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Current Question (ID: 18256)

Question:

$\text{A solution containing one mole per litre each of Cu(NO}_3)_2, \text{AgNO}_3, \text{Hg}_2(\text{NO}_3)_2 \text{ and Mg(NO}_3)_2 \text{ is being electrolyzed by using inert electrodes. The values of standard electrode potentials in volt (reduction potentials) are,}$ $\text{Ag}^+/\text{Ag} = 0.80 \text{ V}, \text{Hg}_2^{2+}/2\text{Hg} = 0.79 \text{ V},$ $\text{Cu}^{2+}/\text{Cu} = +0.34 \text{ V and Mg}^{2+}/\text{Mg} = -2.37 \text{ V}$ $\text{With increasing voltage, the sequence of deposition of metals on the cathode will be:}$

Options:

1. $\text{Ag, Hg, Cu, Mg}$
2. $\text{Mg, Cu, Hg, Ag}$
3. $\text{Ag, Hg, Cu}$
4. $\text{Cu, Hg, Ag}$

Solution:

$\text{HINT: Mg cannot be obtained by electrolysing the aqueous salt solutions.}$ $\text{Also, } E^o_{\text{RP}} \text{ order suggests for Ag, Hg, Cu.}$ $\text{Explanation: More the reduction potential is, less will be reducing power.}$ $\text{Ag}^+/\text{Ag} = 0.80, \text{Hg}_2^{2+}/2\text{Hg} = 0.79,$ $\text{Cu}^{2+}/\text{Cu} = +0.34 \text{ and Mg}^{2+}/\text{Mg} = -2.37$ $\text{i.e. Ag has less reducing power as compared to other elements given here.}$ $\text{The correct order of reducing power will be: Cu > Hg > Ag}$ $\text{Because of less reduction potential of Mg than hydrogen, hydrogen will be evolved here.}$ $\text{Thus, Mg is not deposited from the given solutions.}$ $\text{So, overall order of deposition will be - Ag > Hg > Cu}$

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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
}