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Current Question (ID: 18177)

Question:

$\text{Given:}$ $\text{(i) Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E^\circ = 0.337 \text{ V}$ $\text{(ii) Cu}^+ + e^- \rightarrow \text{Cu} \quad E^\circ = 0.153 \text{ V}$ $\text{Electrode potential, } E^\circ \text{ for the reaction, } \text{Cu}^+ + e^- \rightarrow \text{Cu, will be:}$

Options:

1. 0.52 \text{ V}
2. 0.90 \text{ V}
3. 0.30 \text{ V}
4. 0.38 \text{ V}

Solution:

$\text{HINT: Use Gibbs energy of reaction.}$ $\text{STEP 1:}$ $n = 2$ $\Delta G_1 = -2 F \times E$ $= -2 \times 0.337$ $= -0.674F$ $\Delta G_2 = -nEF, n = 1$ $= -1 \times 0.153 \cdot F$ $\text{STEP 2:}$ $\Delta G_3 = \Delta G_1 - \Delta G_2$ $\Delta G_3 = -0.674F - (-0.153 \cdot F)$ $\Delta G_3 = -0.52F$ $\Delta G_3 = -nFE^\circ_{\text{cell}}$ $E^\circ_{\text{cell}} = \frac{-0.52F}{-1F}$ $= 0.52 \text{ V}$

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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
}