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Current Question (ID: 8631)

Question:

$\text{A solution contains Fe}^{2+}, \text{Fe}^{3+} \text{ and I}^- \text{ ions. This solution was treated with iodine at 35}^\circ\text{C. E}^\circ \text{ for Fe}^{3+}/\text{Fe}^{2+} \text{ is +0.77 V and E}^\circ \text{ for I}_2/2\text{I}^- = 0.536\text{ V.}$ $\text{The favourable redox reaction is:}$

Options:

1. $\text{Fe}^{2+} \text{ will be oxidized to Fe}^{3+}.$
2. $\text{I}_2 \text{ will be reduced to I}^-.$
3. $\text{There will be no redox reaction.}$
4. $\text{I}^- \text{ will be oxidized to I}_2.$

Solution:

$\text{HINT: Write down the reduction half-cell and oxidation half-cell.}$ $\text{The reaction will be favourable when Emf of cell is positive. }(\Delta G^\circ = -n\text{FE}_{\text{cell}}^\circ)$ $\text{The reduction potential of Fe}^{3+}/\text{Fe}^{2+} \text{ is +0.770 V and for I}_2/2\text{I}^- = 0.536\text{ V.}$ $\text{Because of the more reduction potential Fe}^{+3} \text{ will get reduced to Fe}^{+2} \text{ and I}^- \text{ will get oxidised to I}_2.$ $\text{So, the cell reaction will be,}$ $2\text{Fe}^{+3} + 2\text{I}^- \rightarrow \text{I}_2 + 2\text{Fe}^{+2}$ $\text{E}_{\text{cell}}^\circ = \text{E}_{\text{Fe}^{+3}/\text{Fe}^{+2}}^\circ + \text{E}_{2\text{I}^-/\text{I}_2}^\circ$ $= 0.770 - 0.536$ $= 0.234\text{ V}$ $\text{Since E}_{\text{cell}}^\circ \text{ is positive, the reaction is thermodynamically favourable.}$ $\text{Therefore, I}^- \text{ will be oxidized to I}_2.$

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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
}