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Current Question (ID: 18185)

Question:

$\text{Given cell:}$ $\text{Pt(s)}|\text{H}_2(g)(P = 1 \text{ atm})| \text{CH}_3\text{COOH}(0.1 \text{ M}), \text{HCl}(0.1 \text{ M}) || \text{KCl(aq)}|\text{Hg}_2\text{Cl}_2(s)|\text{Hg}$ $\text{EMF of the cell is found to be } 0.045 \text{ V at } 298 \text{ K and the temperature coefficient is } 3.4 \times 10^{-4} \text{ V K}^{-1}. \text{ The cell entropy change of the following cell is:}$ $\text{[Given } K_a \text{ CH}_3\text{COOH} = 10^{-5} \text{ M]}$

Options:

1. $70.8 \text{ J K}^{-1} \text{ mol}^{-1}$
2. $65.2 \text{ J K}^{-1} \text{ mol}^{-1}$
3. $79.2 \text{ J K}^{-1} \text{ mol}^{-1}$
4. $83.5 \text{ J K}^{-1} \text{ mol}^{-1}$

Solution:

$\text{HINT: } \Delta S = nF \times \text{Temp coefficient}$ $\text{Explanation:}$ $\text{Considering the cell reaction}$ $\text{H}_2 \rightarrow 2\text{H}^+ + 2\text{e}^- \quad \text{Hg}_2^{2+} + 2\text{e}^- \rightarrow \text{Hg} \quad (1) + \text{Cl}^- \quad (\text{aq})$ $n = 2$ $\Delta S = nF \times \text{Temp coefficient} = 2 \times 96500 \times 3.4 \times 10^{-4}$ $= 65.223 \text{ J/K/mole}$

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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
}