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Current Question (ID: 18169)

Question:

$\text{The electrode potential for Mg electrode varies according to the equation}$ $E_{\text{Mg}^{2+}/\text{Mg}} = E^\circ_{\text{Mg}^{2+}/\text{Mg}} - \frac{0.059}{2} \log \frac{1}{[\text{Mg}^{2+}]}$ $\text{The graph of } E_{\text{Mg}^{2+}/\text{Mg}} \text{ vs } \log [\text{Mg}^{2+}] \text{ among the following is:}$

Options:

1. $\text{Graph 1}$
2. $\text{Graph 2}$
3. $\text{Graph 3}$
4. $\text{Graph 4}$

Solution:

$\text{Electrode potential for Mg electrode varies according to the equation -}$ $E_{\text{Mg}^{2+}/\text{Mg}} = E^\circ_{\text{Mg}^{2+}/\text{Mg}} - \frac{0.059}{2} \log \left[ \text{Mg}^{2+} \right]$ $E_{\text{Mg}^{2+}/\text{Mg}} = \frac{0.059}{2} \log \left[ \text{Mg}^{2+} \right] + E^\circ_{\text{Mg}^{2+}/\text{Mg}}$ $\text{This equation represents equation of straight line. It can be correlated as}$ $E_{\text{Mg}^{2+}/\text{Mg}} = \frac{0.059}{2} \log \left[ \text{Mg}^{2+} \right] + E^\circ_{\text{Mg}^{2+}/\text{Mg}}$ $\uparrow \quad \uparrow \quad \uparrow \quad \uparrow$ $Y \quad M \quad X \quad +C$ $\text{So, intercept (C) = } E_{\text{Mg}^{2+}/\text{Mg}}$ $\text{Thus, equation can be diagramatically represented as}$ $\text{Slope = } \frac{0.059}{2}$ $C = E^\circ_{\text{Mg}^{2+}/\text{Mg}}$

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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
}