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Current Question (ID: 18530)

Question:
$\text{Brine is electrolysed by using inert electrodes. The reaction at the anode is:}$ $1. \ \text{Cl}^- (\text{aq}) \rightarrow \frac{1}{2} \text{Cl}_2(\text{g}) + e^-; \ E^\circ_{\text{cell}} = 1.36 \ \text{V}$ $2. \ \text{2H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+ + 4e^-; \ E^\circ_{\text{cell}} = 1.23 \ \text{V}$ $3. \ \text{Na}^+(\text{aq}) + e^- \rightarrow \text{Na(s)}; \ E^\circ_{\text{cell}} = 2.71 \ \text{V}$ $4. \ \text{H}^+(\text{aq}) + e^- \rightarrow \frac{1}{2} \text{H}_2(\text{g}); \ E^\circ_{\text{cell}} = 0.00 \ \text{V}$
Options:
  • 1. $\text{Cl}^- (\text{aq}) \rightarrow \frac{1}{2} \text{Cl}_2(\text{g}) + e^-; \ E^\circ_{\text{cell}} = 1.36 \ \text{V}$
  • 2. $\text{2H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+ + 4e^-; \ E^\circ_{\text{cell}} = 1.23 \ \text{V}$
  • 3. $\text{Na}^+(\text{aq}) + e^- \rightarrow \text{Na(s)}; \ E^\circ_{\text{cell}} = 2.71 \ \text{V}$
  • 4. $\text{H}^+(\text{aq}) + e^- \rightarrow \frac{1}{2} \text{H}_2(\text{g}); \ E^\circ_{\text{cell}} = 0.00 \ \text{V}$
Solution:
$\text{Hint: Cl}_2 \text{ produces at anode.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Brine is electrolysed by using inert electrodes. The possible reactions occurring at anode are:}$ $\text{Cl}^- (\text{aq}) \rightarrow \frac{1}{2} \text{Cl}_2(\text{g}) + e^-;$ $\text{2H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+ + 4e^-;$ $E^\circ_{\text{cell}} = 1.36 \ \text{V}$ $E^\circ_{\text{cell}} = 1.23 \ \text{V}$ $\text{STEP 2:}$ $\text{The reaction at anode with lower value of } E^\circ \text{ is preferred and therefore water should get oxidised in preference to Cl}^- (\text{aq}). \text{ However, Cl}_2 \text{ is produced instead of O}_2.$ $\text{This unexpected result is explained on the basis of the fact that water needs greater voltage for oxidation to O}_2 \text{ (as it is kinetically slow process) than that needed for oxidation of Cl}^- \text{ ions to Cl}_2.$

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{
  "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
}