To determine the mole ratio of Na2TeO3, NaI, and HCl in the balanced equation for the oxidation of iodide ions (I-) to iodine (I2) by sodium tellurite (Na2TeO3) in the presence of hydrochloric acid (HCl), we first need to write out the balanced chemical equation for the reaction.
Step-by-Step Balancing of the Reaction
The overall reaction can be represented as follows:
- Oxidation half-reaction: 2 I- → I2 + 2 e-
- Reduction half-reaction: Na2TeO3 + 2 e- → Te + 2 Na+ + 3 O
When combining these half-reactions, we need to ensure that the electrons lost in the oxidation half-reaction equal the electrons gained in the reduction half-reaction. In this case, we can see that 2 moles of iodide ions are oxidized, which corresponds to 1 mole of iodine produced.
Incorporating HCl into the Reaction
Hydrochloric acid (HCl) is necessary for providing the acidic medium and also for balancing the charges. The presence of HCl will contribute to the formation of NaCl as a byproduct. The complete balanced equation can be written as:
Na2TeO3 + 2 NaI + 6 HCl → Te + 3 I2 + 2 NaCl + 3 H2O
Determining the Mole Ratios
From the balanced equation, we can extract the mole ratios of the reactants:
- 1 mole of Na2TeO3
- 2 moles of NaI
- 6 moles of HCl
Thus, the mole ratio of Na2TeO3 : NaI : HCl is:
1 : 2 : 6
Understanding the Reaction Dynamics
This reaction showcases a redox process where iodide ions are oxidized to molecular iodine, while sodium tellurite is reduced to elemental tellurium. The acidic environment provided by HCl is crucial for facilitating the reaction and ensuring that the products are formed correctly. The stoichiometry reflects the conservation of mass and charge, which is fundamental in chemical reactions.
In summary, the balanced equation and the derived mole ratios not only help in understanding the stoichiometry of the reaction but also illustrate the interplay between oxidation and reduction processes in a chemical system.