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Current Question (ID: 18238)

Question:

$\text{Passage of three faradays of charge through an aqueous solution of } \text{AgNO}_3, \text{CuSO}_4, \text{Al(NO}_3)_3, \text{ and NaCl will deposit moles of metals at the cathode in the molar ratio of:}$

Options:

1. $1 : 2 : 3 : 1$
2. $6 : 3 : 2 : 6$
3. $6 : 3 : 0 : 0$
4. $3 : 2 : 1 : 0$

Solution:

$\text{STEP 1: Given solution : } \text{AgNO}_3, \text{CuSO}_4, \text{Al(NO}_3)_3 \text{ and NaCl}$ $\text{In the case of } \text{AgNO}_3, \text{CuSO}_4, \text{Al(NO}_3)_3 \text{ and NaCl, Al and Na will not be deposited at their respective electrodes because their discharge potential is higher than that of hydrogen hence instead of Al, Na, } \text{H}_2 \text{ gas will be evolved at the cathode.}$ $\text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^- $ $2\text{H}_2\text{O} \rightleftharpoons 2\text{H}^+ + 2\text{O}^{2-}$ $\text{At cathode: } \text{Ag}^+ + e^- \rightarrow \text{Ag}$ $(1 \text{ F})$ $\text{Here:}$ $1 \text{ mole } e^- \text{ deposits 1 mole Ag}$ $1\text{F charge deposit } \rightarrow 1 \text{ Mol Ag}$ $3\text{F } \rightarrow x$ $x_1 = 3 \text{ mol}$ $\text{STEP 2:}$ $\text{Similarly for Cu}$ $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ $2\text{F deposites } \rightarrow 1 \text{ mole}$ $3\text{F } \rightarrow x$ $x_2 = \frac{3}{2} \text{ mole}$ $x_1 : x_2$ $= 3 \text{ mol} : \frac{3}{2} \text{ mol}$ $= 3 \times 2 : \frac{3}{2} \times 2$ $= 6 : 3 : 0 : 0 \text{ (multiplied by 2 to remove fraction)}$

Import JSON File

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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
}