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Current Question (ID: 18249)

Question:

$\text{The standard reduction electrode potentials of three metals X, Y, and Z are given as } -1.2 \text{ V}, +0.5 \text{ V}, \text{ and } -3.0 \text{ V}, \text{ respectively. Based on these values, what is the correct order of their reducing power?}$

Options:

1. $Y > X > Z$
2. $Z > X > Y$
3. $X > Y > Z$
4. $Y > Z > X$

Solution:

$\text{Reduction potential } \propto \frac{1}{\text{Reducing power}}$ $\text{Explanation:}$ $E^\circ_X = -1.2 \text{ V}, E^\circ_Y = +0.5 \text{ V}, E^\circ_Z = -3 \text{ V},$ $\therefore \text{Reducing power} = Z > X > Y \; [\because \text{ Higher the reduction potential, lesser the reducing power.}]$

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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
}