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Current Question (ID: 18243)

Question:

$\text{During discharge of a lead storage cell, the density of sulphuric acid in the cell-}$

Options:

1. $\text{Increases.}$
2. $\text{Decreases.}$
3. $\text{Remains unchanged.}$
4. $\text{Initially increases but decreases subsequently.}$

Solution:

$\text{HINT: Pb(s) + PbO}_2\text{(s) + 2H}^+\text{(aq) + 2HSO}_4^-\text{(aq)} \rightarrow \text{2PbSO}_4\text{(s) + 2H}_2\text{O(l)}$ $\text{STEP 1:}$ $\text{This discharge process is driven by the conduction of electrons from the negative plate back into the cell at the positive plate in the external circuit.}$ $\text{Negative plate reaction: Pb(s) + HSO}_4^-\text{(aq)} \rightarrow \text{PbSO}_4\text{(s) + H}^+\text{(aq) + 2e}^- $ $\text{Positive plate reaction: PbO}_2\text{(s) + HSO}_4^-\text{(aq) + 3H}^+\text{(aq) + 2e}^- \rightarrow \text{PbSO}_4\text{(s) + 2H}_2\text{O(l)}$ $\text{Combining these two reactions, one can determine the overall reaction:}$ $\text{Pb(s) + PbO}_2\text{(s) + 2H}^+\text{(aq) + 2HSO}_4^-\text{(aq)} \rightarrow \text{2PbSO}_4\text{(s) + 2H}_2\text{O(l)}$ $\text{STEP 2:}$ $\text{During the discharge of lead storage cell, sulphuric acid is consumed. It's concentration decreases and}$ $\text{therefore, density decreases. Density of H}_2\text{SO}_4 \text{decrease from 1.3 g/mL to 1.13 g/mL}$

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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
}