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Current Question (ID: 21387)

Question:

$\text{Consider the given half-cell reactions:}$ $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad E^0_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.77 \text{ V}}$ $2\text{I}^- \rightarrow \text{I}_2 + 2e^- \quad E^0_{\text{I}_2/\text{I}^-} = 0.54 \text{ V}}$ $\text{The standard electrode potential for the spontaneous reaction in the}$ $\text{cell is } x \times 10^{-2} \text{V at } 298 \text{ K.}$ $\text{The value of } x \text{ is: (Nearest Integer)}$

Options:

1. 32
2. 23
3. 45
4. 17

Solution:

$\text{Hint: } E^0_{\text{Cell}} = (E^0_{\text{RP}})_C - (E^0_{\text{RP}})_A$ $E^0_{\text{Cell}} = (E^0_{\text{RP}})_C - (E^0_{\text{RP}})_A$ $= 0.77 - 0.54$ $= 0.23 \text{ V}$ $= 23 \times 10^{-2} \text{ V}$ $\text{Ans 23}$

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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
}