Help in The Oxidation Number Method ?

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Discussion Overview

The discussion revolves around the application of the Oxidation Number Method for balancing redox reactions, specifically focusing on the reaction between copper and nitric acid. Participants explore the steps involved in balancing the reaction and express confusion regarding the consistency of electron transfer and charge balance throughout the process.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • The original equation presented is Cu + HNO3 = Cu(NO3)2 + NO2 + H2O, and the participant expresses confusion about balancing it using the Oxidation Number Method.
  • One participant suggests balancing the net ionic reaction first to clarify the situation, indicating that there may not be a problem with the approach taken.
  • The participant describes the process of balancing charges by noting that Cu loses 2 electrons while nitrogen gains 1 electron, leading to a multiplication of HNO3 and NO2 to balance the charges.
  • There is a concern raised about the final equation, Cu + 4HNO3 = Cu(NO3)2 + 2NO2 + 2H2O, potentially unbalancing the charges after adjusting the coefficients.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the adjustments made to the coefficients result in an unbalanced equation. There are differing views on the necessity and correctness of the steps taken in the balancing process.

Contextual Notes

The discussion highlights the complexity of balancing redox reactions and the potential for confusion regarding charge balance and electron transfer, particularly when multiple adjustments to coefficients are made.

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Help in "The Oxidation Number Method"?

Homework Statement



I just learned how to balance the redox reactions with this method, but there is something I don't understand ...
Look at this example :
Cu+HNO3 = Cu(NO3)2+NO2+H2O
after making the number of electrons gained and lost equal we get this equation :
Cu+2HNO3 = Cu(NO3)2+2NO2+H2O
Now the number of electrons gained and lost is the same but we must continue balancing the equation and so we get at last :
Cu+4HNO3 = Cu(NO3)2+2NO2+2H2O
Didn't we just ruin the equation again ? ( the number of electrons lost and gained isn't the same ) Or am I wrong ?
Someone explain please

Homework Equations



Cu+HNO3 = Cu(NO3)2+NO2+H2O
 
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Please elaborate on what you are doing, step by step.

My guess is that you should try to balance net ionic reaction first, and it will become obvious what the problem is (or rather - why there is no problem). But I can be wrong, hard to say without details.
 
Borek said:
Please elaborate on what you are doing, step by step.

My guess is that you should try to balance net ionic reaction first, and it will become obvious what the problem is (or rather - why there is no problem). But I can be wrong, hard to say without details.

Ok here is the original equation :
Cu+HNO3 = Cu(NO3)2+NO2+H2O

Now according to "The Oxidation Number Method", I must balance the charges first
and if you look at the equation you will find that :
Cu -> lost 2electrons
N -> gained 1electron

Then balance the charges by multiplying N by 2, so I multiplied HNO3 on the reactant side and multiplied NO2 on the product side
The equation now should be like this :
Cu+2HNO3 = Cu(NO3)2 + 2NO2 + H2O

Now charges are balanced, but the equation isn't, and to do this we should make a little change, it will become like this :
Cu+4HNO3 = Cu(NO3)2 + 2NO2 + 2H2O

Now the equation is fully balanced but we multiplied 2HNO3 again and it became 4HNO3 ...
My question is didn't we - by doing that - unbalanced the charges again ?
Hope I made it clear enough
Thanks.
 
As I signaled earlier - balance net ionic reaction first.
 

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