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Current Question (ID: 18242)

Question:
$\text{The half-cell reaction at the anode during the electrolysis of aqueous sodium chloride solution is represented by:}$
Options:
  • 1. $\text{Na}^+(\text{aq}) + e^- \rightarrow \text{Na}(\text{s}) ; \quad E^\circ_{\text{cell}} = -2.71 \text{ V}$
  • 2. $2\text{H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e^- ; \quad E^\circ_{\text{cell}} = 1.23 \text{ V}$
  • 3. $\text{H}^+(\text{aq}) + e^- \rightarrow \frac{1}{2} \text{H}_2(\text{g}) ; \quad E^\circ_{\text{cell}} = 0.00 \text{ V}$
  • 4. $\text{Cl}^-(\text{aq}) \rightarrow \frac{1}{2} \text{Cl}_2(\text{g}) + e^- ; \quad E^\circ_{\text{cell}} = 1.36 \text{ V}$
Solution:
$\text{HINT: Higher the value of oxidation potential, the higher will be the tendency to oxidize.}$ $\text{Explanation:}$ $\text{In the case of electrolysis of aqueous NaCl, an oxidation reaction occurs at the anode as follows}$ $\text{Cl}^-(\text{aq}) \rightarrow \frac{1}{2} \text{Cl}_2(\text{g}) + e^- ; \quad E^\circ = 1.36 \text{ V}$ $2\text{H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e^- ; \quad E^\circ_{\text{cell}} = 1.23 \text{ V}$ $\text{But due to lower } E^\circ_{\text{cell}} \text{ value water should get oxidized in preference of Cl}^-(\text{aq}).$ $\text{However, the actual reaction taking place in the concentrated solution of NaCl is (d) and not (b) i.e., Cl}_2 \text{ is produced and not O}_2.$ $\text{This unexpected result is explained on the basis of the concept of 'overvoltage', i.e., water needs greater voltage for oxidation to O}_2 \text{ (as it is a kinetically slow process) than that needed for oxidation of Cl}^- \text{ ions to Cl}_2.$ $\text{Thus, the correct option is (4) not (2).}$

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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
}