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Current Question (ID: 21371)

Question:

$\text{Zn}(s) + \text{Cu}^{2+}(0.1\text{M}) \rightarrow \text{Zn}^{2+}(1\text{M}) + \text{Cu}(s)$ $\text{For the redox reaction taking place in a cell, } E_{\text{cell}}^\circ \text{ is 1.10 volt. } E_{\text{cell}} \text{ for the cell will be: } \left(2.303 \frac{RT}{F} = 0.0591\right)$

Options:

1. 2.14 \text{ V}
2. 1.80 \text{ V}
3. 1.07 \text{ V}
4. 0.82 \text{ V}

Solution:

$\text{Hint: Use the relation, that is, } E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{n} \log Q$ $\text{The relation between } E_{\text{cell}} \text{ and } Q \text{ is as follows:}$ $E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{n} \log Q$ $\text{Calculate the value of } E_{\text{cell}} \text{ as follows:}$ $E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{n} \log Q$ $\text{Cu}^{2+}_{0.1\text{M}} + \text{Zn} \rightarrow \text{Zn}^{2+} + \text{Cu}$ $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1}{0.1} = 10$ $E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \log 10$ $= 1.10 - 0.0295$ $= 1.0705 \text{ V}$

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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
}