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Current Question (ID: 8544)

Question:

$\text{The standard reduction potential for Cu}^{2+}/\text{Cu is 0.34 V. The reduction potential at pH = 14 for the above couple given } (K_{sp}[\text{Cu}(\text{OH})_2] = 1 \times 10^{-19}) \text{ is:}$

Options:

1. $-0.22 \text{ V}$
2. $+0.22 \text{ V}$
3. $-0.34 \text{ V}$
4. $+0.34 \text{ V}$

Solution:

$\text{Hint: } K_{sp} = [\text{Cu}^{2+}] \cdot [\text{OH}^-]^2$ $\text{Step 1:}$ $\text{The pH of solution is 14. Calculate the concentration of OH}^- \text{ ion is as follows:}$ $\text{pH} = 14$ $\text{pH} = -\log[\text{H}^+]$ $14 = -\log[\text{H}^+]$ $[\text{H}^+] = 10^{-14}$ $\text{The concentration of OH}^- \text{ ion is}$ $K_w = [\text{H}^+] \times [\text{OH}^-]$ $\frac{10^{-14}}{10^{-14}} = [\text{OH}^-]$ $[\text{OH}^-] = 1$ $\text{Step 2:}$ $\text{Calculate the value of Cu}^{2+} \text{ ion in solution as follows:}$ $\text{The dissociation reaction of Cu(OH)}_2 \text{ is as follows:}$ $\text{Cu}(\text{OH})_2 \rightleftharpoons \text{Cu}^{2+} + 2\text{OH}^-$ $K_{sp} = [\text{Cu}^{2+}] \cdot [\text{OH}^-]^2$ $1 \times 10^{-19} = [\text{Cu}^{2+}] \times (1)^2$ $[\text{Cu}^{2+}] = 1 \times 10^{-19}$ $\text{Step 3:}$ $\text{Calculate the reduction potential of Cu(OH)}_2 \text{ is as follows:}$ $\text{E} = \text{E}^o - \frac{0.059}{n} \log \left[ \frac{1}{[\text{Cu}^{2+}]} \right]$ $\text{Given, E}^o = 0.34$ $\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-$ $\text{E} = 0.34 - \frac{0.059}{2} \log \left[ \frac{1}{[\text{Cu}^{2+}]} \right]$ $\text{E} = 0.34 - \frac{0.059}{2} \log \left[ \frac{1}{10^{-19}} \right]$ $\text{E} = 0.34 - \frac{0.059}{2} \log [10^{19}]$ $\text{E} = 0.34 - \frac{0.059}{2} \times 19$ $\text{E} = 0.34 - 0.56 = -0.22 \text{ V}$

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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
}