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Current Question (ID: 21360)

Question:
$\text{An oxidation-reduction reaction in which 3 electrons are transferred has a } \Delta G^\circ \text{ of 17.37 kJ mol}^{-1} \text{ at 25}^\circ\text{C. The value of } E^\circ_{\text{cell}} \text{ (in V) is A} \times 10^{-2}. \text{ The value of A is-}$ $\text{(1 F = 96,500 C mol}^{-1})$
Options:
  • 1. $-6$
  • 2. $4$
  • 3. $-8$
  • 4. $2$
Solution:
$\text{Hint: } \Delta G^\circ = -nF E^\circ_{\text{cell}}$ $\text{Step 1:}$ $\text{The relation between } \Delta G^\circ \text{ and } E^\circ_{\text{cell}} \text{ is as follows:}$ $\Delta G^\circ = -nF E^\circ_{\text{cell}}$ $\text{The given values are as follows:}$ $n = 3$ $\Delta G^\circ = 17.37 \text{ kJ mol}^{-1} = 17.37 \times 10^3 \text{ J mol}^{-1}$ $F = 96,500 \text{ C mol}^{-1}$ $\text{Step 2:}$ $\text{Calculate the value of } E^\circ_{\text{cell}} \text{ as follows:}$ $17.37 \times 10^3 = -3 \times 96500 \times E^\circ_{\text{cell}}$ $E^\circ_{\text{cell}} = -6 \times 10^{-2} \text{ V}$ $\text{Thus, the value of A is } -6$

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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
}