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Current Question (ID: 21390)

Question:

$\text{In the given reaction sequence, the standard electrode potentials (in volts) are provided for each step as follows:}$ $\text{FeO}_4^{2-} \xrightarrow{+2.0 \text{ V}} \text{Fe}^{3+} \xrightarrow{+0.8 \text{ V}} \text{Fe}^{2+} \xrightarrow{-0.5 \text{ V}} \text{Fe}^0$ $\text{The value of } E^\Theta_{\text{FeO}_4^{2-}/\text{Fe}^{2+}} \text{ is:}$

Options:

1. $1.7 \text{ V}$
2. $1.2 \text{ V}$
3. $2.1 \text{ V}$
4. $1.4 \text{ V}$

Solution:

$\Delta rG = -nFE(\text{cell})$ $\text{FeO}_4^{2-} \xrightarrow{E_1^0=2 \text{ V}} \text{Fe}^{3+} \xrightarrow{E_2^0=0.8 \text{ V}} \text{Fe}^{2+} \xrightarrow{E_3^0=-0.5 \text{ V}} \text{Fe}$ $E_4^0 = ?$ $n_4 = 4$ $\Delta G_4^0 = \Delta G_1^0 + \Delta G_2^0$ $\Rightarrow -n_4FE_4^0 = -n_1FE_1^0 - n_2FE_2^0$ $\Rightarrow +4E_4^0 = 3 \times 2 + (1 \times 0.8)$ $\Rightarrow E_4^0 = \frac{6.8}{4} \text{ V}$ $\Rightarrow E_4^0 = 1.7 \text{ V}$

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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
}