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Current Question (ID: 18228)

Question:

$\text{Al}_2\text{O}_3 \text{ is reduced by electrolysis at low potentials and high currents. If } 4.0 \times 10^4 \text{ A of current is passed through molten } \text{Al}_2\text{O}_3 \text{ for 6 hours, the mass of aluminum produced is:}$ $\text{(Assume 100 \% current efficiency, the atomic mass of Al = 27 g mol}^{-1}\text{)}$

Options:

1. $9.0 \times 10^3 \text{ g}$
2. $8.1 \times 10^4 \text{ g}$
3. $2.4 \times 10^5 \text{ g}$
4. $1.3 \times 10^4 \text{ g}$

Solution:

$\text{HINT: Faraday's first law of electrolysis.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Al}_2\text{O}_3 \text{ ionises as,}$ $\text{Al}_2\text{O}_3 \rightarrow \text{Al}^{3+} + \text{AlO}_3^{3-}$ $\text{Cathode}$ $\text{Anode}$ $\text{At cathode:}$ $\text{Al}^{3+} + 3\text{e}^{-} \rightarrow \text{Al}$ $\text{We know that, to deposit one mole of Al, 3F of electricity is required.}$ $\text{Mass of aluminium deposited by 3 F of electricity = 27 g}$ $\therefore \text{ by unitary method,}$ $\text{Mass of aluminium deposited by } 4.0 \times 10^4 \times 6 \times 3600 \text{ C of electricity}$ $\Rightarrow \frac{27}{3} \times 4.0 \times 10^4 \times 6 \times 3600 \text{ Fg}$ $\Rightarrow \frac{27}{3} \times 4.0 \times 10^4 \times 6 \times 3600 \times 96500 = 8.1 \times 10^4 \text{ g}$

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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
}