Import Question JSON

Current Question (ID: 18175)

Question:

$\text{The value of } E_{\text{cell}} \text{ in the reaction below will be:}$ $\text{Pt}(s)|\text{Br}^- (0.010\ M)|\text{Br}_2(l) || \text{H}^+ (0.030\ M)|\text{H}_2(g)(1\ \text{bar})|\text{Pt}(s)$ $E^\circ_{\text{Br}^- / \text{Br}_2} = -1.09\ \text{V}$

Options:

1. +1.298\ \text{V}
2. -1.398\ \text{V}
3. -1.298\ \text{V}
4. -1.198\ \text{V}

Solution:

$\text{HINT: } E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{P_{\text{H}_2}}{\text{Br}^-^2 \text{H}^+^2}$ $\text{Step 1:}$ $\text{Calculate the value of } E^\circ_{\text{cell}} \text{ as follows:}$ $E^\circ_{\text{cell}} = E^\circ_{\text{Br}^- / \text{Br}_2} + E^\circ_{\text{H}^+ / \text{H}_2}$ $= -1.09 + 0$ $E^\circ_{\text{cell}} = -1.09\ \text{V}$ $\text{Step 2:}$ $\text{Calculate the value of } E_{\text{cell}} \text{ as follows:}$ $E_{\text{cell}} = -1.09 - \frac{0.0591}{2} \log \frac{1}{0.010^2 \times 0.030^2}$ $E_{\text{cell}} = -1.09 - 0.02955 \times \log 0.00000009$ $E_{\text{cell}} = -1.09 - 0.02955 \times \log (1.11 \times 10^7)$ $E_{\text{cell}} = -1.09 - 0.208$ $E_{\text{cell}} = -1.298\ \text{V}$

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
}