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Current Question (ID: 18191)

Question:

$\text{A fuel cell develops an electrical potential from the combustion of butane at 1 bar and 298 K}$ $\text{C}_4\text{H}_{10}(g) + 6.5\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 5\text{H}_2\text{O}(l); \text{E}^\circ \text{ of a cell is:}$ $\text{(Given } \Delta G^\circ = -2746 \text{ kJ/mole)}$

Options:

1. $4.74 \text{ V}$
2. $0.547 \text{ V}$
3. $4.37 \text{ V}$
4. $1.09 \text{ V}$

Solution:

$\text{HINT: } \Delta G^\circ = -nF\text{E}^\circ$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{To find } n \text{ we break the cell reaction into two half cell reduction.}$ $\text{Anode:}$ $13\text{H}_2\text{O}(l) \rightarrow 6.5\text{O}_2(g) + 26\text{H}^+(aq.) + 26e^-$ $\text{Cathode:}$ $4(\text{CO}_2)(g) + 26\text{H}^+(aq.) + 26e^- \rightarrow \text{C}_4\text{H}_{10}(g) + 8\text{H}_2\text{O}(l)$ $\text{STEP 2:}$ $\text{E}^\circ = \frac{-\Delta G^\circ}{nF} = \frac{(-2746) \times 1000}{26 \times 96500} = 1.09\text{V}$

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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
}