What are the oxidation numbers for sulfur in the half reaction SO2 --> H2SO4 ?
+4 on the left; +6 on the right.
http://www.chemteam.info/Redox/Redox-Rules.html
Ok that's what I thought. Thank you!
To determine the oxidation numbers for sulfur in the half reaction SO2 --> H2SO4, we need to analyze the changes in the oxidation states of sulfur.
1. Start with the given compound, SO2. Oxygen (O) generally has an oxidation number of -2, and since there are two oxygen atoms, the total oxidation number from oxygen is -4.
2. Assign an oxidation number to oxygen of -2, since this is the most common oxidation state for oxygen.
3. Since the compound is electrically neutral, the sum of the oxidation numbers of sulfur and oxygen must be zero. Therefore, the oxidation number of sulfur can be calculated by solving the equation:
Oxidation number of sulfur + 2 × Oxidation number of oxygen = 0
Oxidation number of sulfur + 2 × -2 = 0
Oxidation number of sulfur = +4
So, in the compound SO2, the oxidation number of sulfur is +4.
4. Next, let's consider the product compound, H2SO4. Hydrogen (H) generally has an oxidation number of +1, and there are two hydrogen atoms, so the total oxidation number from hydrogen is +2.
5. Assign an oxidation number to hydrogen of +1, since this is the most common oxidation state for hydrogen.
6. Oxygen (O) generally has an oxidation number of -2, and since there are four oxygen atoms, the total oxidation number from oxygen is -8.
7. Since the compound is electrically neutral, the sum of the oxidation numbers of sulfur, hydrogen, and oxygen must be zero. Therefore, the oxidation number of sulfur can be calculated by solving the equation:
Oxidation number of sulfur + 2 × Oxidation number of hydrogen + 4 × Oxidation number of oxygen = 0
Oxidation number of sulfur + 2 × +1 + 4 × -2 = 0
Oxidation number of sulfur = +6
So, in the compound H2SO4, the oxidation number of sulfur is +6.
Hence, the oxidation numbers for sulfur in the half reaction SO2 --> H2SO4 are +4 and +6.