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Current Question (ID: 18159)

Question:

$\text{The maximum work which can be obtained from a Daniel cell}$ $\text{Zn}_{(s)} \mid \text{Zn}^{+2}_{(aq)} \parallel \text{Cu}^{+2}_{(aq)} \mid \text{Cu}_{(s)} \text{ is:}$ $\begin{array}{c|c} E^\circ_{\text{Zn}^{2+}/\text{Zn}} & -0.76 \text{ V} \\ \hline E^\circ_{\text{Cu}^{2+}/\text{Cu}} & 0.34 \text{ V} \end{array}$

Options:

1. +106.15 \text{ kJ}
2. -212.3 \text{ kJ}
3. -424.6 \text{ kJ}
4. +212.3 \text{ kJ}

Solution:

$\text{Hint: } W_{\text{max}} = -\Delta G^\circ = nFE^\circ$ $\text{Explanation:}$ $\text{To calculate the maximum work that can be obtained from a Daniell cell, we}$ $\text{use the following equation:}$ $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ $\text{Step 1: Determine } E^\circ_{\text{cell}}$ $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ $E^\circ_{\text{cell}} = 0.34 \text{ V} - (-0.76 \text{ V}) = 0.34 + 0.76 = 1.10 \text{ V}$ $\text{For the Daniell cell, 2 electrons are transferred (since zinc is oxidized from}$ $\text{Zn to Zn}^{2+}, \text{ and copper is reduced from Cu}^{2+} \text{ to Cu).}$ $n = 2$ $\text{As, Maximum work done is } \Delta G^\circ$ $\text{Using the equation:}$ $W_{\text{max}} = \Delta G^\circ = -nFE^\circ$ $E^\circ = E^\circ_{\text{cathode}} - E_{\text{anode}}$ $= 0.34 - (-0.76) = 1.1 \text{ V}$ $W_{\text{max}} = -2 \times 96500 \times 1.1$ $= -212.3 \text{kJ}$ $\text{The maximum work that can be obtained is } -212.3 \text{kJ/mol, so the correct}$ $\text{answer is:}$ $\text{Option 2: } -212.3 \text{ kJ}$

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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
}