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Current Question (ID: 21358)

Question:

$\text{Emf of the following cell at 298 K in V is } x \times 10^{-2}. \text{ The cell is Zn|Zn}^{2+} (0.1 \text{ M}) || \text{Ag}^{+} (0.01 \text{ M}) | \text{Ag. The value of } x \text{ is- (Rounded off to the nearest integer)}$ $\text{Given; } E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V}$ $E^\circ_{\text{Ag}^{+}/\text{Ag}} = +0.80 \text{ V}; \frac{2.303RT}{F} = 0.059$

Options:

1. 157
2. 147
3. 144
4. 154

Solution:

$\text{Hint: The formula of } E_{\text{cell}} \text{ is } E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \left[ \frac{[\text{Zn}^{2+}]}{[\text{Ag}^{+}]^2} \right]$ $\text{Step 1:}$ $\text{Zn}_{(s)} \rightarrow \text{Zn}^{2+}_{(aq)} + 2e^{-}$ $2 \text{Ag}^{+}_{(aq)} + 2e^{-} \rightarrow 2 \text{Ag}_{(s)}$ $\text{-------------------------------}$ $\text{Zn}_{(s)} + 2 \text{Ag}^{+}_{(aq)} \rightarrow \text{Zn}^{2+}_{(aq)} + 2 \text{Ag}_{(s)}$ $\text{-------------------------------}$ $E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^{+}/\text{Ag}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}}$ $= 0.80 - (-0.76)$ $= 1.56 \text{ V}$ $\text{Step 2:}$ $E_{\text{cell}} = 1.56 - \frac{0.059}{2} \log \left[ \frac{[\text{Zn}^{2+}]}{[\text{Ag}^{+}]^2} \right]$ $E_{\text{cell}} = 1.56 - \frac{0.059}{2} \log \left[ \frac{0.1}{(0.01)^2} \right]$ $= 1.56 - \frac{0.059}{2} \times 3$ $= 1.56 - 0.0885$ $= 1.4715$ $= 147.15 \times 10^{-2}$ $E_{\text{cell}} = 147 \times 10^{-2} \text{ V}$ $\text{Hence, answer is option second.}$

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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
}