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Current Question (ID: 8628)

Question:

$\text{The correct statement about electrolysis of an aqueous solution of CuCl}_2 \text{ with Pt electrode is-}$

Options:

1. $\text{Cu}^{2+} \text{ ion reduced at the cathode; Cl}^- \text{ ion oxidized at the anode}$
2. $\text{Cu}^{2+} \text{ ion reduced at the anode; Cl}^- \text{ ion oxidized at the cathode}$
3. $\text{Cu}^{2+} \text{ ion reduced at the cathode; H}_2\text{O ion oxidized at the anode}$
4. $\text{H}_2\text{O ion reduced at the cathode; Cl}^- \text{ ion oxidized at the anode}$

Solution:

$\text{Hint: Higher is the reduction potential, the higher is the reducing tendency}$ $\text{Explanation: In aqueous solutions, CuCl}_2 \text{ ionizes to give Cu}^{2+} \text{ and Cl}^- \text{ ions}$ $\text{On electrolysis, either Cu}^{2+} \text{ ions or H}_2\text{O molecules can get reduced at the cathode. But the reduction potential of Cu}^{2+} \text{ is more than that of H}_2\text{O molecules.}$ $\text{The reduction potential of Cu}^{2+} \text{ is 0.34 V, and H}_2\text{O is -0.83 V.}$ $\text{Hence, Cu}^{2+} \text{ ions are reduced at the cathode and get deposited.}$ $\text{Similarly, at the anode, either Cl}^- \text{ or H}_2\text{O is oxidized. The oxidation potential of H}_2\text{O higher than that of Cl}^-.$ $\text{Oxidation potential value for Cl}^- \text{ is -1.36 V, H}_2\text{O is -1.23 V.}$ $\text{But oxidation of H}_2\text{O molecules occurs at a lower electrode potential than that of Cl}^- \text{ ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl}^- \text{ ions are oxidized at the anode to liberate Cl}_2 \text{ gas.}$ $\text{The reactions are as follows:}$ $\text{Anode : } 2\text{Cl}^-_{(aq)} \rightarrow \text{Cl}_2_{(g)} + 2e^-$ $\text{Cathode : } \text{Cu}^{2+}_{(aq)} + 2e^- \rightarrow \text{Cu}_{(s)}$

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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
}