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Current Question (ID: 18172)

Question:
$\text{At 298 K the Emf of the following cell is:}$ $\text{Pt|H}_2(1 \text{ atm})|\text{H}^+(0.02 \text{ M}) || \text{H}^+(0.01 \text{ M})|\text{H}_2(1 \text{ atm})|\text{Pt}$
Options:
  • 1. $-0.017 \text{ V}$
  • 2. $0.0295 \text{ V}$
  • 3. $0.1 \text{ V}$
  • 4. $0.059 \text{ V}$
Solution:
$\text{(1) HINT: } E_{\text{cell}} = E^\circ_{\text{cell}} - \left( \frac{0.0591}{n} \log \frac{[\text{H}^+]^2_{\text{anode}}}{[\text{H}^+]^2_{\text{cathode}}} \right)$ $\text{Explanation:}$ $\text{General cell reaction is:}$ $\text{H}_2 \rightleftharpoons 2 \text{H}^+ + 2 \text{e}^-$ $E_{\text{cell}} = E^\circ_{\text{cell}} - \left( \frac{0.0591}{n} \log \frac{[\text{H}^+]^2_{\text{anode}}}{[\text{H}^+]^2_{\text{cathode}}} \right)$ $E_{\text{cell}} = 0 - \left( \frac{0.0591}{2} \log \frac{(0.02)^2}{(0.01)^2} \right)$ $= -\frac{0.0591}{2} \times (0.6)$ $= -0.017 \text{ V}$

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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
}