[ANUABSHF]

Please help me balance this equation in both acidic and basic medium and give the specific name of the type of redox equation it is.


NO2^1- -----> NO3^1- + NO

[stevet]

Firstly, can you identify the oxidation and reduction half reactions? This also might help in describing they type of redox reaction occurring.

[fatcat]

Please help me balance this equation in both acidic and basic medium and give the specific name of the type of redox equation it is.


NO2^1- -----> NO3^1- + NO

First, identify half reactions, using the only reactant there for both and balancing first for major elements then the oxygens and hydrogens using H⁺ ions and water (H₂O). Finally, you balance for the charges and add electrons to each side. Match up the number of electrons by multiplying one or both half reactions.
My finished half reactions after balancing and multiplying reduction half reaction by 3:
Reduction: 3 e^- + 6 H⁺ + 3 NO₂^- ------> 3 NO^- + 3 H₂O
Oxidation: H₂O + NO₂^- ----------> NO₃ + 2 H⁺ + 3 e^-


Combine both together, crossing out intermediate and catalytic reactants for finished reaction, in an acidic environment, of:
Acidic based equation: 4 NO₂^- + 4 H⁺ -----> 3 NO + NO₃ + 2 H₂O

But wait! You said you wanted it in basic form? Well, just add equal parts of hydroxide ions to both sides to cancel out the hydrogen ions! Voila!
Finished reaction, adding 4 OH^- ions to each side:

Basic-based equation: 4 NO₂^- + 2 H₂O -----------> 3 NO + 4 OH^- + NO₃

[stevet]

Please help me balance this equation in both acidic and basic medium and give the specific name of the type of redox equation it is.


NO2^1- -----> NO3^1- + NO

First, identify half reactions, using the only reactant there for both and balancing first for major elements then the oxygens and hydrogens using H⁺ ions and water (H₂O). Finally, you balance for the charges and add electrons to each side. Match up the number of electrons by multiplying one or both half reactions.
My finished half reactions after balancing and multiplying reduction half reaction by 3:
Reduction: 3 e^- + 6 H⁺ + 3 NO₂^- ------> 3 NO^- + 3 H₂O
Oxidation: H₂O + NO₂^- ----------> NO₃ + 2 H⁺ + 3 e^-


Combine both together, crossing out intermediate and catalytic reactants for finished reaction, in an acidic environment, of:
Acidic based equation: 4 NO₂^- + 4 H⁺ -----> 3 NO + NO₃ + 2 H₂O

But wait! You said you wanted it in basic form? Well, just add equal parts of hydroxide ions to both sides to cancel out the hydrogen ions! Voila!
Finished reaction, adding 4 OH^- ions to each side:

Basic-based equation: 4 NO₂^- + 2 H₂O -----------> 3 NO + 4 OH^- + NO₃

This answer is not correct

[fatcat]

Explain? I know I accidentally put a charge on nitrogen oxide but what else?

[stevet]

Yeah, its the charge on NO, and the lack of charge on NO₃^-. This then affects balancing electrons etc...

Also, you shouldn't give answers directly, its not how this forum would like people to answer questions.

[Vidya]

Follow these steps:
identify half equations.
balance each equation separately---1. balance atoms other than H and O 2. add water to balance O 3.add H+ to balance H and in acidic medium or OH - in basic medium  4. add electrons to balance the charge

equalize number of electrons gained or lost and add both half reactions
start with these steps
you can show your working here
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