the following is a redox reaction. show which element(s) are oxidized and reduced. include how many total electrons are gained or lost by each element. 3HNO---->HNO3+H2O+2NO

I am not sure if you have a typo, but for the reaction to be balanced, that should be Nitrous Acid (HNO2) in the reactant side:

3HNO2 -> HNO3 + H2O + 2NO
Note that all the species here are neutral (zero charge). Thus the sum of the individual charges of each element involved should be zero.
H usually has a +1 charge
O usually has a -2 charge
Now we'll check the charge of N in the species containing N (let x = charge of N):
HNO2 :
1 + x + 2(-2) = 0
1 + x - 4 = 0
x = +3

HNO3 :
1 + x + 3(-2) = 0
1 + x - 6 = 0
x = +5

NO :
x + (-2) = 0
x = +2

Therefore, N is both oxidized and reduced. The original is +3 (HNO2), it's oxidized state is +5 (HNO3) and reduced state is +2 (NO). The number of e- lost is 2 (from +3 to +5), and e- gained is 1 (from +3 to +2).

Hope this helps :D

To determine which element(s) are oxidized and reduced in the given redox reaction, we need to assign oxidation numbers to each element in the reactants and products.

First, let's assign oxidation numbers to all the elements:

In HNO, hydrogen (H) has an oxidation number of +1, nitrogen (N) has an oxidation number of +5, and oxygen (O) has an oxidation number of -2.

In HNO3, hydrogen (H) has an oxidation number of +1, nitrogen (N) has an oxidation number of +5, and oxygen (O) also has an oxidation number of -2.

In H2O, hydrogen (H) has an oxidation number of +1, and oxygen (O) also has an oxidation number of -2.

In 2NO, nitrogen (N) has an oxidation number of +2, and oxygen (O) has an oxidation number of -2.

Now, let's compare the oxidation numbers of each element in the reactants (left side of the arrow) and the products (right side of the arrow).

HNO (Reactant):
H: +1
N: +5
O: -2

HNO3 (Product):
H: +1
N: +5
O: -2

H2O (Product):
H: +1
O: -2

2NO (Product):
N: +2
O: -2

From the comparison, we can see that nitrogen (N) goes from an oxidation number of +5 in HNO to +2 in 2NO. This indicates that nitrogen has been reduced because its oxidation number has decreased.

On the other hand, hydrogen (H) and oxygen (O) do not change oxidation numbers, which means they are not oxidized or reduced.

To determine the total number of electrons gained or lost by each element, we need to look at the change in oxidation numbers:

Nitrogen (N):
Oxidation Number Change = Final Oxidation Number - Initial Oxidation Number
= +2 - +5
= -3

Since nitrogen (N) goes from an oxidation number of +5 to +2, it gains 3 electrons.

Hydrogen (H) and oxygen (O) do not change oxidation numbers, so they do not gain or lose any electrons.

Therefore, in the given redox reaction:
- Nitrogen (N) is reduced and gains 3 electrons.
- Hydrogen (H) and oxygen (O) do not undergo any oxidation or reduction and do not gain or lose any electrons.

To determine the elements being oxidized and reduced in the given redox reaction, we need to look at the change in oxidation numbers.

First, let's assign oxidation numbers to each element in the reaction:

3HNO ----> HNO3 + H2O + 2NO

Oxidation number of HNO is +1 for H, +5 for N, and -2 for O.
Oxidation number of HNO3 is +1 for H, +5 for N, and -2 for O.
Oxidation number of H2O is +1 for H and -2 for O.
Oxidation number of NO is +2 for N and -2 for O.

Now, let's compare the oxidation numbers before and after the reaction:

For H:
Before: +1
After: +1 (in HNO3) and +1 (in H2O)
No change in oxidation number for H, so it is not oxidized or reduced.

For N:
Before: +5
After: +5 (in HNO3) and +2 (in NO)
N is reduced from +5 to +2. It gains 3 electrons.

For O:
Before: -2 (in HNO) and 0 (in NO)
After: -2 (in HNO3) and -2 (in H2O) and -2 (in NO)
O is reduced from 0 to -2 in NO. It gains 2 electrons.

In summary:
The element N is reduced, gaining 3 electrons.
The element O is reduced, gaining 2 electrons.
The element H does not undergo any change in oxidation number.

Please note that the reaction should be balanced to accurately determine the electron transfer.