In S2O82-, each S atom has an oxidation number of +5. In SO42-, the oxidation number of S is also +5. There is no change in oxidation number for sulfur when transitioning from S2O82- to SO42-.
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.
None do. Let's look at each compound: BiCl2... this is Bi2+ and Cl- Na2SO4... this is Na+ and SO42- NaCl... this is Na+ and Cl- BiSO4... this is Bi2+ and SO42- So as you can see, nobody is changing its charge, so everyone's oxidation number stayed constant. Of course I am a bit confused by the fact that Bismuth does not ever form a +2 ion (only +3 and +5), so I would have to assume you meant Ba2+. But it would not change what I wrote up there. It would simply be a precipition reaction, not an oxidation-reduction 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 each K is +1. But there are two K's so 1*2 is +2. the charge on the molecule has to equal 0, so the oxidation number of the S must be -2.
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-
None do. Let's look at each compound: BiCl2... this is Bi2+ and Cl- Na2SO4... this is Na+ and SO42- NaCl... this is Na+ and Cl- BiSO4... this is Bi2+ and SO42- So as you can see, nobody is changing its charge, so everyone's oxidation number stayed constant. Of course I am a bit confused by the fact that Bismuth does not ever form a +2 ion (only +3 and +5), so I would have to assume you meant Ba2+. But it would not change what I wrote up there. It would simply be a precipition reaction, not an oxidation-reduction 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 each K is +1. But there are two K's so 1*2 is +2. the charge on the molecule has to equal 0, so the oxidation number of the S must be -2.
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.