Import Question JSON

Current Question (ID: 21374)

Question:

$\text{What is the nearest integer value of } x \text{ for the Gibbs free energy change at } 298 \text{ K, expressed as } x \times 10^{-1} \text{ kJmol}^{-1} \text{ for the given reaction.}$ $\text{Cu}(s) + \text{Sn}^{2+}(0.001M) \rightarrow \text{Cu}^{2+}(0.01M) + \text{Sn}(s)$ $\text{Given: } E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \text{ V}; \; E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \text{ V}; \; F = 96500 \text{ C mol}^{-1}$

Options:

1. 873
2. 983
3. 1002
4. 911

Solution:

$\text{Hint: } \Delta G = -nF E_{\text{cell}}$ $\text{STEP 1: The given reaction is:}$ $\text{Cu}(s) + \text{Sn}^{2+}(0.001M) \rightarrow \text{Cu}^{2+}(0.01M) + \text{Sn}(s)$ $\text{Reaction at cathode:}$ $\text{Sn}^{2+}(0.001 \text{ M}) + 2e^- \rightarrow \text{Sn}(s); \; E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \text{ V}$ $\text{Reaction at anode:}$ $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(0.01 \text{ M}) + 2e^-; \; E^\circ_{\text{Cu}/\text{Cu}^{2+}} = -0.34 \text{ V}$ $\text{Thus, } E^\circ_{\text{cell}} = (-0.14 - 0.34) \text{ V} = -0.48 \text{ V}$ $\text{STEP 2: Use Nernst Equation to find the cell potential:}$ $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \left[ \frac{[\text{Cu}^{2+}]}{[\text{Sn}^{2+}]} \right]$ $E_{\text{cell}} = -0.48 - \frac{0.059}{2} \log \frac{0.01}{0.001}$ $E_{\text{cell}} = -0.48 - \frac{0.059}{2} \log 10$ $E_{\text{cell}} = -0.48 - 0.0295$ $\text{or, } E_{\text{cell}} = -0.509$ $\text{STEP 3: We know that, } \Delta G = -nF E_{\text{cell}}$ $\text{So, } \Delta G = -2 \times 96500 \times (-0.5095) \text{ J mol}^{-1}$ $\text{or, } \Delta G = 98333.5 \text{ J mol}^{-1} = 98.333 \text{ kJ mol}^{-1}$ $\text{or, } \Delta G = 983.33 \times 10^{-1} \text{ kJ mol}^{-1}$ $\text{So, option 2 is the correct answer.}$

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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
}