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Current Question (ID: 18236)

Question:

$\text{During electrolysis of conc. } \text{H}_2\text{SO}_4, \text{ perdisulphuric acid } (\text{H}_2\text{S}_2\text{O}_8), \text{ and } \text{O}_2 \text{ form in equimolar amount. The amount of } \text{H}_2 \text{ that will form simultaneously will be:}$

Options:

1. $\text{Thrice that of } \text{O}_2 \text{ in moles.}$
2. $\text{Twice that of } \text{O}_2 \text{ in moles.}$
3. $\text{Equal to that of } \text{O}_2 \text{ in moles.}$
4. $\text{Half of that of } \text{O}_2 \text{ in moles.}$

Solution:

$\text{HINT: } 2\text{H}_2\text{SO}_4 + 8\text{H}_2\text{O} \rightarrow \text{H}_2\text{S}_2\text{O}_8 + \text{O}_2 + 3\text{H}_2 + 6\text{H}^+ + 6\text{OH}^- $ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Anode:}$ $2\text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_8 + 2\text{H}^+ + 2\text{e}^- $ $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^- $ $\text{Cathode: } (2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^-) \times 3.$ $\text{STEP 2:}$ $\text{Overall reaction:}$ $2\text{H}_2\text{SO}_4 + 8\text{H}_2\text{O} \rightarrow \text{H}_2\text{S}_2\text{O}_8 + \text{O}_2 + 3\text{H}_2 + 6\text{H}^+ + 6\text{OH}^- $ $\text{The ratio of } \text{H}_2 \text{ and } \text{O}_2 \text{ is } 1:3.$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}