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Current Question (ID: 8639)

Question:

$\text{From the following, identify the reaction having the top position in the EMF series (standard reduction potential) according to their electrode potential at 298 K.}$

Options:

1. $\text{Mg}^{2+} + 2\text{e}^- \rightarrow \text{Mg}_{(\text{s})}$
2. $\text{Fe}^{2+} + 2\text{e}^- \rightarrow \text{Fe}_{(\text{s})}$
3. $\text{Au}^{3+} + 3\text{e}^- \rightarrow \text{Au}_{(\text{s})}$
4. $\text{K}^+ + \text{e}^- \rightarrow \text{K}_{(\text{s})}$

Solution:

$\text{According to electrode potential series, Au ion is having higher tendency to get converted into atomic form, that's why they are known as inert metal}$ $\text{From the electrode potential table provided:}$ $\text{F}_2 + 2\text{e}^- \rightarrow 2\text{F}^- \text{ has E}^0 = +2.87 \text{ V (topmost)}$ $\text{Au}^{3+} + 3\text{e}^- \rightarrow \text{Au} \text{ has E}^0 = +1.50 \text{ V}$ $\text{Fe}^{2+} + 2\text{e}^- \rightarrow \text{Fe} \text{ has E}^0 = -0.440 \text{ V}$ $\text{Among the given options, Au}^{3+}/\text{Au has the highest standard reduction potential (+1.50 V)}$ $\text{Higher the reduction potential, higher is the tendency to get reduced and occupy top position in EMF series}$ $\text{Therefore, Au}^{3+} + 3\text{e}^- \rightarrow \text{Au}_{(\text{s})} \text{ has the topmost position among the given reactions}$

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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
}