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Current Question (ID: 18188)

Question:

$\text{The equilibrium constant of a 2 electron redox reaction at 298 K is } 3.8 \times 10^{-3}. \text{ The cell potential } E^\circ \text{ (in V) and the free energy change } \Delta G^\circ \text{ (in kJ mol}^{-1}\text{) for this equilibrium respectively, are -}$

Options:

1. $-0.071, -13.8$
2. $-0.071, 13.8$
3. $0.71, -13.8$
4. $0.071, -13.8$

Solution:

$\text{HINT: } \Delta G^\circ = -nFE^\circ = -RT \ln K$ $\text{STEP 1:}$ $\Delta G^\circ = -RT \ln K$ $= -8.3 \text{ J/mol} \times 298 \text{ K} \ln (3.8 \times 10^{-3})$ $= 13.786 \text{ kJ/mol}$ $\text{STEP 2:}$ $\Delta G^\circ = -2FE^\circ$ $\frac{-\Delta G^\circ}{2F} = E^\circ$ $E^\circ = \frac{13.768 \times 10^3}{2 \times 96500}$ $E^\circ = -0.071$

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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
}