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Current Question (ID: 18254)

Question:

$\text{The most stable oxidized species among the following is:}$ $E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \text{ V};$ $E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \text{ V}$ $E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51 \text{ V};$ $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \text{ V}$

Options:

1. $\text{Cr}^{3+}$
2. $\text{MnO}_4^-$
3. $\text{Cr}_2\text{O}_7^{2-}$
4. $\text{Mn}^{2+}$

Solution:

$\text{HINT: Lower the value of reduction potential, the higher will be the stability of oxidized species.}$ $\text{Explanation:}$ $E^\circ_{\text{Cr}^{3+}/\text{Cr}} \text{ has most - ve value equal to -0.74 among the given four choices.}$ $\text{So, Cr}^{3+} \text{ is the most stable oxidized species.}$

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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
}