What is the overall stoichiometry for the reaction between iodide and peroxydisulfate ions?
S2O82-(aq) + 2I-(aq)==> 2SO42-(aq) + I2(aq)
To determine the overall stoichiometry of a reaction between iodide (I-) and peroxydisulfate ions (S2O8 2-), we need to know the balanced chemical equation for the reaction. The balanced equation shows the molar ratios between reactants and products.
1. Start by writing the chemical equation for the reaction:
I- + S2O8 2- ⟶ Products
2. Determine the oxidation state of each element in the reactants and products. In this case, Iodide (I-) has an oxidation state of -1, and peroxydisulfate (S2O8 2-) has an oxidation state of +6.
3. Since iodide is being oxidized and peroxydisulfate is being reduced, we can write the following half-reactions:
I- ⟶ I2 (oxidation, gain of electrons)
S2O8 2- ⟶ 2SO4 2- (reduction, loss of electrons)
4. Balance the half-reactions to ensure the same number of atoms on both sides:
I- ⟶ I2 + 2e-
S2O8 2- + 2e- ⟶ 2SO4 2-
5. To combine the two half-reactions, we need to multiply them by appropriate coefficients to ensure the electrons cancel out:
2I- + S2O8 2- ⟶ I2 + 2SO4 2-
Therefore, the overall stoichiometry for the reaction between iodide and peroxydisulfate ions is 2:1. For every 2 moles of iodide (I-) and 1 mole of peroxydisulfate ions (S2O8 2-), we get 1 mole of iodine (I2) and 2 moles of sulfate ions (SO4 2-).