need for a significant number of electrons, which leads to confusion about the balancing process.
  • Another participant confirms the need for 12 electrons in the reaction involving C2H4 and MnO4-, indicating agreement on that point.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding the balancing of redox equations. There is no consensus on the correct approach or final answers, as multiple interpretations and methods are presented.

Contextual Notes

Some participants highlight limitations in their understanding of the balancing process, particularly regarding charge calculations and the necessity of electrons in half-reactions. There are unresolved questions about specific steps and methods used in balancing.

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

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

What is charge on the left? What is charge on the right? Are they identical?
 
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.
 
Yes, it needs 12 electrons.
 

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