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Current Question (ID: 8640)

Question:

$\text{The correct statement about the electrolysis of an aqueous solution of AgNO}_3 \text{ with Ag electrode is:}$

Options:

1. $\text{Ag}^+ \text{ ion gets oxidised at cathode; Ag(s) is reduced at anode.}$
2. $\text{H}_2\text{O gets reduced at cathode; H}_2\text{O gets oxidised at anode.}$
3. $\text{Ag}^+ \text{ ion gets reduced at cathode; H}_2\text{O is oxidised at anode.}$
4. $\text{Ag}^+ \text{ ion gets reduced at cathode; Ag(s) is oxidised at anode.}$

Solution:

$\text{Hint: Higher is the reduction potential, the higher is the reducing tendency.}$ $\text{Explanation:}$ $\text{AgNO}_3 \text{ ionizes in aqueous solutions to form Ag}^+ \text{ and NO}_3^- \text{ ions}$ $\text{On electrolysis, either Ag}^+ \text{ ions or H}_2\text{O Ag}^+ \text{ molecules can be reduced at the cathode. But the reduction potential of}$ $\text{ions are higher than that of H}_2\text{O}.$ $\text{Hence, } \text{Ag}^+ \text{ ions are reduced at the cathode.}$ $\text{Similarly, Ag metal or H}_2\text{O molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H}_2\text{O molecules.}$ $\text{Therefore, Ag metal gets oxidized at the anode.}$ $\text{The reactions are as follows:}$ $\text{Cathode : Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s})$ $\text{Anode : Ag}(\text{s}) \rightarrow \text{Ag}^+(\text{aq}) + \text{e}^-$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}