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Current Question (ID: 21362)

Question:

$\text{For the given cell:}$ $\text{Cu(s)}|\text{Cu}^{2+}(\text{C}_1\text{M})||\text{Cu}^{2+}(\text{C}_2\text{M})|\text{Cu(s)}$ $\text{change in Gibbs energy (}\Delta G\text{) is negative, if:}$

Options:

1. $\text{C}_1 = 2\text{C}_2$
2. $\text{C}_2 = \frac{\text{C}_1}{\sqrt{2}}$
3. $\text{C}_1 = \text{C}_2$
4. $\text{C}_2 = \sqrt{2}\text{C}_1$

Solution:

$\text{Hint: } E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$ $\text{Step 1:}$ $\Delta G = -nFE_{\text{cell}}$ $\text{The } \Delta G \text{ value is negative hence, } E_{\text{cell}} \text{ value must be positive.}$ $\text{The overall cell representation is as follows:}$ $\text{Cu(s)}|\text{Cu}^{2+}(\text{C}_1\text{ M})||\text{Cu}^{2+}(\text{C}_2\text{ M})|\text{Cu(s)}$ $\text{The reaction at the anode and at the cathode is as follows:}$ $\text{Anode: Cu(s)} \rightarrow \text{Cu}^{+2}(\text{C}_1) + 2e^- : E^\circ$ $\text{Cathode: Cu}^{+2}(\text{C}_2) + 2e^- \rightarrow \text{Cu(S)}: -E^\circ$ $\text{Cell reaction: Cu(s)} \rightarrow \text{Cu}^{+2}(\text{C}_1)E_{\text{cell}} = 0$ $\text{Step 2:}$ $\text{The Nernst equation for Cu(s)}||\text{Cu}^{2+}(\text{C}_1 \text{ M})||\text{Cu}^{2+}(\text{C}_2 \text{ M})||\text{Cu(s)} \text{ is as follows:}$ $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$ $E_{\text{cell}} = 0 - \frac{0.0591}{n} \log \left( \frac{\text{C}_1}{\text{C}_2} \right)$ $E_{\text{cell}} > 0; \text{ if } \frac{\text{C}_1}{\text{C}_2} < 1 \Rightarrow \text{C}_1 < \text{C}_2$ $\text{Thus, the only in option fourth } \text{C}_2 > \text{C}_1 , \text{ that is, } \text{C}_2 = \sqrt{2}\text{C}_1.$

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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
}