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Current Question (ID: 18232)

Question:

$\text{Three electrolytic cells A, B, C containing solutions of ZnSO}_4, \text{ AgNO}_3, \text{ and CuSO}_4, \text{ respectively are connected in series.}$ $\text{A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell B. The current flow time is-}$

Options:

1. $\text{14 minutes}$
2. $\text{25 minutes}$
3. $\text{20 minutes}$
4. $\text{11 minutes}$

Solution:

$\text{Use Faraday's law of electrolysis.}$ $\text{Step 1: Calculate the charge deposited on one mole of Ag}$ $\text{According to the reaction:}$ $\text{Ag}^+ (aq) + e^- \rightarrow \text{Ag}(aq)$ $108g$ $\text{i.e., 108 g of Ag is deposited by 96487 C.}$ $\text{Therefore, 1.45 g of Ag is deposited by}= \frac{96487}{108} \times 1.45 \text{ C} = 1295.43 \text{ C}$ $\text{Step 2:}$ $\text{Use current and charge relationships to find time}$ $\text{Given,}$ $\text{Current} = 1.5 \text{ A}$ $\text{We know that, Charge} = \text{current} \times \text{time}$ $\text{Time} = \frac{\text{charge}}{\text{current}}$ $\text{Time} = \frac{1295.43}{1.5} ; \text{ 1A = 1 C s}^{-1}$ $= 863.6 \text{ s}$ $= 14.4 \text{ min}$

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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
}