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Current Question (ID: 18868)

Question:

$\text{Which of the following has the highest negative } (\text{M}^{2+}/\text{M}) \text{ standard electrode potential?}$

Options:

1. $\text{Ti}$
2. $\text{Cu}$
3. $\text{Mn}$
4. $\text{Zn}$

Solution:

$\text{HINT: Ti}^{+2}/\text{Ti reduction potential value is max negative.}$ $\text{Explanation:}$ $\text{STEP 1: The reduction potential value of given elements are as follows:}$ $\text{Mn} = -1.18\text{V} ; \text{Zn} = 0.24\text{V} ; \text{Ti} = -1.63\text{V} ; \text{Cu} = +0.34\text{V}$ $\text{STEP 2: The general trend of reduction potential for transition elements is:}$ $\text{Across the series, the tendency to form divalent cations decreases}$ $\text{indicating higher stability of the reduced state. There is no regular trend for } \text{E}^\circ.$ $\text{Hence, Ti has the lowest reduction potential value / highest negative}$ $\text{reduction potential value.}$

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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
}