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Current Question (ID: 21380)

Question:

$\text{For a given reaction in an acidic medium,}$ $5e^- + 8\text{H}^+ + \text{MnO}_4^- (0.001\ M) \rightarrow \text{Mn}^{2+}(0.1\ M) + 4\text{H}_2\text{O}$ $\text{(Given values are: } E^\circ_{\text{cell}} = 1.54\ \text{V}, E_{\text{cell}} = 1.2832\ \text{V})$ $\text{The } pH \text{ of the solution is:}$

Options:

1. $2.46$
2. $3.52$
3. $6.38$
4. $1.02$

Solution:

$\text{Hint: } E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \times \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][\text{H}^+]^8} \right)$ $\text{Explanation:}$ $\text{Step 1: The given reaction is:}$ $5e^- + 8\text{H}^+ + \text{MnO}_4^- (0.001\ M) \rightarrow \text{Mn}^{2+}(0.1\ M) + 4\text{H}_2\text{O}$ $E^\circ_{\text{cell}} = 1.54\ \text{V}, E_{\text{cell}} = 1.2832\ \text{V}$ $\text{The Nernst equation will be: } E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \times \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][\text{H}^+]^8} \right)$ $\text{Step 2: Temperature = 298\text{K}, n= 5, F= 96500\ \text{Cmol}^{-1} \text{ and } R = 8.314\ \text{Jmol}^{-1}\text{K}^{-1}.}$ $1.2832 = 1.54 - \frac{8.314 \times 298}{5 \times 96500} \times \log \left( \frac{0.1}{0.001 \times [\text{H}^+]^8} \right)$ $1.2832 = 1.54 - \frac{0.0591}{5} \log \frac{10^{-1}}{(10^{-3})(\text{H}^+)^8}$ $-0.2568 = -\frac{0.0591}{5} \left( \log 10^2 - 8 \log \text{H}^+ \right)$ $21.72 = 2 + 8 \times pH$ $19.72 = 8 \times pH$ $pH = \frac{19.72}{8} = 2.46$ $\text{So, the correct answer is option 1.}$

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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
}