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Current Question (ID: 21377)

Question:

$\text{Consider the following cell representation:}$ $\text{Pt/H}_2(1 \text{ atm})/\text{H}^+(1 \text{ M}) \parallel \text{Fe}^{3+}/\text{Fe}^{2+}$ $\text{The ratio of the concentration of Fe}^{2+} \text{ to Fe}^{3+} \text{ is-}$ $[\text{Given } E_{\text{cell}} = 0.712, E^0_{\text{cell}} = 0.771]$

Options:

1. $12$
2. $10$
3. $16$
4. $8$

Solution:

$\frac{1}{2} \text{H}_2(g) + \text{Fe}^{3+}(aq) \rightarrow \text{H}^+(aq) + \text{Fe}^{2+}(aq)$ $E = E^\circ - \frac{0.059}{1} \log \left[ \frac{\text{Fe}^{2+}}{\text{Fe}^{3+}} \right]$ $\Rightarrow 0.712 = (0.771 - 0) - \frac{0.059}{1} \log \left[ \frac{\text{Fe}^{2+}}{\text{Fe}^{3+}} \right]$ $\Rightarrow \log \left[ \frac{\text{Fe}^{2+}}{\text{Fe}^{3+}} \right] = \frac{(0.771 - 0.712)}{0.059} = 1$ $\Rightarrow \left[ \frac{\text{Fe}^{2+}}{\text{Fe}^{3+}} \right] = 10$

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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
}