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Current Question (ID: 21375)

Question:

$\text{For the following cell with hydrogen electrodes at two different pressures } p_1 \text{ and } p_2$ $\text{Pt(H}_2\text{)} \mid \text{H}^+(\text{aq}) \mid \text{Pt (H}_2\text{)}$ $p_1 \quad 1\text{M} \quad p_2$ $\text{emf is given by:}$

Options:

1. $\frac{RT}{F} \log_e \frac{P_1}{p_2}$
2. $\frac{RT}{2F} \log_e \frac{p_2}{P_1}$
3. $\frac{RT}{F} \log_e \frac{p_1}{p_2}$
4. $\frac{RT}{2F} \log_e \frac{P_2}{p_1}$

Solution:

$\text{Hint: Use nernst equation, } E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \log_e Q$ $\text{Step 1:}$ $\text{The reactions in the two half cell is as follows:}$ $\text{LHS half cell:}$ $\text{H}_2(\text{g}) \; P_1 \rightarrow 2\text{H}^+ (1\text{M}) + 2\text{e}^-$ $\text{RHS cell is}$ $2\text{H}^+ (1\text{M}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) \; P_2$ $\text{Step 2:}$ $\text{The concentration of H}^+ \text{ ions at the cathode and anode are equal.}$ $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \log_e Q$ $E^\circ_{\text{cell}} = 0.00 \text{ V}, Q = \frac{p_2 \times [\text{H}^+(\text{anode})]^2}{p_1 \times [\text{H}^+(\text{cathode})]^2}, n = 2$ $= 0 - \frac{RT}{2F} \log_e \frac{p_2}{p_1}$ $E_{\text{cell}} = \frac{RT}{2F} \log_e \frac{p_1}{p_2}$

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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
}