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Current Question (ID: 21370)

Question:

$\text{The standard emf of a cell, involving one electron change is found to be } 0.591 \text{ V at } 25^\circ\text{C.}$ $\text{The equilibrium constant of the reaction is:}$ $\text{(Given F = 96500 C mol}^{-1}\text{)}$

Options:

1. $1.0 \times 10^1$
2. $1.0 \times 10^5$
3. $1.0 \times 10^{10}$
4. $1.0 \times 10^{30}$

Solution:

$\text{Hint: } E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_{\text{equ}}$ $\text{Step 1:}$ $\text{Use the formula given below:}$ $E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_{\text{equ}}$ $\text{The given values are as follows:}$ $E^\circ_{\text{cell}} = 0.591 \text{ V}$ $n = 1$ $\text{Step 2:}$ $\text{Calculate the equilibrium constant value as follows:}$ $0.591 = \frac{0.0591}{1} \log K_{\text{equ}}$ $10 = \log K_{\text{equ}}$ $K_{\text{equ}} = 10^{10}$

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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
}