As SO42- has an overall -2 charge, The Oxygen has a -2 oxidation state, so to balance and give an overall -2 charge, the Sulfur has to have a +6 oxidation state. (-2 x 4) + (s) = -2 s = +6
In an iodometric titration experiment, the oxidation number of sulfur changes from -2 in the thiosulfate ion (S2O32-) to +4 in the sulfate ion (SO42-) as sulfur gains oxygen atoms. This change indicates the transfer of electrons and oxidation of sulfur during the reaction.
The element that decreases its oxidation number in this reaction is iodine (I). In the reactant bisulfate ion (HSO4-), iodine is assigned an oxidation state of +7 in bismuthate ion (BiSO4-). In the product sodium bisulfate (NaHSO4), iodine is assigned an oxidation state of +5 in sodium iodide (NaI). Thus, iodine undergoes a decrease in its oxidation number in this reaction.
To determine the oxidation number of sulfur (S) in the polyatomic ion S4O6^2-, we can set up an equation where the sum of the oxidation numbers equals the charge of the ion. In this case, the total charge is -2. Each oxygen atom has an oxidation number of -2, so the total oxidation number contributed by oxygen is -12. To solve for sulfur, we set up the equation: 4x + 6(-2) = -2, where x is the oxidation number of sulfur. By solving this equation, we find that the oxidation number of sulfur in S4O6^2- is +5.
The oxidation number of potassium (K) in K2SO4 is +1. This is because alkali metals like potassium typically have an oxidation number of +1 in compounds.
As SO42- has an overall -2 charge, The Oxygen has a -2 oxidation state, so to balance and give an overall -2 charge, the Sulfur has to have a +6 oxidation state. (-2 x 4) + (s) = -2 s = +6
In an iodometric titration experiment, the oxidation number of sulfur changes from -2 in the thiosulfate ion (S2O32-) to +4 in the sulfate ion (SO42-) as sulfur gains oxygen atoms. This change indicates the transfer of electrons and oxidation of sulfur during the reaction.
The oxidation state of S in SO42- is +6. Each oxygen atom has an oxidation state of -2, so the total charge of -2 for sulfate ion requires sulfur to have an oxidation state of +6 to balance the charge.
In this ion the oxidation state of sulfur is 6+ and the oxidation state of each oxygen is 2-
The element that decreases its oxidation number in this reaction is iodine (I). In the reactant bisulfate ion (HSO4-), iodine is assigned an oxidation state of +7 in bismuthate ion (BiSO4-). In the product sodium bisulfate (NaHSO4), iodine is assigned an oxidation state of +5 in sodium iodide (NaI). Thus, iodine undergoes a decrease in its oxidation number in this reaction.
To determine the oxidation number of sulfur (S) in the polyatomic ion S4O6^2-, we can set up an equation where the sum of the oxidation numbers equals the charge of the ion. In this case, the total charge is -2. Each oxygen atom has an oxidation number of -2, so the total oxidation number contributed by oxygen is -12. To solve for sulfur, we set up the equation: 4x + 6(-2) = -2, where x is the oxidation number of sulfur. By solving this equation, we find that the oxidation number of sulfur in S4O6^2- is +5.
The oxidation number of potassium (K) in K2SO4 is +1. This is because alkali metals like potassium typically have an oxidation number of +1 in compounds.
The balanced redox reaction in acid solution is: 6 FeSO4 + Cr2O7^2- + 14 H+ -> 3 Fe2(SO4)3 + 2 Cr^3+ + 7 H2O
The oxidation state of sulfur in SO4^2- is +6. This is because oxygen typically has an oxidation state of -2, and there are 4 oxygen atoms in SO4^2-. Since the overall charge of the ion is -2, the oxidation state of sulfur must be +6 to balance the charges.
In FeSO4, the oxidation number of Fe is +2, and the oxidation number of oxygen is -2. To find the oxidation number of S, we can set up an equation: 2(-2) + x + 4(-2) = 0 (overall charge of the compound is zero). Solving this equation gives the oxidation number of S as +6.
The sulfate ion is SO42 -. The oxidation state of the sulfur is +6 (work it out!); therefore, the ion is more properly named the sulfate(VI) ion. The sulfite ion is SO32-. The oxidation state of the sulfur is +4.
The thermal decomposition of ammonium persulfate involves a redox reaction where the persulfate ion (S2O82-) breaks down into sulfate ions (SO42-) and oxygen gas (O2), releasing energy in the form of heat. This reaction is usually initiated by heat or other suitable energy sources.