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Current Question (ID: 18239)

Question:

$\text{Pairs of reactants (R) and product (P) are given below. The pair which can be used in a fuel cell is:}$ $1. \ R = \text{H}_2(g), \ \text{O}_2(g); \ P = \text{H}_2\text{O}_2(g)$ $2. \ R = \text{H}_2(g), \ \text{O}_2(g); \ P = \text{H}_2\text{O}(\ell)$ $3. \ R = \text{H}_2(g), \ \text{O}_2(g), \ \text{Cl}_2(g); \ P = \text{HClO}_4(aq)$ $4. \ R = \text{H}_2(g), \ \text{N}_2(g); \ P = \text{NH}_3(aq)$

Options:

1. $R = \text{H}_2(g), \ \text{O}_2(g); \ P = \text{H}_2\text{O}_2(g)$
2. $R = \text{H}_2(g), \ \text{O}_2(g); \ P = \text{H}_2\text{O}(\ell)$
3. $R = \text{H}_2(g), \ \text{O}_2(g), \ \text{Cl}_2(g); \ P = \text{HClO}_4(aq)$
4. $R = \text{H}_2(g), \ \text{N}_2(g); \ P = \text{NH}_3(aq)$

Solution:

$\text{HINT: Water is obtained in fuel cell.}$ $\text{Explanation:}$ $\text{Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells.}$ $\text{The electrode reactions are given below:}$ $\text{Cathode:}$ $\text{O}_2 \ (g) + 2\text{H}_2\text{O}(\ell) + 4e^- \rightarrow 4\text{OH}^-(aq)$ $\text{Anode:}$ $2\text{H}_2 \ (g) + 4\text{OH}^-(aq) \rightarrow 4\text{H}_2\text{O}(\ell) + 4e^-$

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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
}