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Current Question (ID: 8627)

Question:

$\text{Based on standard electrode potentials given above, the correct arrangement for increasing order of reducing power of elements is:}$ $\text{(a) } E^o_{K^+/K} = -2.93 \text{ V}; E^o_{Ag^+/Ag} = 0.80 \text{ V}$ $\text{(b) } E^o_{Hg^{2+}/Hg} = 0.79 \text{ V}; E^o_{Mg^{2+}/Mg} = -2.37 \text{ V}$ $\text{(c) } E^o_{Cr^{3+}/Cr} = -0.74 \text{ V}$

Options:

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

Solution:

$\text{Hint: Reducing and oxidising power depends on reduction potential.}$ $\text{The lower the electrode potential, the stronger is the reducing agent. More positive the reduction potential value, more easily it can get reduced and oxidise other species.}$ $\text{The increasing order of reducing power is : Ag < Hg < Cr < Mg < K}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}