Question

Balance the following redox equation, identifying the element oxidized and the element reduced. Show all of the work used to solve the problem.

HNO3 + P yields H3PO4 + NO

Answer 1

@nincompoop ?

Answer 2

@mayankdevnani ?

Answer 3

@Abmon98 ?

Answer 4

@superhelp101 i know how to balance it (shortcut) but i am extremely sorry that i can't explain you effectively by using comments....

Answer 5

ok may i show you my work and you tell me if it is right?

Answer 6

So it this right ?
1. HNO3 + P -->H3PO4 + NO
balnce H
2. 3HNO3 + P -->H3PO4 + NO
balance N
3. 3HNO3 + P -->H3PO4 + 3NO cant balance O

4. 3HNO3 + P +O2 -->H3PO4 + 3NO

Answer 7

your method is wrong:-
this is redox reaction so first balance all atoms other than O and H and then use OXIDATION NUMBER METHOD to balance it.

Answer 8

you know what is OXIDATION NUMBER METHOD ?

Answer 9

no can u teach me the steps ?

Answer 10

see, scroll up and see what my first comment was?
see this is no easy to teach redox reaction balance method by using comments

Answer 11

Here is link you can go through :-
https://www.youtube.com/results?search_query=oxidation+number+method+balancing

Answer 12

https://www.google.co.in/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=oxidation%20number%20method%20for%20balancing%20redox%20equations

Answer 13

okay thanks! how about I did it again :)
HNO3 → NO
P → H3PO4

We need to determine the oxidation state for these species.

HNO3 => H+ + NO3- => in NO3-, each O is -2, so 3*(-2) = -6, then subtract the charge on the ion (-1), -6 - (-1) = -6 +1 = -5, so N must be +5

NO => O = -2, so N = +2

So N goes from +5 to +2 how? By gaining electrons, N^5+ + 3e- → N^2+
Gain Electrons, Reduced so N is being reduced

P → H3PO4

P => elemental P is in the 0 oxidation state by definition
H3PO4 => 3H+ + PO3^3- => in PO3^3-, each O is -2, so 3*(-2) = -6, then subtract the charge on the ion (-3), -6 - (-3) = -6 + 3 = -3, so P must be +3

So P goes from 0 to +3 how? By losing electrons, P^0 → P^3+ 3e-
Lose Electrons, Oxidized so P is being oxidized

Balancing the equation:

HNO3 + P → H3PO4 + NO
Do an atom inventory:

Left side ............... Right side
1H, 1N, 1 P, 3O......3H, 1N, 1P, 5O

Because these numbers don't seem to lend themselves to a common multiple (this is something you pick up on after working enough of these types of problems), let's work on the half reactions and then combine them.

HNO3 → NO
HNO3 + 3e- → NO (from what we found on oxidation numbers)
HNO3 + 3e- → NO + 3OH-
(we need negative charges on the right side to balance the electrons of the left, and OH- also supplies us with H that we need on the right, too)
HNO3 + H2O + 3e- → NO + 3OH-
Check the balance:
Left side 3H, 1N, 4O, (-3) charge → Right side 3H, 1N, 4O, (-3) charge => checks out

P → H3PO4
P → H3PO4 + 3e- (from what we found on oxidation numbers)
P + 3OH- → H3PO4 + 3e- (balances the negatives and adds H to the left))
P + 3OH- + H2O → H3PO4 + 3e- (adds one more O needed on the left)
P + 3OH- + H2O → H3PO4 + 3e- + H2

Check the balance:
Left side 5H, 1P, 4O, (-3) charge → Right side 5H, 1P, 4O, (-3) charge => checks out

Now add the two half-reactions together

HNO3 + H2O + 3e- → NO + 3OH-
P + 3OH- + H2O → H3PO4 + 3e- + H2
HNO3 + H2O + 3e- +P + 3OH- + H2O → NO + 3OH- + H3PO4 + 3e- + H2

Cancel out the like terms (3e-, 3OH-) on both sides:

HNO3 + H2O + P + H2O → NO + H3PO4 + H2

The two H2O on the left can be combined, so we have as the final balanced equation:

HNO3 + P + 2H2O → NO + H3PO4 + H2

Answer 14

extremely sorry if i hurts you. :(

Answer 15

what hurt me ??

Answer 16

that i can't teach you

Answer 17

oh no it is fine!

Answer 18

GREAT JOB... @superhelp101 you are correct

Answer 19

Yaaaasssssss thanks! for checking

Answer 20

this method is known as HALF REACTION METHOD

Answer 21

yep thanks again!

Answer 22

welcome :)