In reaction, 2Na2S2O3 + I2 → A + 2NaI; A will be _________.
Na2S4O6
Understanding the Reaction of Sodium Thiosulfate and Iodine
The question asks to identify the product A in the given chemical reaction:
$$2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{A} + 2\text{NaI}$$
This is a classic redox reaction between sodium thiosulfate ($\text{Na}_2\text{S}_2\text{O}_3$) and iodine ($\text{I}_2$). In this reaction, sodium thiosulfate acts as a reducing agent and iodine acts as an oxidizing agent.
Let's analyze the reaction and determine the formula for product A by balancing the atoms on both sides of the equation.
The balanced equation needs to have the same number of atoms of each element on the reactant side and the product side.
Reactant side:
- Sodium (Na): $2 \times 2 = 4$ atoms
- Sulfur (S): $2 \times 2 = 4$ atoms
- Oxygen (O): $2 \times 3 = 6$ atoms
- Iodine (I): $2$ atoms
Product side (so far):
- Sodium (Na) in $2\text{NaI}$: $2 \times 1 = 2$ atoms
- Iodine (I) in $2\text{NaI}$: $2 \times 1 = 2$ atoms
Comparing the reactant and product sides, we can find the atoms that must be present in product A:
Atoms required for A = (Atoms on reactant side) - (Atoms in $2\text{NaI}$)
- Sodium (Na): $4 - 2 = 2$ atoms
- Sulfur (S): $4 - 0 = 4$ atoms
- Oxygen (O): $6 - 0 = 6$ atoms
- Iodine (I): $2 - 2 = 0$ atoms
So, product A must contain 2 sodium atoms, 4 sulfur atoms, and 6 oxygen atoms. The chemical formula for A is therefore $\text{Na}_2\text{S}_4\text{O}_6$. This compound is known as sodium tetrathionate.
The complete balanced reaction is:
$$2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI}$$
Let's check the given options:
- $\text{Na}_2\text{S}_2\text{O}_3\text{.H}_2\text{O}$ (Sodium thiosulfate pentahydrate - This is a hydrated form of the reactant)
- $\text{Na}_2\text{S}_4\text{O}_6$ (Sodium tetrathionate - Matches our derived formula for A)
- $\text{Na}_2\text{SO}_4$ (Sodium sulfate - Not formed in this specific reaction)
- $\text{Na}_2\text{SO}_3$ (Sodium sulfite - Not formed in this specific reaction)
Based on our analysis, the product A is sodium tetrathionate, $\text{Na}_2\text{S}_4\text{O}_6$.
Important Questions from Redox Reactions
Nitric acid is reduced to nitrogen dioxide in which of the following reactions?
The role of H3PO4 in the estimation of Fe(II) with K2Cr2O7 using diphenylamine sulphonate as indicator is to
- In the following redox reaction, which compound is acting as a reducing agent ?
$N_2H_4 + 2H_2O_2 \rightarrow N_2 + 4H_2O$ Number of electrons (x) involved in the following reaction is :
\(N \equiv N+ 8H^{+}+x \rightarrow 2N H^{3}+ H_{2}\)- Consider the following pairs of reactions and properties :
Reactions Properties I. $\text{Fe} \, (s) + 2\text{HCl} \, (aq) \rightarrow \text{FeCl}_2 \, (aq) + \text{H}_2 \, (g)$ $\text{Fe}$ is reductant II. $\text{Zn} \, (s) + \text{CuSO}_4 \, (aq) \rightarrow \text{ZnSO}_4 \, (aq) + \text{Cu} \, (s)$ $\text{Cu}$ is being reduced III. $\text{Br}_2 \, (l) + 2\text{I}^- \, (aq) \rightarrow 2\text{Br}^- \, (aq) + \text{I}_2 \, (s)$ $\text{Br}_2$ is reductant IV. $\text{CH}_4 \, (g) + 2\text{O}_2 \, (g) \rightarrow \text{CO}_2 \, (g) + 2\text{H}_2\text{O} \, (l)$ $\text{C}$ atom in $\text{CH}_4$ is oxidized
Which of the above pairs are correct?
