write an oxidation- reduction reaction for the reactants OCl- and S2O3-2
To write an oxidation-reduction (redox) reaction, we need to first determine the oxidation numbers of each element in the reactants OCl- (hypochlorite) and S2O3-2 (thiosulfate).
In OCl-, oxygen has an oxidation number of -2, and since there is one oxygen atom in the molecule, the overall oxidation number contributed by oxygen is -2. Since the overall charge of the hypochlorite ion is -1, chlorine (Cl) must have an oxidation number of +1.
In S2O3-2, since there are two oxygen atoms contributing a total of -4 to the overall charge of -2, sulfur (S) must have an oxidation number of +2. The oxidation number for oxygen is still -2, so the remaining oxygen atom has an oxidation number of -2.
Now that we have determined the oxidation numbers for each element, we can write the redox reaction. In a redox reaction, electrons are transferred from one species to another. The species that loses electrons is oxidized, while the one that gains electrons is reduced.
In this case, the oxidation state of chlorine changes from +1 in OCl- to a different oxidation state in the product, which means it is undergoing oxidation. On the other hand, the oxidation state of sulfur goes from +2 in S2O3-2 to a different oxidation state in the product, indicating that it is undergoing reduction.
Thus, the oxidation-reduction reaction for the reactants OCl- and S2O3-2 can be written as follows:
2 OCl- + S2O3-2 → 2 Cl- + 2 SO4-2
Here, chlorine is being oxidized from an oxidation state of +1 to -1, while sulfur is being reduced from +2 to +6.
It is important to note that this is a balanced equation. If you want to balance the equation yourself, you would need to ensure that the number of atoms on each side of the reaction is the same while also conserving charge.