Balance the following redox reaction in acidic solution?

#\sf{P_4(s)+NO_3^(-)\toH_2PO_4^(-)(aq)+NO(g)}#

I have the oxidation, but not the reduction.
Answer key tells me there is #\tt{2H_2O}# but that doesn't match with the #\tt{NO_3^(-)}# on the other end of the arror.

Plus, why add #\tt{3e^-}#? Bounded oxygen typically has a charge of -2, right? So three oxygens equals six electrons...?

Answer 1

And so elemental phosphorus #P(0)# is oxidized to phosphoric acid #P(V)#, a five electron oxidation....

#1/4P_4(s)+ 4H_2O rarr HPO_4^(2-)+7H^+ +5e^(-)# #(i)#

Since the mass and charge are balanced, this is kosher.

Nitrate #N(+V)# is REDUCED to #NO# #N(+II)#.. . (Oxygen is slightly more electronegative than nitrogen, so it FORMALLY gets the electrons of the bonds. What is the oxidation number of oxygen in the REAL molecule, #OF_2#?)

#NO_3^(-) + +4H^+ +3e^(-) rarr NO + 2H_2O# #(ii)#

We add #3xx(i)+5xx(ii)# to get..

#3/4P_4(s)+ 5NO_3^(-) +20H^+ +15e^(-)+12H_2O rarr 3HPO_4^(2-)+21H^+ +15e^(-)+5NO + 10H_2O#

...and proceed to cancel...

#3/4P_4(s)+ 5NO_3^(-) +2H_2O rarr 3HPO_4^(2-) +5NO+1H^+#

Since anyone can manipulate the arifmetik, is this balanced? If not, should it be? We know that an unbalanced representation of mass and charge cannot be taken seriously as accurate depiction of reality.

Phosphorus and nitrogen alone are involved in redox in this situation; oxygen is not. In reality, all I have done is assign oxidation states.

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Answer 2

Avery Adams

Final balanced equation:
#3P_4(s)+20NO_3^(-)(aq)+8H^+(aq) +8H_2O(l)->20NO(g)+12H_2PO_4^(-)(aq)#

Let's get started:

I can carefully insert water and hydrogen ions wherever they are needed in the equation when it is in an acidic solution:

#P_4(s)+16H_2O->4H_2PO_4^-)+20e^- )+24H^+#

#NO_3^-)+4H^(+)+ 3e^(- )->NO+2H_2O#

Why add #3e^-#? Because #N^(+5)# is reduced to #N^(+2)#

Each side's electrons must be equal:

#3(P_4(s)+16H_2O->4H_2PO_4^(-)+20e^(- )+24H^+)# #20(NO_3^-)+4H^(+)+ 3e^(- )->NO+2H_2O)#

#3P_4(s)+48H_2O->12H_2PO_4^-)+60e^- )+72H^+# #20NO_3^(-)+80H^(+)+ 60e^(- )->20NO+40H_2O#

I am left with: #3P_4(s)+20NO_3^(-)(aq)+8H^+(aq) +8H_2O(l)->20NO(g)+12H_2PO_4^(-)(aq)#

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Answer 3

Luna Chen

Please provide the redox reaction that you would like me to balance in acidic solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.