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Current Question (ID: 18189)

Question:

$\text{The cell in which the following reactions occurs:}$ $2\text{Fe}^{3+}_{(aq)} + 2\text{I}^-_{(aq)} \rightarrow 2\text{Fe}^{2+}_{(aq)} + \text{I}_2_{(s)} \text{ has } E^\circ_{\text{cell}} = 0.236 \text{ V at } 298 \text{ K.}$ $\text{The equilibrium constant of the cell reaction is:}$

Options:

1. $9.57 \times 10^7$
2. $8.43 \times 10^6$
3. $3.68 \times 10^{-7}$
4. $1.74 \times 10^5$

Solution:

$\text{HINT: } \Delta_r G^\circ = -nFE^\circ_{\text{cell}} \text{ and } \Delta_r G^\circ = -2.303RT \log K_c$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Here, } n = 2, E^\circ_{\text{cell}} = 0.236 \text{ V, } T = 298 \text{ K}$ $\text{We know that:}$ $\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$ $= -2 \times 96487 \times 0.236$ $= -45541.864 \text{ J mol}^{-1}$ $= -45.54 \text{ kJ mol}^{-1}$ $\text{STEP 2:}$ $\text{Again, } \Delta_r G^\circ = -2.303RT \log K_c$ $\Rightarrow \log K_c = \frac{-\Delta_r G^\circ}{2.303 \cdot RT}$ $= \frac{-45.54 \times 10^3}{2.303 \times 8.314 \times 298}$ $= 7.981$ $\therefore K_c = \text{Antilog } (7.981)$ $= 9.57 \times 10^7$

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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
}