use oxidation states to identify the element that is being oxidized and the element that is being reduced in the redox reaction
Sn + 4HNO3 ---> SnO2 + 4NO2 + 2H2O
I see that SN is oxidized 0 --> 4+ and book says N is reduced +5 --> +4 I don't see that. Please help
HNO3.
H = +1.
O = 3*-2=-6
N must be +5 for HNO3 to be zero.
See the other post for NO2 of +4 for N.
Well, looks like we have a redox reaction here! Let's break it down and identify the elements being oxidized and reduced.
In the reaction Sn + 4HNO3 → SnO2 + 4NO2 + 2H2O, we can assign oxidation states to each element involved.
Starting with Sn (tin), its oxidation state changes from 0 to +4 in SnO2. This means that tin is being oxidized.
Now, let's focus on nitrogen (N) in the nitric acid (HNO3). In HNO3, nitrogen has an oxidation state of +5. In NO2, nitrogen has an oxidation state of +4. Hence, nitrogen is being reduced from +5 to +4.
So, to summarize:
- Tin (Sn) is being oxidized from an oxidation state of 0 to +4.
- Nitrogen (N) is being reduced from an oxidation state of +5 to +4.
Hope that clears it up for you!
To identify the element that is being oxidized and the one that is being reduced in a redox reaction, you need to determine the changes in oxidation states.
In the given reaction: Sn + 4HNO3 ---> SnO2 + 4NO2 + 2H2O
Notice that the oxidation state of Sn changes from 0 to +4, while the oxidation state of N changes from +5 to +4.
To see this, let's assign oxidation states to the elements in the reaction:
- The oxidation state of Sn in its elemental form is 0.
- The oxidation state of H in compounds is +1, and there are 4H atoms, so the total oxidation state of H is +4.
- In HNO3, the overall molecule is neutral, so the sum of the oxidation states must be zero. Since we have 4H atoms with an oxidation state of +1 each and one N atom whose oxidation state we are trying to find, we can set up the equation as follows:
4(+1) + X + 3(-2) = 0 (where X represents the oxidation state of N)
Simplifying the equation: 4 + X - 6 = 0
X - 2 = 0
X = +2
Therefore, the oxidation state of N in HNO3 is +5.
Now, let's consider the products:
- SnO2 is a compound, and the overall molecule is neutral, so the sum of the oxidation states must be zero. Since we have one Sn atom and two O atoms with an oxidation state of -2 each, we can set up the equation as follows:
X + 2(-2) = 0 (where X represents the oxidation state of Sn)
Simplifying the equation: X - 4 = 0
X = +4
Therefore, the oxidation state of Sn in SnO2 is +4.
- Each NO2 molecule has a total charge (oxidation state) of zero. Since we have four NO2 molecules, we need to divide the oxidation state equally among them. Therefore, each NO2 molecule has an oxidation state of +4/4 = +1. However, we are looking for the change in oxidation state of N. Since its initial oxidation state is +5 in HNO3, and now it is +1 in NO2, the change is: +5 - +1 = +4. Hence, the oxidation state of N has been reduced from +5 to +4.
Based on these calculations, the element that is being oxidized in the redox reaction is Sn (from 0 to +4), and the element that is being reduced is N (from +5 to +4).
To identify the element that is being oxidized and the element that is being reduced in a redox reaction, you can examine the changes in the oxidation states of each element.
In the given reaction:
Sn + 4HNO3 -> SnO2 + 4NO2 + 2H2O
Let's analyze the changes in oxidation states:
1. Tin (Sn): Sn is initially in its elemental form, and its oxidation state is 0. In the product SnO2, the oxidation state of Sn is +4. Therefore, the oxidation state of Sn has increased from 0 to +4. This indicates that Sn has been oxidized.
2. Nitrogen (N): Nitrogen is initially in the +5 oxidation state in nitric acid (HNO3). In the product NO2, the oxidation state of N is +4. Therefore, the oxidation state of N has decreased from +5 to +4. This indicates that N has been reduced.
It seems there was an error in the book's statement. The correct change in oxidation state for nitrogen is from +5 to +4, indicating reduction. The element being oxidized is Sn, and the element being reduced is N.
To determine the oxidation states, it is helpful to know common oxidation states for elements and the rules for assigning them. You can refer to a periodic table or an oxidation state table for reference.