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Question:
$\text{The oxidising agent and reducing agent in the given reaction are:}$ $\text{Cl}_2\text{O}_{7(g)} + 4\text{H}_2\text{O}_{2(aq)} + 2\text{OH}^{-}_{(\text{aq})} \rightarrow 2\text{ClO}_{2(aq)}^{-} + 4\text{O}_{2(g)} + 5\text{H}_2\text{O}_{(l)}$
Options:
  • 1. $\text{Oxidizing agent } = \text{H}_2\text{O}_2\text{; Reducing agent } = \text{Cl}_2\text{O}_7$
  • 2. $\text{Oxidizing agent } = \text{Cl}_2\text{O}_7\text{; Reducing agent } = \text{H}_2\text{O}_2$
  • 3. $\text{Oxidizing agent } = \text{H}_2\text{O}_2\text{; Reducing agent } = \text{H}_2\text{O}_2$
  • 4. $\text{None of the above}$
Solution:
$\text{Hint: A left-hand side and right-hand side of the reaction overall charged must be balanced.}$ $\text{Let's determine the oxidation numbers of the elements involved in the reaction:}$ $\text{Reactants:}$ $\text{Cl}_2\text{O}_7\text{:}$ $\text{Oxygen typically has an oxidation number of } -2\text{.}$ $2 \times (\text{O.N. of Cl}) + 7 \times (-2) = 0$ $2 \times (\text{O.N. of Cl}) - 14 = 0$ $2 \times (\text{O.N. of Cl}) = +14$ $\text{O.N. of Cl} = +7$ $\text{H}_2\text{O}_2 \text{ (Hydrogen Peroxide):}$ $\text{Hydrogen is } +1\text{.}$ $2 \times (+1) + 2 \times (\text{O.N. of O}) = 0$ $2 + 2 \times (\text{O.N. of O}) = 0$ $2 \times (\text{O.N. of O}) = -2$ $\text{O.N. of O} = -1$ $\text{OH}^{-}\text{:}$ $\text{Hydrogen is } +1\text{.}$ $\text{O.N. of O} + (+1) = -1$ $\text{O.N. of O} = -2$ $\text{Products:}$ $\text{ClO}_2^{-}\text{:}$ $\text{O.N. of Cl} + 2 \times (-2) = -1$ $\text{O.N. of Cl} - 4 = -1$ $\text{O.N. of Cl} = +3$ $\text{O}_2\text{:}$ $\text{O.N. of O} = 0 \text{ (elemental form)}$ $\text{H}_2\text{O}$: $\text{Hydrogen is } +1\text{.}$ $\text{Oxygen is } -2\text{.}$ $\text{Summary of changes in oxidation numbers:}$ $\text{For Cl: From } +7 \text{ in } \text{Cl}_2\text{O}_7 \text{ to } +3 \text{ in } \text{ClO}_2^{-}\text{.}$ $\text{The oxidation number of Cl decreases by 4 (from } +7 \text{ to } +3\text{). This is a reduction.}$ $\text{Therefore, } \text{Cl}_2\text{O}_7 \text{ is the oxidizing agent.}$ $\text{For O: From } -1 \text{ in } \text{H}_2\text{O}_2 \text{ to } 0 \text{ in } \text{O}_2\text{.}$ $\text{The oxidation number of O increases by 1 (from } -1 \text{ to } 0\text{). This is an oxidation.}$ $\text{Therefore, } \text{H}_2\text{O}_2 \text{ is the reducing agent.}$ $\text{The oxidation state of oxygen in } \text{OH}^{-} \text{ and } \text{H}_2\text{O} \text{ remains } -2 \text{, so they are not directly involved in the redox change.}$ $\text{Conclusion:}$ $\text{Oxidizing agent } = \text{Cl}_2\text{O}_7$ $\text{Reducing agent } = \text{H}_2\text{O}_2$

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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
}