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Current Question (ID: 18167)

Question:

$\text{Consider the following cell reaction}$ $2\text{Fe(s)} + \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + 2\text{H}_2\text{O(l)}$ $E^\circ = 1.67 \text{ V}, \text{ At } [\text{Fe}^{2+}] = 10^{-3} \text{ M}, P_{\text{O}_2} = 0.1 \text{ atm and pH} = 3, \text{ the cell potential at } 25\, ^\circ \text{C is:}$

Options:

1. 1.27 \text{ V}
2. 1.77 \text{ V}
3. 1.87 \text{ V}
4. 1.57 \text{ V}

Solution:

$\text{HINT: } E_{\text{cell}} = E^\circ_{\text{cell}} + \frac{0.059}{4} \log \frac{P_{\text{O}_2} \times [\text{H}^+]^4}{[\text{Fe}^{2+}]^2}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{The given reaction is as follows:}$ $2\text{Fe(s)} + \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + 2\text{H}_2\text{O(l)}$ $\text{The Nernst equation for the above equation is as follows:}$ $2\text{Fe} \rightarrow 2\text{Fe}^{2+} + 4\text{e}^- $ $4\text{e}^- + \text{O}_2 + 4\text{H}^+ \rightarrow 2\text{H}_2\text{O}$ $\text{Therefore, } E_{\text{cell}} = E^\circ_{\text{OP}_{\text{Fe}}} + \frac{0.059}{4} \log [\text{Fe}^{2+}]^2 + E^\circ_{\text{RP}_{\text{O}_2}} + \frac{0.059}{4} \log \frac{P_{\text{O}_2} \times [\text{H}^+]^4}{[\text{Fe}^{2+}]^2}$ $\text{Step 2:}$ $\text{Calculate } E_{\text{cell}} \text{ as follows:}$ $= E^\circ_{\text{cell}} + \frac{0.059}{4} \log \frac{P_{\text{O}_2} \times [\text{H}^+]^4}{[\text{Fe}^{2+}]^2}$ $= 1.67 + \frac{0.059}{4} \log \frac{0.1 \times (10^{-3})^4}{(10^{-3})^2}$ $= 1.67 + \frac{0.059}{4} \log 10^{-7}$ $= 1.67 + \frac{-0.059 \times (-7)}{4}$ $= 1.57 \text{ V}$

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{
  "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
}