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Current Question (ID: 18146)

Question:

$\text{The standard electrode potential for } \text{Sn}^{4+}/\text{Sn}^{2+} \text{ couple is } +0.15 \text{ V and that for the } \text{Cr}^{3+}/\text{Cr} \text{ couple is } -0.74 \text{ V. These two couples in their standard state are connected to make a cell. The cell potential will be:}$

Options:

1. $+0.89 \text{ V}$
2. $+0.18 \text{ V}$
3. $+1.83 \text{ V}$
4. $+1.199 \text{ V}$

Solution:

$\text{HINT: For working of cell, } E^\circ_{\text{cell}} \text{ should be positive.}$ $\text{Explanation:}$ $\text{STEP 1: Given-}$ $E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15 \text{ V and } E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \text{ V}$ $\text{STEP 2: The reduction potential of } \text{Sn}^{4+} \text{ to } \text{Sn}^{2+} \text{ is more thus it will act as cathode.}$ $\text{So, the cell will be: } \text{Sn}^{4+} + \text{Cr} \rightarrow \text{Sn}^{2+} + \text{Cr}^{3+}$ $\text{And,}$ $E^\circ_{\text{cell}} = E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} - E^\circ_{\text{Cr}^{3+}/\text{Cr}}$ $E^\circ_{\text{cell}} = +0.15 - (-0.74) \text{ V} = +0.89 \text{ V}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}