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Current Question (ID: 8614)

Question:
$\text{In the reactions given below, thiosulphate reacts differently with iodine than with bromine.}$ $2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^{-}$ $\text{S}_2\text{O}_3^{2-} + 2\text{Br}_2 + 5\text{H}_2\text{O} \rightarrow 2\text{SO}_4^{2-} + 2\text{Br}^{-} + 10\text{H}^{+}$ $\text{Choose the statements among the following that best describe the above dual behaviour of thiosulphate.}$
Options:
  • 1. $\text{Bromine is a stronger oxidant than iodine.}$
  • 2. $\text{Bromine is a weaker oxidant than iodine.}$
  • 3. $\text{Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.}$
  • 4. $\text{Bromine undergoes oxidation and iodine undergoes reduction in these reactions.}$
Solution:
$\text{Hint: Gain of electron is reduction and loss of electron is oxidation.}$ $\text{The two reactions are as follows:}$ $2\text{S}_2\text{O}_3^{2-}_{(\text{aq})} + \text{I}_{2(\text{s})} \rightarrow \text{S}_4\text{O}_6^{2-}_{(\text{aq})} + 2\text{I}^{-}_{(\text{aq})}$ $\text{S}_2\text{O}_3^{2-}_{(\text{aq})} + 2\text{Br}_{2(l)} + 5\text{H}_2\text{O}_{(l)} \rightarrow 2\text{SO}_4^{2-}_{(\text{aq})} + 2\text{Br}^{-}_{(\text{aq})} + 10\text{H}^{+}_{(\text{aq})}$ $\text{Let's analyze the change in oxidation states:}$ $\text{In the reaction with Iodine:}$ $\text{For S in } \text{S}_2\text{O}_3^{2-} \text{: oxidation state is } +2 \text{.}$ $\text{For S in } \text{S}_4\text{O}_6^{2-} \text{: oxidation state is } +2.5 \text{. (Change: } +2 \rightarrow +2.5 \text{, Oxidation)}$ $\text{For I in } \text{I}_2 \text{: oxidation state is } 0 \text{.}$ $\text{For I in } \text{I}^{-} \text{: oxidation state is } -1 \text{. (Change: } 0 \rightarrow -1 \text{, Reduction)}$ $\text{In the reaction with Bromine:}$ $\text{For S in } \text{S}_2\text{O}_3^{2-} \text{: oxidation state is } +2 \text{.}$ $\text{For S in } \text{SO}_4^{2-} \text{: oxidation state is } +6 \text{. (Change: } +2 \rightarrow +6 \text{, Oxidation)}$ $\text{For Br in } \text{Br}_2 \text{: oxidation state is } 0 \text{.}$ $\text{For Br in } \text{Br}^{-} \text{: oxidation state is } -1 \text{. (Change: } 0 \rightarrow -1 \text{, Reduction)}$ $\text{In both reactions, the halogen acts as an oxidising agent and oxidizes } \text{S}_2\text{O}_3^{2-} \text{ while itself getting reduced.}$ $\text{The change in the oxidation state of sulfur in the case of } \text{Br}_2 \text{ is from } +2 \text{ to } +6 \text{ (a change of } +4 \text{).}$ $\text{The change in the oxidation state of sulfur in the case of } \text{I}_2 \text{ is from } +2 \text{ to } +2.5 \text{ (a change of } +0.5 \text{).}$ $\text{Since bromine causes a greater increase in the oxidation state of sulfur (from } +2 \text{ to } +6 \text{) compared to iodine (from } +2 \text{ to } +2.5 \text{), this indicates that bromine is a stronger oxidizing agent than iodine.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}