Question

The thiosulfate ion \(\left(S_{2} O_{3}^{2-}\right)\) is oxidized by iodine as follows: $$2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+2 \mathrm{I}^{-}(a q)$$ In a certain experiment, \(7.05 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} .\) Calculate the rate of production of iodide ion.

Short answer

The rate of consumption of the thiosulfate ion (S₂O₃²⁻) is -6.41 x 10⁻⁴ mol/(L·s) and the rate of production of the iodide ion (I⁻) is 6.41 x 10⁻⁴ mol/(L·s).

Step-by-step solution

Calculate the rate of consumption for the thiosulfate ion(S₂O₃²⁻)

The rate of consumption is calculated as the change in concentration over time. \[Rate = \frac{Δ[S_{2}O_{3}^{2-}]}{Δt}\] Where Δ[S₂O₃²⁻] is the change in concentration of the thiosulfate ion, and Δt is the change in time. Since the concentration is consumed, the change in concentration, Δ[S₂O₃²⁻], will be negative. The Initial concentration of S₂O₃²⁻ = 7.05 × 10⁻³ mol/L Δt = 11.0 seconds We have the reaction: \[2S_{2}O_{3}^{2-}(aq) + I_{2}(aq) \longrightarrow S_{4}O_{6}^{2-}(aq) + 2I^{-}(aq)\] Now, let's plug in the values and calculate the rate of consumption of S₂O₃²⁻. \[Rate_{S_{2}O_{3}^{2-}} = \frac{-7.05 \times 10^{-3}\,\text{mol/L}}{11.0\,\text{s}} = -6.41 \times 10^{-4}\,\frac{\text{mol}}{\text{L} \cdot \text{s}}\]

Calculate the rate of production for the iodide ion (I⁻)

From the balanced reaction equation, we can observe the stoichiometric relationship between S₂O₃²⁻ and I⁻: \[2S_{2}O_{3}^{2-}(aq) \longrightarrow 2I^{-}(aq)\] Notice that the molar ratio between S₂O₃²⁻ and I⁻ in the reaction is 2:2 or 1:1. This means that for every 1 mole of S₂O₃²⁻ reacted, 1 mole of I⁻ is produced. Using this stoichiometric relationship, the rate of production for the iodide ion can be determined as follows: \[Rate_{I^{-}} = \frac{1}{2} \times Rate_{S_{2}O_{3}^{2-}}\] By plugging the previously calculated rate of consumption for S₂O₃²⁻, we can determine the rate of production for I⁻: \[Rate_{I^{-}} = \frac{1}{2} \times \left(-6.41 \times 10^{-4}\,\frac{\text{mol}}{\text{L} \cdot \text{s}}\right) = 6.41 \times 10^{-4}\,\frac{\text{mol}}{\text{L} \cdot \text{s}}\] The rate of production of I⁻ is 6.41 x 10⁻⁴ mol/(L·s).

Key concepts

Thiosulfate Ion

Thiosulfate ions, represented as \( S_{2}O_{3}^{2-} \), are known for their ability to undergo oxidation in various chemical reactions. In the reaction given, thiosulfate ions are consumed as they react with iodine \(I_2\) to form tetrathionate ions \(S_4O_6^{2-}\) and iodide ions \(I^-\). The thiosulfate ion can act as a reducing agent, which means it donates electrons to other substances, such as iodine in this reaction.
In the realm of chemistry, thiosulfate ions are particularly interesting because they serve as a central component in processes like photographic fixing and in determining the concentration of iodine in solutions. Their role as a reducing agent can be observed and measured through the rate at which they are consumed in reactions, providing insight into the dynamics of chemical interactions.

Iodide Ion

Iodide ions \((I^-)\) are the products of the reduction of iodine \((I_2)\) in this reaction. In the oxidation-reduction (redox) process, iodine molecules gain electrons, which are donated by the thiosulfate ions, transforming into iodide ions. This transformation is significant in many applications, including iodine-based titration and biological processes.
Understanding the production of iodide ions is crucial, especially since they serve as indicators of reaction completion in various analytical procedures. In the discussed reaction, calculating the rate of iodide ion production helps demonstrate the relationship between the reactants and products. It also showcases how products are generated over time in response to reactant consumption, offering a clear picture of the reaction kinetics.

Stoichiometry

Stoichiometry plays a fundamental role in understanding and calculating the quantities involved in chemical reactions. In the context of the provided equation, stoichiometry refers to the precise molar relationships between the reactants and products. Here, the balanced chemical equation shows a 2:2 ratio between the thiosulfate ions \((S_{2}O_{3}^{2-})\) and iodide ions \((I^-)\).
By observing this stoichiometric ratio, we can determine that for every 2 moles of thiosulfate consumed, 2 moles of iodide ions are produced. This one-to-one relationship simplifies calculations, allowing us to directly relate the rate of consumption of thiosulfate ions to the rate of production of iodide ions. Grasping stoichiometry is essential for chemists to predict product formation and optimize conditions for maximum yield in reactions.

Oxidation-Reduction

Oxidation-reduction reactions, or redox reactions, involve the transfer of electrons between substances, altering their oxidation states. In the discussed reaction, thiosulfate ions are oxidized; they lose electrons, while iodine is reduced; it gains electrons.
These processes occur simultaneously, establishing the principle of conservation of charge. Oxidation can be remembered as an increase in oxidation state, while reduction leads to a decrease. In our reaction, iodine's transformation into iodide illustrates reduction, as it gains electrons and decreases in oxidation state. Meanwhile, thiosulfate's conversion shows oxidation, contributing electrons to the iodine.

• Oxidation: Loss of electrons (Thiosulfate becomes tetrathionate)
• Reduction: Gain of electrons (Iodine becomes iodide)

Understanding redox reactions are vital for fields ranging from energy storage and production to industrial synthesis and biological systems.