December 11, 2023, 06:09:44 PM
Forum Rules: Read This Before Posting

### Topic: Weird Redox reaction  (Read 11076 times)

0 Members and 1 Guest are viewing this topic.

#### Faisal

• Regular Member
• Posts: 36
• Mole Snacks: +0/-1
##### Weird Redox reaction
« on: May 12, 2008, 03:00:24 PM »
Here is the weird question:Balance the following equation using the electron transfer method. You may wish to assign oxidation numbers and write the electronic equations for oxidation and reduction to help balance the overall reaction.
__ HNO3  +  __ P  +  __ H2O ----> __ H3PO4  +  __ NO
What is the sum of the coefficients in this balanced equation?
A. 5
B. 8
C. 10
D. 12
E. 18
What is the answer for this question?
I tried separating the oxidation and reducction reactions and ignoring the H2O.And then tried following the usual steps but I end up getting coefficients of 4 and 5.So I multiply each reaction by 4 and 5(to cancel the electrons) and end up getting huge coefficients like 20.
Can anyone tell me wht I am doing wrong?

#### tflint89

• Regular Member
• Posts: 12
• Mole Snacks: +0/-0
##### Re: Weird Redox reaction
« Reply #1 on: May 12, 2008, 03:33:03 PM »
You might simply have assigned oxidation numbers incorrectly!

The nitrogen is reduced from +5 (HNO3) to +2 (NO) whereas phosphorous is oxidised from 0 (P) to +5 (H3PO4). Therefore there is a 5:3 ratio of N:P in this equation. Use the water to balance up remaining elements.

If you don't want to bypass the half-equation method just write the 2 down remembering that this is in acid solution so H+ can be used to balance the charge of the electrons lost/gained. Do not ignore water as it has to appear in both of these and does in most redox 1/2 eqns involving acid or alkali.

#### Faisal

• Regular Member
• Posts: 36
• Mole Snacks: +0/-1
##### Re: Weird Redox reaction
« Reply #2 on: May 12, 2008, 03:48:25 PM »
k check my work...I think I dunno something

half rx:HNO3->NO
P->H3PO4
The N and P are already balanced so I balance my O and H.
6H+ + HNO3->NO+3H2O
4H2O+P->H3PO4+5H+
is this correct?

#### tflint89

• Regular Member
• Posts: 12
• Mole Snacks: +0/-0
##### Re: Weird Redox reaction
« Reply #3 on: May 12, 2008, 03:53:48 PM »
I always put electrons in my 1/2 eqns to check charges, the P one is correct but think about electron transfer and H+ balancing for N, the eqn is not properly balanced as you have written it.

#### Faisal

• Regular Member
• Posts: 36
• Mole Snacks: +0/-1
##### Re: Weird Redox reaction
« Reply #4 on: May 12, 2008, 04:02:29 PM »
oh yeah...its 3H+ + HNO3->NO+2H20
Got it...its E...Thx it was juz a silly mistake...god I thought I din knw something

#### Borek

• Mr. pH
• Deity Member
• Posts: 27487
• Mole Snacks: +1788/-409
• Gender:
• I am known to be occasionally wrong.
##### Re: Weird Redox reaction
« Reply #5 on: May 12, 2008, 04:30:46 PM »
k check my work...I think I dunno

Thx it was juz a silly mistake...god I thought I din knw

we hafta

ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

• Chemist
• Full Member
• Posts: 690
• Mole Snacks: +55/-9
• Gender:
• postdoc at ANL
##### Re: Weird Redox reaction
« Reply #6 on: May 12, 2008, 04:35:58 PM »
Always use half reactions:
Reduction:
HNO3 {oxidation nr. of N = 5} + H2O + 3e- --> NO {oxidation nr. of N = 2} + 3OH-

difference in oxidation nr is 3, hence 3 electron reaction

Oxidation:
P {oxidation nr. = 0} + 5 OH- --> H3PO4 {oxidation nr. of P = 5} + H2O + 5 e-

difference in oxidation nr. is 5, hence 5 electron reaction.

Note: a half reaction is only balanced if the charges are also balanced. Electrons are a necessary part of a redox half-reaction

Water/OH- is just used to balance O and H (water+ H+ is also good to balance of course)

after that just balance the two and you will indeed see the answer is E.

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7981
• Mole Snacks: +555/-93
• Gender:
##### Re: Weird Redox reaction
« Reply #7 on: May 13, 2008, 04:24:38 AM »
Corect reaction should contain P4
__ HNO3   +  __ P4  +  __ H2O ----> __ H3PO4  +  __ NO

But this is no problem - just divide sum of all coefficients by 4  (during division treat P4 as 4P, hence 3P4 = 12P)
Balance of electron gained (HNO3 - 3) and lost (P4 - 4 times 5)
x 3 = y 20
==>
20 times 3 = 3 times 20

20HNO3   +  3P4  +  __H2O ----> 12H3PO4  +  20NO

Balancing hydrogen and oxygen atoms you will find a number of water molecules (=8)

AWK