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Current Question (ID: 18148)

Question:

$\text{Given the standard electrode potentials:}$ $\begin{array}{l} \text{K}^+/\text{K} = -2.93\, \text{V} \\ \text{Ag}^+/\text{Ag} = 0.80\, \text{V} \\ \text{Hg}^{2+}/\text{Hg} = 0.79\, \text{V} \\ \text{Mg}^{2+}/\text{Mg} = -2.37\, \text{V} \\ \text{Cr}^{3+}/\text{Cr} = -0.74\, \text{V} \end{array}$ $\text{The correct increasing order of reducing power of the metals is:}$

Options:

1. $\text{Cr} < \text{Mg} < \text{K} < \text{Ag} < \text{Hg}$
2. $\text{Mg} < \text{K} < \text{Ag} < \text{Hg} < \text{Cr}$
3. $\text{K} < \text{Ag} < \text{Hg} < \text{Cr} < \text{Mg}$
4. $\text{Ag} < \text{Hg} < \text{Cr} < \text{Mg} < \text{K}$

Solution:

$\text{HINT: The lower the reduction potential, the higher is the reducing power.}$ $\text{Explanation:}$ $\text{The given standard electrode potentials increase in the order of}$ $\text{K}^+/\text{K} < \text{Mg}^{2+}/\text{Mg} < \text{Cr}^{3+}/\text{Cr} < \text{Hg}^{2+}/\text{Hg} < \text{Ag}^+/\text{Ag}.$ $\text{Hence, the reducing power of the given metals increases in the following order:}$ $\text{Ag} < \text{Hg} < \text{Cr} < \text{Mg} < \text{K}$

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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
}