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Current Question (ID: 18160)

Question:

$\text{The standard reduction potential for } \text{Fe}^{2+}|\text{Fe} \text{ and } \text{Sn}^{2+}|\text{Sn} \text{ electrodes are } -0.44 \text{ V and } -0.14 \text{ V respectively. For the cell reaction,}$ $\text{Fe}^{2+} + \text{Sn} \rightarrow \text{Fe} + \text{Sn}^{2+}, \text{ the standard Emf is -}$

Options:

1. $+0.30 \text{ V}$
2. $0.58 \text{ V}$
3. $+0.58 \text{ V}$
4. $-0.30 \text{ V}$

Solution:

$\text{HINT: } E^\circ_{\text{cell}} = E^\circ_{\text{oxidation}} + E^\circ_{\text{reduction}}$ $E^\circ_{\text{cell}} = E^\circ_{\text{OP}_{\text{Sn}}} + E^\circ_{\text{RP}_{\text{Fe}}} = 0.14 + (-0.44)$ $= -0.30 \text{ V}$ $\text{because } \text{Fe}^{+2} \text{ is undergoing reduction and Sn is undergoing oxidation.}$

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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
}