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Current Question (ID: 21367)

Question:

$\text{Given below are the half-cell reactions:}$ $\text{Mn}^{2+} + 2\text{e}^- \rightarrow \text{Mn}; \ E^\circ = -1.18 \ \text{V}$ $2 \left( \text{Mn}^{3+} + \text{e}^- \rightarrow \text{Mn}^{2+} \right); \ E^\circ = +1.51 \ \text{V}$ $\text{The } E^\circ \text{ for } 3\text{Mn}^{2+} \rightarrow \text{Mn} + 2\text{Mn}^{3+} \text{ will be:}$

Options:

1. $-2.69 \ \text{V}; \text{ the reaction will occur}$
2. $-0.33 \ \text{V}; \text{ the reaction will not occur}$
3. $-0.33 \ \text{V}; \text{ the reaction will occur}$
4. $-2.69 \ \text{V}; \text{ the reaction will not occur}$

Solution:

$\text{Hint: } E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$ $\text{Step 1:}$ $\text{Reaction at cathode:}$ $\text{Mn}^{2+} + 2\text{e}^- \rightarrow \text{Mn} \quad E^\circ_{\text{red.}} = -1.18 \ \text{V}$ $\text{Reaction at anode:}$ $2\text{Mn}^{2+} \rightarrow 2\text{Mn}^{3+} + 2\text{e}^- \quad E^\circ_{\text{red.}} = 1.51 \ \text{V}$ $\text{Overall reaction}$ $3\text{Mn}^{2+} \rightarrow 2\text{Mn}^{3+} + \text{Mn}$ $E^\circ = E_{\text{cathode}} - E_{\text{anode}}$ $= -1.18 - 1.51 = -2.69 \ \text{V}$ $\text{Since value of } E^\circ \text{ is negative therefore the reaction will not occur.}$

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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
}