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Current Question (ID: 18173)

Question:

$\text{Given the following cell:}$ $\text{Pt}(s)|\text{Br}_2(l)|\text{Br}^- (0.010 \ M) ||\text{H}_2(g)(1 \ \text{bar})|\text{H}^+(0.030 \ M)|\text{Pt}(s)$ $\text{If the concentration of } \text{Br}^- \text{ becomes 2 times and the concentration of } \text{H}^+ \text{ becomes half of the initial value, then emf of the cell will become:}$

Options:

1. $\text{Two times.}$
2. $\text{Four times.}$
3. $\text{Eight times.}$
4. $\text{Remains the same.}$

Solution:

$\text{HINT - } E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{1}{[\text{Br}^-]^2[\text{H}^+]^2}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Nernst equation is,}$ $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$ $\text{Reactions involved are,}$ $\text{At cathode:}$ $2\text{H}^+(1\text{M}) + 2e^- \rightarrow \text{H}_2(1 \ \text{bar})$ $\text{Net reaction is:}$ $2\text{Br}^-(1\text{M}) + 2\text{H}^+(1\text{M}) \rightarrow \text{Br}_2(l) + \text{H}_2(1 \ \text{bar})$ $\text{STEP 2:}$ $\text{Here,}$ $\text{Nernst equation for the given cell, } E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{1}{[\text{Br}^-]^2[\text{H}^+]^2}$ $\text{If } [\text{Br}^-]' = 2[\text{Br}^-] \text{ and } [\text{H}^+]' = \frac{[\text{H}^+]}{2}$ $\text{STEP 3:}$ $\text{Then ,}$ $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{1}{([\text{Br}^-]')^2 ([\text{H}^+])^2}$ $\Rightarrow E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{1}{4} [\text{Br}^-]^2 \times \frac{[\text{H}^+]^2}{4}$ $\Rightarrow E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{1}{[\text{Br}^-]^2[\text{H}^+]^2}$ $\text{i.e. No effect on the Emf of cell is observed.}$

Import JSON File

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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
}