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Current Question (ID: 21359)

Question:

$\text{The Gibbs energy change (in J) for the given reaction at } [\text{Cu}^{2+}] = [\text{Sn}^{2+}] = 1 \text{ M and } 298 \text{ K is-}$ $\text{Cu(s)} + \text{Sn}^{2+}(\text{aq.}) \rightarrow \text{Cu}^{2+}(\text{aq.}) + \text{Sn(s)}$ $E^0_{\text{Sn}^{2+}/\text{Sn}} = -0.16 \text{ V}, \quad E^0_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \text{ V}$ $\text{Take } F = 96500 \text{ C mol}^{-1}$

Options:

1. 97850 \text{ J}
2. 3500 \text{ J}
3. 45660 \text{ J}
4. 96500 \text{ J}

Solution:

$\text{Hint: } \Delta G = -nF E_{\text{cell}}$ $\text{Step 1:}$ $\text{First, calculate the value of } E_{\text{cell}} \text{ using the Nernst equation}$ $E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Cu}^{2+}]}{[\text{Sn}^{2+}]}$ $\text{The reaction is as follows:}$ $\text{Cu(s)} + \text{Sn}^{2+}(\text{aq.}) \rightarrow \text{Cu}^{2+}(\text{aq.}) + \text{Sn(s)}$ $n = 2$ $E_{\text{cell}} = (E^0_{\text{oxidation}} + E^0_{\text{reduction}}) - \frac{0.0591}{2} \log \frac{[1]}{[1]}$ $E_{\text{cell}} = (E^0_{\text{oxidation}} + E^0_{\text{reduction}})$ $E_{\text{cell}} = -0.34 + (-0.16)$ $E_{\text{cell}} = -0.5 \text{ V}$ $\text{Step 2:}$ $\text{Calculate the value of Gibbs energy change as follows:}$ $\Delta G = -nF E_{\text{cell}}$ $= -2 \times 96500 \times (-0.5)$ $= 96500 \text{ J}$ $\text{Hence, option 4th is the correct answer}$

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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
}