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Current Question (ID: 21363)

Question:

$\text{A solution of Ni(NO}_3\text{)}_2 \text{ is electrolysed between platinum electrodes using 0.1 Faraday electricity. The number of moles of Ni that will be deposited at the cathode are:}$

Options:

1. $0.10$
2. $0.05$
3. $0.20$
4. $0.15$

Solution:

$\text{Hint: The charge on one mole of electrons is equal to 1 faraday or 1F.}$ $\text{The reaction occur during the electrolysis of Ni(NO}_3\text{)}_2 \text{ is as follows:}$ $\text{Ni}^{2+} + 2\text{e}^- \rightarrow \text{Ni}$ $\text{1 mole of Ni}^{2+} \text{ required 2F charge.}$ $\text{Hence, 2F is required for 1 mole of Ni}$ $\text{1F is required for } \frac{1}{2} \text{ mol of Ni}$ $0.1\text{F is required for } \frac{1}{2} \times 0.1 \text{ mole of Ni}$ $= 0.05 \text{ mol of Ni}$

Import JSON File

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
}